001_introduction.Rmd

# Introduction

```{r echo=FALSE, warning=FALSE, message=FALSE}
library(tidyverse)
library(latex2exp)
library(patchwork)
```

## Historical Introduction

### Exercises I 1.4 {#exr1.4}

**1.a** Modify the region in Figure I.3 by assuming that the ordinate at each $x$ is $2x^2$ instead of $x^2$. Draw the new figure. Check through the principal steps in the foregoing section and find what effect this has on the calculation of the area.

```{r echo=FALSE, warning=FALSE, fig.height=6, fig.width=7, fig.align='center'}

plot_integral_approx <- function(fun, 
                                 approx_type = c("below", "above"),
                                 y_labels = NULL,
                                 y_brake_init = 2,
                                 y_limit_inf = 0,
                                 fill_bars) {
    
    limit_inf <- 0
    limit_sup <- 1
    n_breaks  <- 6
    
    # Specify the type of approximation to apply
    approx_type <- switch(approx_type[[1]],
        "below" = 1,
        "above" = -1)
    
    # Specify bin width
    bin_width <- (limit_sup - limit_inf) / (n_breaks - 1)
    
    # Data
    tbl_data <- tibble::tibble(x = seq(from = {{limit_inf}}, 
                               to = {{limit_sup}}, 
                               length.out = {{n_breaks}})) %>% 
        # Define y-axis using the function fun
        dplyr::mutate(y = fun(x))
    
    # Plot 
    ggplot2::ggplot() +
        ggplot2::geom_col(data = tbl_data, aes(x, y), 
                 width = bin_width, 
                 # Approximation
                 ## from below:  1*bin_width
                 ## from above: -1*bin_width
                 position = position_nudge(x = (approx_type*bin_width) / 2),
                 color = "black",
                 fill  = fill_bars) + 
        # Plot the continuous function fun
        ggplot2::geom_function(fun = fun) +
        ggplot2::scale_x_continuous(breaks = seq(from = limit_inf, 
                                        to   = limit_sup,
                                        by   = bin_width), 
                           labels = c(latex2exp::TeX("$0$"),
                                      latex2exp::TeX("$\\frac{b}{n}$"),
                                      latex2exp::TeX("$\\frac{2b}{n}$"),
                                      latex2exp::TeX("$\\cdots$"),
                                      latex2exp::TeX("$\\frac{b(n-1)}{n}$"),
                                      latex2exp::TeX("$b")), 
                           limits = c(limit_inf, limit_sup), 
                           expand = c(0, 0)) + 
        ggplot2::scale_y_continuous(breaks = tbl_data$y[y_brake_init:length(tbl_data$y)],
                           labels = y_labels, 
                           expand = c(0, 0), 
                           limits = c(y_limit_inf, NA))  +
        ggplot2::theme(axis.text.y = element_text(hjust = 1),
              axis.text.x = element_text(vjust = 0.5), 
              axis.title  = element_blank(), 
              axis.line   = element_line(color = "black"),
              panel.background = element_blank())
}

parabola_2x2 <- function(x) {
    return(2*x^2)
}

approx_below_2x2 <- plot_integral_approx(fun = parabola_2x2, 
                     approx_type = "below", 
                     y_labels = c(expression(paste(2,bgroup("(", frac(b,n), ")")^2)),
                                  expression(paste(2,bgroup("(", frac(paste("2b"),n), ")")^2)),
                                  TeX("$\\cdots$"),
                                  expression(paste(2,bgroup("(", frac(paste("b(n-1)"),n), ")")^2)),
                                  TeX("$2b^2$")), 
                     y_brake_init = 2, 
                     fill_bars = "#a1d76a")

approx_above_2x2 <- plot_integral_approx(fun = parabola_2x2, 
                     approx_type = "above", 
                     y_labels = c(expression(paste(2,bgroup("(", frac(b,n), ")")^2)),
                                  expression(paste(2,bgroup("(", frac(paste("2b"),n), ")")^2)),
                                  TeX("$\\cdots$"),
                                  expression(paste(2,bgroup("(", frac(paste("b(n-1)"),n), ")")^2)),
                                  TeX("$2b^2$")), 
                     y_brake_init = 2, 
                     fill_bars = "#d8b365")

approx_below_2x2 + approx_above_2x2  
```

First we need to calculate the area of each rectangle:

$$\frac{b}{n} \mathbf{2}\left[\frac{kb}{n}\right]^2 = \mathbf{2}\frac{k^2b^3}{n^3}$$

Second we need to calculate $s_n$ 

$$\begin{split}
  s_n & = \sum_{k=1}^{n-1} \mathbf{2}\frac{k^2b^3}{n^3} \\
      & = \mathbf{2}\frac{b^3}{n^3} \sum_{k=1}^{n-1} k^2 \\
      & = \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} \right]
  \end{split}$$
  
Third we need to calculate $S_n$ 

$$\begin{split}
  S_n & = \sum_{k=1}^{n} \mathbf{2}\frac{k^2b^3}{n^3} \\
      & = \mathbf{2}\frac{b^3}{n^3} \sum_{k=1}^{n} k^2 \\
      & = \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right]
  \end{split}$$
  
Fourth we use the following inequalities where $n \geq 1$

$$\begin{split}
  \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} & < \frac{n^3}{3} & < \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \\
  \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}\right] & < \mathbf{2}\frac{b^3}{3} & < \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right] \\
  s_n & < \mathbf{2}\frac{b^3}{3} & < S_n
  \end{split}$$

Fifth we assume that there is some $A$ such that $s_n < A < S_n$ and prove that $A = \mathbf{2}\frac{b^3}{3}$ using the following inequalities: $\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} < \frac{n^3}{3}$ and $\frac{n^3}{3} < \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$

$$\begin{split}
  \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} < \frac{n^3}{3} & \iff \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} < \frac{n^3}{3} + n^2 \\
  & \iff \mathbf{2}\frac{b^3}{n^3}\left[\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right] < \mathbf{2}\frac{b^3}{n^3}\left[\frac{n^3}{3} + n^2\right] \\
  & \iff S_n < \mathbf{2}\left[\frac{b^3}{3} + \frac{b^3}{n}\right]
  \end{split}$$
  
  $$\begin{split}
  \frac{n^3}{3} < \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} & \iff \frac{n^3}{3} - n^2 < \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} \\
  & \iff \mathbf{2}\frac{b^3}{n^3}\left[\frac{n^3}{3} - n^2\right] < \mathbf{2}\frac{b^3}{n^3}\left[\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}\right] \\
  & \iff \mathbf{2}\left[\frac{b^3}{3} - \frac{b^3}{n}\right] < s_n \\
  \end{split}$$

Therefore we have also that $\mathbf{2}\left[\frac{b^3}{3} - \frac{b^3}{n}\right] < A < \mathbf{2}\left[\frac{b^3}{3} + \frac{b^3}{n}\right]$. If $A > \mathbf{2}\frac{b^3}{3}$ then $A - \mathbf{2}\frac{b^3}{3} > 0$ so using $A < \mathbf{2}\left[\frac{b^3}{3} + \frac{b^3}{n}\right]$ we can conclude that $n < \frac{\mathbf{2}b^3}{A - \mathbf{2}\frac{b^3}{3}}$ but $n \geq 1$ so there are some values of $n$ where $n < \frac{\mathbf{2}b^3}{A - \mathbf{2}\frac{b^3}{3}}$ is false. If $A < \mathbf{2}\frac{b^3}{3}$ then $0 < \mathbf{2}\frac{b^3}{3} - A$ so using $\mathbf{2}\left[\frac{b^3}{3} - \frac{b^3}{n}\right] < A$ we can also conclude that $n < \frac{\mathbf{2}b^3}{\mathbf{2}\frac{b^3}{3} - A}$ and again this is false. The only possibility is that $A = \mathbf{2}\frac{b^3}{3}$ so there is a unique value $A$ such that $s_n < A < S_n$.            

**1.b** Do the same if the ordinate at each $x$ is $3x^2$

```{r echo=FALSE, warning=FALSE, fig.height=6, fig.width=7, fig.align='center'}

parabola_3x2 <- function(x) {
    return(3*x^2)
}

approx_below_3x2 <- plot_integral_approx(fun = parabola_3x2, 
                                     approx_type = "below", 
                                     y_labels = c(expression(paste(3,bgroup("(", frac(b,n), ")")^2)),
                                  expression(paste(3,bgroup("(", frac(paste("2b"),n), ")")^2)),
                                  TeX("$\\cdots$"),
                                  expression(paste(3,bgroup("(", frac(paste("b(n-1)"),n), ")")^2)),
                                  TeX("$3b^2$")), 
                                     y_brake_init = 2, 
                                     fill_bars = "#a1d76a")

approx_above_3x2 <- plot_integral_approx(fun = parabola_3x2, 
                                     approx_type = "above", 
                                     y_labels = c(expression(paste(3,bgroup("(", frac(b,n), ")")^2)),
                                  expression(paste(3,bgroup("(", frac(paste("2b"),n), ")")^2)),
                                  TeX("$\\cdots$"),
                                  expression(paste(3,bgroup("(", frac(paste("b(n-1)"),n), ")")^2)),
                                  TeX("$3b^2$")), 
                                     y_brake_init = 2, 
                                     fill_bars = "#d8b365")

approx_below_3x2 + approx_above_3x2
```

The process is similar as in **1.a** where we only need to change $\mathbf{2}$ with $\mathbf{3}$.  

**1.c** Do the same if the ordinate at each $x$ is $\frac{1}{4}x^2$

```{r echo=FALSE, warning=FALSE, fig.height=6, fig.width=7, fig.align='center'}

parabola_1_4x2 <- function(x) {
    return((1/4)*x^2)
}

approx_below_1_4x2 <- plot_integral_approx(fun = parabola_1_4x2, 
                                     approx_type = "below", 
                                     y_labels = c(expression(paste(frac(paste("1"),paste("4")),bgroup("(", frac(b,n), ")")^2)),
                                  expression(paste(frac(paste("1"),paste("4")),bgroup("(", frac(paste("2b"),n), ")")^2)),
                                  TeX("$\\cdots$"),
                                  expression(paste(frac(paste("1"),paste("4")),bgroup("(", frac(paste("b(n-1)"),n), ")")^2)),
                                  TeX("$\\frac{1}{4}b^2$")), 
                                     y_brake_init = 2, 
                                     fill_bars = "#a1d76a")

approx_above_1_4x2 <- plot_integral_approx(fun = parabola_1_4x2, 
                                     approx_type = "above", 
                                     y_labels = c(expression(paste(frac(paste("1"),paste("4")),bgroup("(", frac(b,n), ")")^2)),
                                  expression(paste(frac(paste("1"),paste("4")),bgroup("(", frac(paste("2b"),n), ")")^2)),
                                  TeX("$\\cdots$"),
                                  expression(paste(frac(paste("1"),paste("4")),bgroup("(", frac(paste("b(n-1)"),n), ")")^2)),
                                  TeX("$\\frac{1}{4}b^2$")), 
                                     y_brake_init = 2, 
                                     fill_bars = "#d8b365")

approx_below_1_4x2 + approx_above_1_4x2
```

The process is similar as in **1.a** where you only need to change $\mathbf{2}$ with $\mathbf{\frac{1}{4}}$.

**1.d** Do the same if the ordinate at each $x$ is $2x^2 + 1$

```{r echo=FALSE, warning=FALSE, fig.height=6, fig.width=7, fig.align='center'}

parabola_2x2_1 <- function(x) {
    return(2*x^2 + 1)
}

approx_below_2x2_1 <- plot_integral_approx(fun = parabola_2x2_1, 
                                     approx_type = "below", 
                                     y_labels = c(TeX("$1$"),
                                                  expression(paste(2,bgroup("(", frac(b,n), ")")^2) + 1),
                                                  expression(paste(2,bgroup("(", frac(paste("2b"),n), ")")^2) + 1),
                                                  TeX("$\\cdots$"),
                                                  expression(paste(2,bgroup("(", frac(paste("b(n-1)"),n), ")")^2) + 1),
                                                  TeX("$2b^2 + 1$")), 
                                     y_brake_init = 1, 
                                     y_limit_inf = 0,   
                                     fill_bars = "#a1d76a")

approx_above_2x2_1 <- plot_integral_approx(fun = parabola_2x2_1, 
                                     approx_type = "above", 
                                     y_labels = c(TeX("$1$"),
                                                  expression(paste(2,bgroup("(", frac(b,n), ")")^2) + 1),
                                                  expression(paste(2,bgroup("(", frac(paste("2b"),n), ")")^2) + 1),
                                                  TeX("$\\cdots$"),
                                                  expression(paste(2,bgroup("(", frac(paste("b(n-1)"),n), ")")^2) + 1),
                                                  TeX("$2b^2 + 1$")), 
                                     y_brake_init = 1, 
                                     y_limit_inf = 0,  
                                     fill_bars = "#d8b365")

approx_below_2x2_1 + approx_above_2x2_1
```

First we need to calculate the area of each rectangle:

$$\frac{b}{n} \left[\mathbf{2}\left[\frac{kb}{n}\right]^2 + \mathbf{1}\right] = \mathbf{2}\frac{k^2b^3}{n^3} + \mathbf{1}\frac{b}{n}$$

Second we need to calculate $s_n$ 

$$\begin{split}
  s_n & = \sum_{k=1}^{n-1} \left[\mathbf{2}\frac{k^2b^3}{n^3} + \mathbf{1}\frac{b}{n}\right] \\
      & = \mathbf{2}\frac{b^3}{n^3} \sum_{k=1}^{n-1} k^2 + \mathbf{1}\frac{b(n-1)}{n} \\
      & = \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} \right] + \mathbf{1}\frac{b(n-1)}{n}
  \end{split}$$
  
Third we need to calculate $S_n$ 

$$\begin{split}
  S_n & = \sum_{k=1}^{n} \left[\mathbf{2}\frac{k^2b^3}{n^3} + \mathbf{1}\frac{b}{n}\right] \\
      & = \mathbf{2}\frac{b^3}{n^3} \sum_{k=1}^{n} k^2 + \mathbf{1}b \\
      & = \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \right] + \mathbf{1}b
  \end{split}$$
  
Fourth we use the following inequalities where $n \geq 1$

$$\begin{split}
  \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} & < \frac{n^3}{3} < \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} \\
  \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}\right] + \mathbf{1}b & < \mathbf{2}\frac{b^3}{3} + \mathbf{1}b < \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right] + \mathbf{1}b \\
  \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}\right] + \mathbf{1}\frac{b(n-1)}{n} & < \mathbf{2}\frac{b^3}{3} + \mathbf{1}b < \mathbf{2}\frac{b^3}{n^3} \left[\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right] + \mathbf{1}b
  s_n & < \mathbf{2}\frac{b^3}{3} + \mathbf{1}b < S_n
  \end{split}$$

Fifth we assume that there is some $A$ such that $s_n < A < S_n$ and prove that $A = \mathbf{2}\frac{b^3}{3} + \mathbf{1}b$ using the following inequalities: $\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} < \frac{n^3}{3}$ and $\frac{n^3}{3} < \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$

$$\begin{split}
  \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} < \frac{n^3}{3} & \iff \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} < \frac{n^3}{3} + n^2 \\
  & \iff \mathbf{2}\frac{b^3}{n^3}\left[\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}\right] + \mathbf{1}b < \mathbf{2}\frac{b^3}{n^3}\left[\frac{n^3}{3} + n^2\right] + \mathbf{1}b \\
  & \iff S_n < \mathbf{2}\left[\frac{b^3}{3} + \frac{b^3}{n}\right] + \mathbf{1}b
  \end{split}$$
  
  $$\begin{split}
  \frac{n^3}{3} < \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} & \iff \frac{n^3}{3} - n^2 < \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6} \\
  & \iff \mathbf{2}\frac{b^3}{n^3}\left[\frac{n^3}{3} - n^2\right] + \mathbf{1}b < \mathbf{2}\frac{b^3}{n^3}\left[\frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}\right] + \mathbf{1}b \\
  & \iff \mathbf{2}\left[\frac{b^3}{3} - \frac{b^3}{n}\right] + \mathbf{1}b < s_n \\
  \end{split}$$

Therefore we have also that $\mathbf{2}\left[\frac{b^3}{3} - \frac{b^3}{n}\right] + \mathbf{1}b < A < \mathbf{2}\left[\frac{b^3}{3} + \frac{b^3}{n}\right] + \mathbf{1}b$. If $A > \mathbf{2}\frac{b^3}{3} + \mathbf{1}b$ then $A - \mathbf{2}\frac{b^3}{3} - \mathbf{1}b > 0$ so using $A < \mathbf{2}\left[\frac{b^3}{3} + \frac{b^3}{n}\right] + \mathbf{1}b$ we can conclude that $n < \frac{\mathbf{2}b^3}{A - \mathbf{2}\frac{b^3}{3} - \mathbf{1}b}$ but $n \geq 1$ so there are some values of $n$ where $n < \frac{\mathbf{2}b^3}{A - \mathbf{2}\frac{b^3}{3} - \mathbf{1}b}$ is false. If $A < \mathbf{2}\frac{b^3}{3} + \mathbf{1}b$ then $0 < \mathbf{2}\frac{b^3}{3} + \mathbf{1}b - A$ so using $\mathbf{2}\left[\frac{b^3}{3} - \frac{b^3}{n}\right] + \mathbf{1}b < A$ we can also conclude that $n < \frac{\mathbf{2}b^3}{\mathbf{2}\frac{b^3}{3} + \mathbf{1}b - A}$ and again this is false. The only possibility is that $A = \mathbf{2}\frac{b^3}{3} + \mathbf{1}b$ so there is a unique value $A$ such that $s_n < A < S_n$.

**1.e** Do the same if the ordinate at each $x$ is $ax^2 + c$

In this part we are going to assume that $a > 0$ and $c > 0$. If $a = 0$ then it is not possible to calculate the area of the rectangles and if $a < 0$ we would need to define the concept of a negative area. Also if $c = 0$ it is a case we already cover and if $c < 0$ we would need again to define the concept of a negative area.    

```{r echo=FALSE, warning=FALSE, fig.height=6, fig.width=7, fig.align='center'}

parabola_ax2_c <- function(x) {
    return(2*x^2 + 1)
}

approx_below_ax2_c <- plot_integral_approx(fun = parabola_ax2_c, 
                                     approx_type = "below", 
                                     y_labels = c(TeX("$c$"),
                                                  expression(paste(a,bgroup("(", frac(b,n), ")")^2) + c),
                                                  expression(paste(a,bgroup("(", frac(paste("2b"),n), ")")^2) + c),
                                                  TeX("$\\cdots$"),
                                                  expression(paste(a,bgroup("(", frac(paste("b(n-1)"),n), ")")^2) + c),
                                                  TeX("$ab^2 + c$")), 
                                     y_brake_init = 1, 
                                     y_limit_inf = 0,   
                                     fill_bars = "#a1d76a")

approx_above_ax2_c <- plot_integral_approx(fun = parabola_ax2_c, 
                                     approx_type = "above", 
                                     y_labels = c(TeX("$c$"),
                                                  expression(paste(a,bgroup("(", frac(b,n), ")")^2) + c),
                                                  expression(paste(a,bgroup("(", frac(paste("2b"),n), ")")^2) + c),
                                                  TeX("$\\cdots$"),
                                                  expression(paste(a,bgroup("(", frac(paste("b(n-1)"),n), ")")^2) + c),
                                                  TeX("$ab^2 + c$")), 
                                     y_brake_init = 1, 
                                     y_limit_inf = 0,  
                                     fill_bars = "#d8b365")

approx_below_ax2_c + approx_above_ax2_c
```

The process is similar as in **1.d** where we only need to change $\mathbf{2}$ with $\mathbf{a}$ and $\mathbf{1}$ with $\mathbf{c}$.

**2** Modify the region in Figure I.3 by assuming that the ordinate at each $x$ is $x^3$ instead of $x^2$. Draw a new figure.

```{r echo=FALSE, warning=FALSE, fig.height=8, fig.width=7, fig.align='center'}

x3 <- function(x) {
    return(x^3)
}

approx_below_x3 <- plot_integral_approx(fun = x3, 
                                     approx_type = "below", 
                                     y_labels = c(expression(bgroup("(", frac(b,n), ")")^3),
                                                  expression(bgroup("(", frac(paste("2b"),n), ")")^3),
                                                  TeX("$\\cdots$"),
                                                  expression(bgroup("(", frac(paste("b(n-1)"),n), ")")^3),
                                                  TeX("$b^3$")), 
                                     y_brake_init = 2, 
                                     y_limit_inf = 0,   
                                     fill_bars = "#a1d76a")

approx_above_x3 <- plot_integral_approx(fun = x3, 
                                     approx_type = "above", 
                                     y_labels = c(expression(bgroup("(", frac(b,n), ")")^3),
                                                  expression(bgroup("(", frac(paste("2b"),n), ")")^3),
                                                  TeX("$\\cdots$"),
                                                  expression(bgroup("(", frac(paste("b(n-1)"),n), ")")^3),
                                                  TeX("$b^3$")), 
                                     y_brake_init = 2, 
                                     y_limit_inf = 0,  
                                     fill_bars = "#d8b365")

approx_below_x3 + approx_above_x3
```

**2a** Use a construction similar to that illustrated in Figure I.3 and show that the outer and inner sums $S_n$ and $s_n$ are given by

$$S_n = \frac{b^4}{n^4}\left[1^3 + 2^3 + \cdots + n^3\right] \; , \; s_n = \frac{b^4}{n^4}\left[1^3 + 2^3 + \cdots + (n-1)^3\right]$$

First we need to calculate the area of each rectangle:

$$\frac{b}{n} \left[\frac{kb}{n}\right]^3 = \frac{k^3b^4}{n^4}$$

Second we need to calculate $s_n$ 

$$\begin{split}
  s_n & = \sum_{k=1}^{n-1} \frac{k^3b^4}{n^4} \\
      & = \frac{b^4}{n^4} \sum_{k=1}^{n-1} k^3
  \end{split}$$
  
Third we need to calculate $S_n$ 

$$\begin{split}
  S_n & = \sum_{k=1}^{n} \frac{k^3b^4}{n^4} \\
      & = \frac{b^4}{n^4} \sum_{k=1}^{n} k^3
  \end{split}$$

**2b** Use the inequalities (which can be proved by mathematical induction; see Section I4.2)

$$1^3 + 2^3 + \cdots + (n-1)^3 < \frac{n^4}{4} < 1^3 + 2^3 + \cdots + n^3$$

to show that $s_n < \frac{b^4}{4} < S_n$ for every $n$, and prove that $\frac{b^4}{4}$ is the only number which lies between $s_n$ and $S_n$ for every $n$

First we use the following inequality where $n \geq 1$

$$\begin{split}
  1^3 + 2^3 + \cdots + (n-1)^3 & < \frac{n^4}{4} & < 1^3 + 2^3 + \cdots + n^3 \\
  \frac{b^4}{n^4} \sum_{k=1}^{n-1} k^3 & < \frac{b^4}{4} & < \frac{b^4}{n^4} \sum_{k=1}^{n} k^3 \\
  s_n & < \frac{b^4}{4} & < S_n
  \end{split}$$

Second we assume that there is some $A$ such that $s_n < A < S_n$ and prove that $A = \frac{b^4}{4}$ using the following inequalities: $s_n < \frac{b^4}{4}$ and $\frac{b^4}{4} < S_n$

If $A > \frac{b^4}{4}$ and $s_n < \frac{b^4}{4}$ then $S_n > \frac{b^4}{4}$ and $0 < \frac{b^4}{4} - s_n$. Therefore because $S_n - s_n > \frac{b^4}{4} - s_n$ this implies that $\frac{b^4}{\frac{b^4}{4} - s_n} > n$ but $n \geq 1$ so there are some values of $n$ where $\frac{b^4}{\frac{b^4}{4} - s_n} > n$ is false. 

If $A < \frac{b^4}{4}$ and $\frac{b^4}{4} < S_n$ then $\frac{b^4}{4} > s_n$ and $0 < S_n - \frac{b^4}{4}$. Therefore because $S_n - \frac{b^4}{4} < S_n - s_n$ this implies that $n < \frac{b^4}{S_n - \frac{b^4}{4}}$ but $n \geq 1$ so there are some values of $n$ where $n < \frac{b^4}{S_n - \frac{b^4}{4}}$ is false.

**2c** What numbers takes the place of $\frac{b^4}{4}$ if the ordinate at each $x$ is $ax^3 + c$?

The process to answer this question is similar to the answer in **2b** and **2a** 

First we need to calculate the area of each rectangle:

$$\frac{b}{n} \left[ a\left[\frac{kb}{n}\right]^{\mathbf{3}} + c\right] = a\frac{k^{\mathbf{3}}b^{\mathbf{4}}}{n^{\mathbf{4}}} + c \frac{b}{n}$$

Second we need to calculate $s_n$ 

$$\begin{split}
  s_n & = \sum_{k=1}^{n-1} \left[a\frac{k^{\mathbf{3}}b^{\mathbf{4}}}{n^{\mathbf{4}}} + c \frac{b}{n}\right] \\
      & = a\frac{b^{\mathbf{4}}}{n^{\mathbf{4}}} \sum_{k=1}^{n-1} k^{\mathbf{3}} + c\frac{b(n-1)}{n}
  \end{split}$$
  
Third we need to calculate $S_n$ 

$$\begin{split}
  S_n & = \sum_{k=1}^{n} \left[a\frac{k^{\mathbf{3}}b^{\mathbf{4}}}{n^{\mathbf{4}}} + c \frac{b}{n}\right] \\
      & = a\frac{b^{\mathbf{4}}}{n^{\mathbf{4}}} \sum_{k=1}^{n} k^{\mathbf{3}} + cb
  \end{split}$$

Fourth we use the following inequality where $n \geq 1$

$$\begin{split}
  1^{\mathbf{3}} + 2^{\mathbf{3}} + \cdots + (n-1)^{\mathbf{3}} & < \frac{n^{\mathbf{4}}}{{\mathbf{4}}} < 1^{\mathbf{3}} + 2^{\mathbf{3}} + \cdots + n^{\mathbf{3}} \\
  a\frac{b^{\mathbf{4}}}{n^{\mathbf{4}}} \sum_{k=1}^{n-1} k^{\mathbf{3}} + cb & < a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb < a\frac{b^{\mathbf{4}}}{n^{\mathbf{4}}} \sum_{k=1}^{n} k^{\mathbf{3}} + cb \\
  a\frac{b^{\mathbf{4}}}{n^{\mathbf{4}}} \sum_{k=1}^{n-1} k^{\mathbf{3}} + c\frac{b(n-1)}{n} & < a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb < a\frac{b^{\mathbf{4}}}{n^{\mathbf{4}}} \sum_{k=1}^{n} k^{\mathbf{3}} + cb \\
  s_n & < a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb < S_n
  \end{split}$$

Sixth we assume that there is some $A$ such that $s_n < A < S_n$ and prove that $A = a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb$ using the following inequalities: $s_n < a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb$ and $a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb < S_n$

If $A > a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb$ and $s_n < a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb$ then $S_n > a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb$ and $0 < a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb - s_n$. Therefore because $S_n - s_n > a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb - s_n$ this implies that $\frac{ab^{\mathbf{4}} + cb}{a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb - s_n} > n$ but $n \geq 1$ so there are some values of $n$ where $\frac{ab^{\mathbf{4}} + cb}{a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb - s_n} > n$ is false. 

If $A < a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb$ and $a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb < S_n$ then $a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} + cb > s_n$ and $0 < S_n - a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} - cb$. Therefore because $S_n - a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} - cb < S_n - s_n$ this implies that $n < \frac{ab^4 + cb}{S_n - a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} - cb}$ but $n \geq 1$ so there are some values of $n$ where $n < \frac{ab^{\mathbf{4}} + cb}{S_n - a\frac{b^{\mathbf{4}}}{{\mathbf{4}}} - cb}$ is false.

**3** The inequalities (I.5) and (I.12) are special cases of the more general inequalities

$$1^{\mathbf{m}} + 2^{\mathbf{m}} + \cdots + (n-1)^{\mathbf{m}} < \frac{n^{\mathbf{m + 1}}}{\mathbf{m + 1}} < 1^{\mathbf{m}} + 2^{\mathbf{m}} + \cdots + n^{\mathbf{m}}$$

that are valid for every integer $n \geq 1$ and every integer ${\mathbf{m}} \geq 1$. Asumme the validity of (I.13) and generalize the results of Exercise 2.

We can generalize this result to $ax^{\mathbf{m}} + c$ where the process is similar as in **2.c**. We only need to change $\mathbf{3}$ with $\mathbf{m}$ and $\mathbf{4}$ with $\mathbf{m + 1}$.

## Some Basic Concepts of the Theory of Sets

### Exercises I 2.5 {#exrI2.5}

**1** Use the roster notation to designate the following sets of real numbers.

$$A = \{x | x^2 - 1 = 0 \} = \{ -1, 1 \}$$

$$B = \{x | (x - 1)^2 = 0 \} = \{ 1 \}$$

$$C = \{x | x + 8 = 9 \} = \{ 1 \}$$

In this part $x^3 - 2x^2 + x = 2 \Longleftrightarrow (x^2 + 1)(x - 2) = 0$. In that sense there are also 2 complex roots $x = -i$ and $x = i$ but they are not real numbers (Remember that $S = \mathbb{R}$)

$$D = \{x | x^3 - 2x^2 + x = 2 \} = \{ 2 \}$$

$$E = \{x | (x + 8)^2 = 9^2 \} = \{ -17, 1 \}$$

$$F = \{x | (x^2 + 16x)^2 = 17^2 \} = \{ -17, -8 - \sqrt{47}, 1, \sqrt{47} - 8 \}$$

**2** For the sets in Exercise 1, note that $B \subseteq A$. List all the inclusion relations $\subseteq$ that hold among the sets $A$, $B$, $C$, $D$, $E$, $F$. 

$B \subseteq A$, $C \subseteq A$, $C \subseteq B$, $B \subseteq C$, $E \subseteq B$, $E \subseteq C$, $E \subseteq F$ 

**3** Let $A = \{ 1 \}$, $B = \{ 1, 2 \}$. Discuss the validity of the following statements (prove the ones that are true and explain why the others are not true).

**3a** $A \subset B$

It is true because $1 \in A$ and $1 \in B$ but $2 \notin A$ and $2 \in B$ so $A \subseteq B$ and $A \neq B$

**3b** $A \subseteq B$

It is true because $1 \in A$ and $1 \in B$ but $2 \notin A$ and $2 \in B$ so $A \subset B$ or $A = B$ is true

**3c** $A \in B$

It is false because ${1}$ is not an element of $B$. Remember that $1$ is not the same as $\{ 1 \}$

**3d** $1 \in A$

It is true because $1$ is an element of $A$. 

**3e** $1 \subseteq A$

It is false because the subsets of $A$ are $\emptyset$ and $\{ 1 \}$

**3f** $1 \subset B$

It is false because the proper subsets of $B$ are $\emptyset$, $\{ 1 \}$ and $\{ 2 \}$

**4** Solve Exercise 3 if $A = \{ 1 \}$ and $B = \{ \{ 1 \}, 1 \}$. 

**4a** $A \subset B$

It is true because $1 \in A$ and $1 \in B$ but $\{ 1 \} \notin A$ and $\{ 1 \} \in B$ so $A \subseteq B$ and $A \neq B$

**4b** $A \subseteq B$

It is true because $1 \in A$ and $1 \in B$ but $\{ 1 \} \notin A$ and $\{ 1 \} \in B$ so $A \subset B$ or $A = B$ is true

**4c** $A \in B$

It is true because $\{ 1 \} \in B$ and $A = \{ 1 \}$ so $A \in B$ 

**4d** $1 \in A$

It is true because $1$ is an element of $A$. 

**4e** $1 \subseteq A$

It is false because the subsets of $A$ are $\emptyset$ and $\{ 1 \}$

**4f** $1 \subset B$

It is false because the proper subsets of $B$ are $\emptyset$, $\{ \{1 \} \}$ and $\{ 1 \}$

**5** Given the set $S = \{ 1, 2, 3, 4 \}$. Display all subsets of $S$. There are 16 altogether, counting $\emptyset$ and $S$.

$\emptyset$, $\{ 1 \}$, $\{ 2 \}$, $\{ 3 \}$, $\{ 4 \}$, $\{ 1, 2 \}$, $\{ 1, 3 \}$, $\{ 1, 4 \}$, $\{ 2, 3 \}$, $\{ 2, 4 \}$, $\{ 3, 4 \}$, $\{ 1, 2, 3 \}$, $\{ 1, 2, 4 \}$, $\{ 1, 3, 4 \}$, $\{ 2, 3, 4 \}$, $\{ 1, 2, 3, 4 \}$

**6** Given the following four sets 

$$A = \{ 1, 2 \}, \;\; B = \{ \{ 1 \}, \{ 2 \} \}, \;\; C = \{ \{ 1 \}, \{ 1, 2 \} \}, \;\; D = \{ \{ 1 \}, \{ 2 \}, \{ 1, 2 \} \}$$  

discuss the validity of the following statements (prove the ones that are true and explain why the others are not true).

**6a** $A = B$

It is false because $1 \in A$ and $1 \notin B$

**6b** $A \subseteq B$

It is false because $1 \in A$ and $1 \notin B$

**6c** $A \subset C$

It is false because $1 \in A$ and $1 \notin C$

**6d** $A \in C$

It is true because $\{ 1, 2 \} \in C$ and $A = \{ 1, 2 \}$ so $A \in C$ 

**6e** $A \subset D$

It is false because $1 \in A$ and $1 \notin D$

**6f** $B \subset C$

It is false because $\{ 2 \} \in B$ and $\{ 2 \} \notin C$

**6g** $B \subset D$

It is true because $\{ 1 \} \in B$, $\{ 2 \} \in B$ and $\{ 1 \} \in D$, $\{ 2 \} \in D$ but $\{ 1, 2 \} \notin B$ and $\{ 1, 2 \} \in D$

**6h** $B \in D$

It is not true because the elements of $D$ are $\{ 1 \}$, $\{ 2 \}$ and $\{ 1, 2 \}$ so $\{ \{ 1 \}, \{ 2 \} \}$ is not an element of $D$

**6i** $A \in D$

It is true because $\{ 1, 2 \} \in D$ and $A = \{ 1, 2 \}$ so $A \in D$

**7** Prove the following properties of set equality.

**7a** $\{ a, a \} = \{ a \}$

$$\begin{split}
  a \in \{ a, a \} & \Longleftrightarrow a \in \{ a \}
  \end{split}$$

**7b** $\{ a, b \} = \{ b, a \}$

$$\begin{split}
  a \in \{ a, b \} \land b \in \{ a, b \} & \Longleftrightarrow b \in \{ a, b \} \land a \in \{ a, b \} \\
  & \Longleftrightarrow b \in \{ b, a \} \land a \in \{ b, a \}
  \end{split}$$

**7c** $\{ a \} = \{ b, c \}$ if and only if $a = b = c$

$$\begin{split}
  \{ a \} = \{ b, c \} & \Longleftrightarrow a \in \{ b, c \} \land (b \in \{ a \} \land c \in \{ a \}) \\
  & \Longleftrightarrow (a = b \lor a = c) \land (b = a \land c = a) \\
  & \Longleftrightarrow (a = b \lor a = c) \land (a = b = c) \\
  & \Longleftrightarrow a = b = c
  \end{split}$$

Prove the set relations in Exercises 8 through 19. (Sample proofs are given at the end of this section)

**8** *Commutative laws:* $A \cup B = B \cup A$, $A \cap B = B \cap A$

$$\begin{split}
  x \in A \cup B & \Longleftrightarrow x \in A \lor x \in B \\
  & \Longleftrightarrow x \in B \lor x \in A \\
  & \Longleftrightarrow \in B \cup A
  \end{split}$$

$$\begin{split}
  x \in A \cap B & \Longleftrightarrow x \in A \land x \in B \\
  & \Longleftrightarrow x \in B \land x \in A \\
  & \Longleftrightarrow \in B \cap A
  \end{split}$$

**9** *Associative laws:* $A \cup (B \cup C) = (A \cup B) \cup C$, $A \cap (B \cap C) = (A \cap B) \cap C$

$$\begin{split}
  x \in A \cup (B \cup C) & \Longleftrightarrow x \in A \lor x \in (B \cup C) \\
  & \Longleftrightarrow x \in A \lor (x \in B \lor x \in C) \\
  & \Longleftrightarrow (x \in A \lor x \in B) \lor x \in C \\
  & \Longleftrightarrow x \in A \cup B \lor x \in C \\
  & \Longleftrightarrow x \in (A \cup B) \cup C \\
  \end{split}$$

$$\begin{split}
  x \in A \cap (B \cap C) & \Longleftrightarrow x \in A \land x \in (B \cap C) \\
  & \Longleftrightarrow x \in A \land (x \in B \land x \in C) \\
  & \Longleftrightarrow (x \in A \land x \in B) \land x \in C \\
  & \Longleftrightarrow x \in A \cap B \land x \in C \\
  & \Longleftrightarrow x \in (A \cap B) \cap C \\
  \end{split}$$

**10** *Distributive laws:* $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

$$\begin{split}
  x \in A \cap (B \cup C) & \Longleftrightarrow x \in A \land x \in (B \cup C) \\
  & \Longleftrightarrow x \in A \land (x \in B \lor x \in C) \\
  & \Longleftrightarrow (x \in A \land x \in B) \lor (x \in A \land x \in C) \\
  & \Longleftrightarrow x \in A \cap B \lor x \in A \cap C \\
  & \Longleftrightarrow x \in (A \cap B) \cup (A \cap C) \\
  \end{split}$$
  
$$\begin{split}
  x \in A \cup (B \cap C) & \Longleftrightarrow x \in A \lor x \in (B \cap C) \\
  & \Longleftrightarrow x \in A \lor (x \in B \land x \in C) \\
  & \Longleftrightarrow (x \in A \lor x \in B) \land (x \in A \lor x \in C) \\
  & \Longleftrightarrow x \in A \cup B \land x \in A \cup C \\
  & \Longleftrightarrow x \in (A \cup B) \cap (A \cup C) \\
  \end{split}$$

**11** $A \cup A = A$, $A \cap A = A$

$$\begin{split}
  x \in A \cup A & \Longleftrightarrow x \in A \lor x \in A \\
  & \Longleftrightarrow x \in A
  \end{split}$$
  
$$\begin{split}
  x \in A \cap A & \Longleftrightarrow x \in A \land x \in A \\
  & \Longleftrightarrow x \in A
  \end{split}$$
  
**12** $A \subseteq A \cup B$, $A \cap B \subseteq A$
 
$$\begin{split}
  x \in A & \Longrightarrow x \in A \lor x \in B \\
  & \Longrightarrow x \in A \cup B
  \end{split}$$
  
In this part we also have $A \cap B \subseteq B$ but because $A$ and $B$ are arbitrary sets the proof related to $A \cap B \subseteq A$ include this case. 
  
$$\begin{split}
  x \in A \cap B & \Longrightarrow x \in A \land x \in B \\
  & \Longrightarrow x \in A
  \end{split}$$
  
**13** $A \cup \emptyset = A$, $A \cap \emptyset = \emptyset$

In this part remember that $x \in \emptyset$ is false but $x \in A \lor x \in \emptyset$ is true if and only if $x \in A$  

$$\begin{split}
  x \in A \cup \emptyset & \Longleftrightarrow x \in A \lor x \in \emptyset \\
  & \Longleftrightarrow x \in A
  \end{split}$$

In this part remember that $\emptyset$ is always a subset of every set and that $\emptyset$ is a set containing no elements whatsoever. Taking into account these true statements we can prove that $A \cap \emptyset = \emptyset$.  

First, $A \cap \emptyset \supseteq \emptyset$ is true because $\emptyset$ is always a subset of every set.

Second, if $x \in A \cap \emptyset$ then $x \in A \land x \in \emptyset$ but $x \notin \emptyset$ so $x \in A \cap \emptyset$ is false. Therefore for any $x$ we have that $x \notin A \cap \emptyset$ must be true. Also let's assume that $A \cap \emptyset \nsubseteq \emptyset$ where this would mean that there is some $y$ such that $y \in A \cap \emptyset \land y \notin \emptyset$ but this is false because $A \cap \emptyset$ is a set containing no elements whatsoever. Therefore $A \cap \emptyset \nsubseteq \emptyset$ is false and $A \cap \emptyset \subseteq \emptyset$ must be true.

Third, because $A \cap \emptyset \supseteq \emptyset$ and $A \cap \emptyset \subseteq \emptyset$ are true we have that $A \cap \emptyset = \emptyset$. 

**14** $A \cup (A \cap B) = A$, $A \cap (A \cup B) = A$

First we prove $A \cup (A \cap B) = A$ in 2 parts:

$$\begin{split}
  x \in A \cup (A \cap B) & \Longleftrightarrow x \in A \lor x \in A \cap B \\
  & \Longleftrightarrow x \in A \lor (x \in A \land x \in B) \\
  & \Longleftrightarrow (x \in A \lor x \in A) \land (x \in A \lor x \in B) \\
  & \Longleftrightarrow x \in A \land x \in (A \cup B) \\
  & \Longrightarrow x \in A \\
  x \in A & \Longrightarrow x \in A \lor x \in A \cap B \\
  & \Longleftrightarrow x \in A \cup (A \cap B)
  \end{split}$$
  
$$\begin{split}
  x \in A & \Longrightarrow x \in A \lor x \in A \cap B \\
  & \Longleftrightarrow x \in A \cup (A \cap B)
  \end{split}$$

Second we prove $A \cap (A \cup B) = A$ using $A \cup (A \cap B) = A$:

$$\begin{split}
  A \cap (A \cup B) & = (A \cap A) \cup (A \cap B) \\
  & = A \cup (A \cap B) \\
  & = A
  \end{split}$$
  
**15** If $A \subseteq C$ and $B \subseteq C$, then $A \cup B \subseteq C$

$$\begin{split}
  (A \subseteq C \land B \subseteq C) \land x \in A \cup B & \Longleftrightarrow (A \subseteq C \land B \subseteq C) \land (x \in A \lor x \in B) \\
  & \Longrightarrow (A \subseteq C \land x \in A) \lor (B \subseteq C \land x \in B) \\
  & \Longrightarrow x \in C
  \end{split}$$

**16** If $C \subseteq A$ and $C \subseteq B$, then $C \subseteq A \cap B$

$$\begin{split}
  (C \subseteq A \land C \subseteq B) \land x \in C & \Longrightarrow (C \subseteq A \land x \in C) \land (C \subseteq B \land x \in C) \\
  & \Longrightarrow x \in A \land x \in B \\
  & \Longrightarrow x \in A \cap B
  \end{split}$$

**17a** If $A \subset B$ and $B \subset C$, prove that $A \subset C$

$$\begin{split}
  (A \subset B \land B \subset C) \land x \in A & \Longrightarrow (x \in B \land B \subset C) \land \exists y: y \notin A \land y \in B \\
  & \Longrightarrow x \in C \land \exists y: y \notin A \land y \in C
  \end{split}$$  

**17b** If $A \subseteq B$ and $B \subseteq C$, prove that $A \subseteq C$

$$\begin{split}
  (A \subseteq B \land B \subseteq C) \land x \in A & \Longrightarrow x \in B \land B \subseteq C \\
  & \Longrightarrow x \in C
  \end{split}$$

**17c** What can you conclude if $A \subset B$ and $B \subseteq C$?

First, if $B = C$ we can conclude that $A \subset C$.

Second, if $B \subset C$ using the proof in **17a** we can conclude again that $A \subset C$.

**17d** If $x \in A$ and $A \subseteq B$, is it necessarily true that $x \in B$?

Yes, because $A \subseteq B$ means that $\forall y: y \in A \Longrightarrow y \in B$ and if $x \in A$ is true also $x \in B$ must be true.

**17e** If $x \in A$ and $A \in B$ is it necessarily true that $x \in B$?

No, for example in **6** a counterexample was presented where $A = \{ 1, 2 \}$ and $C = \{ \{ 1\}, \{ 1, 2 \} \}$. In this case we have that $1 \in A$, $A \in C$ and $1 \notin C$.

**18** $A - (B \cap C) = (A - B) \cup (A - C)$

$$\begin{split}
  x \in A - (B \cap C) & \Longleftrightarrow x \in A \land x \notin B \cap C \\
  & \Longleftrightarrow x \in A \land (x \notin B \lor x \notin C) \\
  & \Longleftrightarrow (x \in A \land x \notin B) \lor (x \in A \land x \notin C) \\
  & \Longleftrightarrow x \in A - B \lor x \in A - C \\
  & \Longleftrightarrow x \in A - B \cup A - C
  \end{split}$$

**19** Let $\mathscr{F}$ be a class of sets. Then 

$$B - \bigcup_{A \in \mathscr{F}} A = \bigcap_{A \in \mathscr{F}} (B - A) \text{ and } B - \bigcap_{A \in \mathscr{F}} A = \bigcup_{A \in \mathscr{F}} (B - A)$$

$$\begin{split}
  x \in B - \bigcup_{A \in \mathscr{F}} A & \Longleftrightarrow x \in B \land x \notin \bigcup_{A \in \mathscr{F}} A \\
  & \Longleftrightarrow x \in B \land \forall A: A \in \mathscr{F} \land x \notin A \\
  & \Longleftrightarrow (\forall A: A \in \mathscr{F} \land x \in B) \land (\forall A: A \in \mathscr{F} \land x \notin A) \\
  & \Longleftrightarrow \forall A: A \in \mathscr{F} \land (x \in B \land x \notin A) \\
  & \Longleftrightarrow \forall A: A \in \mathscr{F} \land x \in B - A \\
  & \Longleftrightarrow x \in \bigcap_{A \in \mathscr{F}} B - A
  \end{split}$$

$$\begin{split}
  x \in B - \bigcap_{A \in \mathscr{F}} A & \Longleftrightarrow x \in B \land x \notin \bigcap_{A \in \mathscr{F}} A \\
  & \Longleftrightarrow x \in B \land \exists A: A \in \mathscr{F} \land x \notin A \\
  & \Longleftrightarrow (\exists A: A \in \mathscr{F} \land x \in B) \land (\exists A: A \in \mathscr{F} \land x \notin A) \\
  & \Longleftrightarrow \exists A: A \in \mathscr{F} \land (x \in B \land x \notin A) \\
  & \Longleftrightarrow \exists A: A \in \mathscr{F} \land x \in B - A \\
  & \Longleftrightarrow x \in \bigcup_{A \in \mathscr{F}} B - A
  \end{split}$$
  
**20a** Prove that one of the following two formulas is always right and the other one is sometimes wrong:

$$\text{(i) } A - (B - C) = (A - B) \cup C$$

$\text{(i)}$ is false because $A \cap C = C$ is not always true:

$$\begin{split}
  x \in A - (B - C) & \Longleftrightarrow x \in A \land x \notin B - C \\
  & \Longleftrightarrow x \in A \land (x \notin B \lor x \in C) \\
  & \Longleftrightarrow (x \in A \land x \notin B) \lor (x \in A \land x \in C) \\
  & \Longleftrightarrow x \in A - B \lor x \in A \cap C \\
  & \Longleftrightarrow x \in (A - B) \cup (A \cap C)
  \end{split}$$

$$\text{(ii) } A - (B \cup C) = (A - B) - C$$

$\text{(ii) }$ is true because:

$$\begin{split}
  x \in A - (B \cup C) & \Longleftrightarrow x \in A \land x \notin B \cup C \\
  & \Longleftrightarrow x \in A \land (x \notin B \land x \notin C) \\
  & \Longleftrightarrow (x \in A \land x \notin B) \land x \notin C \\
  & \Longleftrightarrow x \in A - B \land x \notin C \\
  & \Longleftrightarrow x \in (A - B) - C
  \end{split}$$

**20b** State an additional necessary and sufficient condition for the formula which is sometimes incorrect to be always right.

For $\text{(i)}$ to be true we need that $A \cap C = C$. Also because $A \cap C = C \Longleftrightarrow C \subseteq A$ another alternative is to point out that $C \subseteq A$ is a necessary and sufficient condition for $\text{(i)}$ to be true.   

## A Set of Axioms for the Real-Number System

### Exercises I 3.3 {#exrI3.3}

**1** Prove Theorems I.5 through I.15, using Axioms 1 through 6 and Theorems I.1 through I.4.

**Theorem I.5** $a(b - c) = ab - ac$

$$\begin{split}
  a(b - c) & = a(b  + (-c)) & \text{ Theorem I.3} \\
  & = ab  + a(-c) & \text{ Axiom 3} \\
  & = (ab + a(-c)) + 0 & \text{ Axiom 4} \\
  & = (ab + a(-c)) + (ac + (-ac)) & \text{ Theorem I.2} \\
  & = ab + (a(-c) + (ac + (-ac))) & \text{ Axiom 2} \\
  & = ab + ((a(-c) + ac) + (-ac)) & \text{ Axiom 2} \\
  & = ab + (a((-c) + c) + (-ac)) & \text{ Axiom 3} \\
  & = ab + (a( c + (-c)) + (-ac)) & \text{ Axiom 1} \\
  & = ab + (a0 + (-ac)) & \text{ Theorem I.2} \\
  & = (ab + a0) + (-ac) & \text{ Axiom 2} \\
  & = a(b + 0) + (-ac) & \text{ Axiom 3} \\
  & = ab + (-ac) & \text{ Axiom 4} \\
  & = ab -ac & \text{ Theorem I.3}
  \end{split}$$

**Theorem I.6** $0a = a0 = 0$

First 

$$0a = a0 \text{ Axiom 1}$$

Second

$$\begin{split}
  a0 & = a(0 + 0) & \text{ Axiom 4} \\
  & = a0 + a0 & \text{ Axiom 3}    
  \end{split}$$

Third 

$$\begin{split}
  a0 & = a0 + 0 \text{ Axiom 4}
  \end{split}$$

Fourth

$$\begin{split}
  a0 + a0 & = a0 + 0 \Longrightarrow a0 = 0  \text{ Theorem I.1}
  \end{split}$$

Therefore $0a = a0 = 0$

**Theorem I.7** CANCELLATION LAW OF MULTIPLICATION. If $ab = ac$ and $a \neq 0$, then $b = c$. (In particular, this shows that the number $1$ of Axiom 4 is unique)

First

$$\begin{split}
  a \neq 0 \Longrightarrow \exists y: y \in \mathbb{R} \land ay = 1 \text{ Axiom 6}
  \end{split}$$

Second

$$\begin{split}
  b & = 1b & \text{ Axiom 4} \\
  & = (ay)b \\
  & = (ya)b & \text{ Axiom 1} \\
  & = y(ab) & \text{ Axiom 2} \\
  & = y(ac) \\
  & = c & \text{ Axioms 2, 1, 4} \\
  \end{split}$$

**Theorem I.8** POSSIBILITY OF DIVISION. Given $a$ and $b$ with $a \neq 0$, there is exactly one $x$ such that $ax = b$. This $x$ is denoted by $b/a$ or $\frac{a}{b}$ and is called the quotient of $b$ and $a$. In particular, $1/a$ is also written $a^{-1}$ and is called the reciprocal of a.

First

$$\begin{split}
  a \neq 0 \Longrightarrow \exists y: y \in \mathbb{R} \land ay = 1 \text{ Theorem I.3}
  \end{split}$$

Second

$$\begin{split}
  b & = 1b & \text{ Axiom 4} \\
  & = (ay)b \\
  & = a(yb) & \text{ Axiom 2} \\
  & = ax & \text{ where } x = yb 
  \end{split}$$

Third, if $ax = b$ and $az = b$ then by **Theorem I.7** $x = z$ so $x$ is unique.

**Theorem I.9** If $a \neq 0$, then $b/a = ba^{-1}$

First

$$\begin{split}
  a \neq 0 \land 1 \in \mathbb{R} \Longrightarrow \exists! a^{-1}: a^{-1} \in \mathbb{R} \land aa^{-1} = 1 \text{ Theorem I.8}
  \end{split}$$

Second

$$\begin{split}
  a \neq 0 \land b \in \mathbb{R} \Longrightarrow \exists! b/a: b/a \in \mathbb{R} \land ab/a = b \text{ Theorem I.8}
  \end{split}$$

Third

$$\begin{split}
  b & = 1b & \text{ Axiom 4} \\
  & = (aa^{-1})b \\
  & = a(a^{-1}b) & \text{ Axiom 2} \\
  & = a(ba^{-1}) & \text{ Axiom 1} \\
  \end{split}$$

Fourth

$$\begin{split}
  ab/a = b \land b = a(ba^{-1}) \Longrightarrow b/a = ba^{-1} \text{ Theorem I.7}
  \end{split}$$

**Theorem I.10** If $a \neq 0$, then $(a^{-1})^{{-1}} = a$

First, if $a \neq 0$ we can use **Theorem I.8**

Second

$$\begin{split}
  1 & = a^{-1}(a^{-1})^{{-1}} \text{ Theorem I.8} \\
  \end{split}$$

Third

$$\begin{split}
  1 & = aa^{-1} & \text{ Theorem I.8} \\
  & = a^{-1}a & \text{ Axiom 1}
  \end{split}$$

Fourth

$$\begin{split}
  1 = a^{-1}(a^{-1})^{{-1}} \land 1 = a^{-1}a  \Longrightarrow (a^{-1})^{{-1}} = a \text{ Theorem I.7}
  \end{split}$$

**Theorem I.11** If $ab = 0$, then $a = 0$ or $b = 0$

To prove this statement we are going to use the method of *reductio ad absurdum* and the equivalence between the statements $A \Longrightarrow B$ and $\neg(A \land (\neg B))$. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First

$$\begin{split}
  (ab = 0 \Longrightarrow (a = 0 \lor b = 0)) & \Longleftrightarrow \neg(ab = 0 \land \neg(a = 0 \lor b = 0)) \\
  & \Longleftrightarrow \neg(ab = 0 \land (a \neq 0 \land b \neq 0))
  \end{split}$$

Second, we are going to assume $(ab = 0 \land (a \neq 0 \land b \neq 0)$, find a contradiction and conclude that $\neg(ab = 0 \land (a \neq 0 \land b \neq 0)$ is true where this means that the original statement is true: $ab = 0 \Longrightarrow (a = 0 \lor b = 0)$

Third

$$\begin{split}
  a0 = 0 \text{ Theorem I.6}
  \end{split}$$

Fourth

$$\begin{split}
  a0 = 0 \land ab = 0 \Longrightarrow b = 0 \text{ Theorem I.7}
  \end{split}$$

Fifth, there is a contradiction because we assume initially that $b \neq 0$ so $\neg(ab = 0 \land (a \neq 0 \land b \neq 0)$ is true by the method of *reductio ad absurdum*. Therefore the original statement is true: $ab = 0 \Longrightarrow (a = 0 \lor b = 0)$

**Theorem I.12** $(-a)b = -(ab)$ and $(-a)(-b) = ab$

In the case of $(-a)b = -(ab)$:

First

$$\begin{split}
  0 & = (-a)b + (-(-a)b) & \text{ Theorem I.8} \\
  & = (-a)b + (ab) & \text{ Theorem I.4} \\
  & = (ab) + (-a)b & \text{ Axiom 1}
  \end{split}$$

Second

$$\begin{split}
  0 & = (ab) + (-(ab)) \text{ Theorem I.8}
  \end{split}$$

Third

$$\begin{split}
  0 = (ab) + (-a)b \land 0 = (ab) + (-(ab)) \Longrightarrow (-a)b = (-(ab))  \text{ Theorem I.1}
  \end{split}$$

Fourth

$$\begin{split}
  -(ab) & = 0 - (ab) & \text{ Theorem I.2} \\
  & = 0 + (-(ab)) & \text{ Theorem I.3} \\
  & = (-(ab)) + 0 & \text{ Axiom 1} \\
  & = (-(ab)) & \text{ Axiom 4}
  \end{split}$$

Fifth

$$\begin{split}
  (-a)b & = (-(ab)) \\
  & = -(ab)
  \end{split}$$

In the case of $(-a)(-b) = ab$ we use the first part in **Theorem I.12**:

$$\begin{split}
  (-a)(-b) & = -(a(-b)) \\
  & = -((-b)a) & \text{ Axiom 1} \\
  & = (-(-b))a \\
  & = -(-b)a & \text{ Theorem I.2, Theorem I.3, Axiom 4} \\
  & = ba & \text{ Theorem I.4} \\
  & = ab & \text{ Axiom 1}
  \end{split}$$

**Theorem I.13** $(a / b) + (c / d) = (ad + bc) / (bd)$ if $b \neq 0$ and $d \neq 0$

First we prove that $bd(b^{-1}d^{-1}) = 1$

$$\begin{split}
  bd(b^{-1}d^{-1}) & = bb^{-1}(dd^{-1}) \text{ Axiom 2, Axiom 1} \\
  & = 1\cdot1 \text{ Theorem I.8} \\
  & = 1 \text{ Axiom 4}
  \end{split}$$

Second we prove that $(b^{-1}d^{-1}) = (bd)^{-1}$ taking into account that $bd(bd)^{-1} = 1$ by **Theorem I.8** 

$$\begin{split}
  bd(b^{-1}d^{-1}) = 1 \land bd(bd)^{-1} = 1 \Longrightarrow (b^{-1}d^{-1}) = (bd)^{-1} \text{ Theorem I.11, Theorem I.7}  
  \end{split}$$


Third

$$\begin{split}
  (a / b) + (c / d) & = (ab^{-1}) + (cd^{-1}) & \text{ Theorem I.9} \\
  & = 1(ab^{-1}) + 1(cd^{-1}) & \text{ Axiom 4} \\
  & = dd^{-1}(ab^{-1}) + bb^{-1}(cd^{-1}) & \text{ Theorem I.8, Theorem I.9} \\
  & = (b^{-1}d^{-1})ad + (b^{-1}d^{-1})bc & \text{ Axiom 2, Axiom 1} \\
  & = (ad + bc)(b^{-1}d^{-1}) & \text{ Axiom 3, Axiom 1} \\
  & = (ad + bc)(bd)^{-1} \\
  & = (ad + bc) / (bd) & \text{  Theorem I.9}
  \end{split}$$
  
**Theorem I.14** $(a / b)(c / d) = (ac) / (bd)$ if $b \neq 0$ and $d \neq 0$

First, check out the proof of **Theorem I.13** where we prove that $(b^{-1}d^{-1}) = (bd)^{-1}$

Second

$$\begin{split}
  (a / b)(c / d) & = (ab^{-1})(cd^{-1}) & \text{ Theorem I.9} \\
  & = (ac)(b^{-1}d^{-1}) & \text{ Axiom 2, Axiom 1} \\
  & = (ac)(bd)^{-1} & \text{ Axiom 2, Axiom 1} \\
  & = (ac) / (bd) & \text{ Theorem I.9}
  \end{split}$$

**Theorem I.15** $(a / b) / (c / d) = (ad) / (bc)$ if $b \neq 0$, $c \neq 0$ and $d \neq 0$ 

First, check out the proof of **Theorem I.13** where we prove that $(b^{-1}d^{-1}) = (bd)^{-1}$. Using this result we can also prove that $(b^{-1}(d^{-1})^{-1}) = (bd^{-1})^{-1}$ where $b$ and $d$ are any real numbers different from $0$

Second

$$\begin{split}
  (a / b) / (c / d) = (ad) / (bc) & = (ab^{-1}) / (cd^{-1}) & \text{ Theorem I.9} \\
  & = (ab^{-1})(cd^{-1})^{-1} & \text{ Theorem I.9} \\
  & = (ab^{-1})(c^{-1}(d^{-1})^{-1}) \\
  & = (ab^{-1})(c^{-1}d) & \text{ Theorem I.10} \\
  & = (ad)(b^{-1}c^{-1}) & \text{ Axiom 2, Axiom 1} \\
  & = (ad)(bc)^{-1} \\
  & = (ad) / (bc) & \text{ Theorem I.9}
  \end{split}$$

In Exercises 2 through 10, prove the given statements or establish the given questions. You may use Axioms 1 through 6 and Theorems I.1 through I.15.

**2** $-0 = 0$

$$\begin{split}
  -0 \land 0 & \Longleftrightarrow 0 + (-0) = 0 \land 0 + 0 = 0 & \text{ Axiom 4, Theorem I.2} \\
  & \Longleftrightarrow (-0) = 0 & \text{ Theorem I.1} \\
  & \Longleftrightarrow -0 = 0 & \text{ Theorem I.3, Theorem I.2, Axiom 4}
  \end{split}$$

**3** $1^{-1} = 1$

$$\begin{split}
  1^{-1} \land 1 & \Longleftrightarrow 1\cdot 1^{-1} = 1 \land 1 \cdot 1 = 1 & \text{ Axiom 4, Axiom 6, Theorem I.8} \\
  & \Longleftrightarrow 1^{-1} = 1 & \text{ Theorem I.7}
  \end{split}$$

**4** Zero has no reciprocal

To prove this statement we are going to use the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First, let's assume that $x$ is the reciprocal of $0$. Then by **Theorem 1.8** $0 \cdot x = 1$.

Second by **Theorem 1.6** $0 \cdot x = 0$ but we are assuming that $0 \cdot x = 1$ so $0 = 1$.

Third, there is a contradiction because by **Axiom 4** $0$ and $1$ are distinct.

Fourth $0 \cdot x = 1$ is false and $0 \cdot x \neq 1$ is true so zero has no reciprocal.

**5** $-(a + b) = -a - b$

$$\begin{split}
  -(a + b) & = -1(a + b) & \text{ Axiom 4} \\
  & = -1a + (-1b) & \text{ Axiom 3} \\
  & = -a + (-b) & \text{ Axiom 4} \\
  & = -a - b & \text{ Theorem I.3} \\
  \end{split}$$

**6** $-(a - b) = -a + b$

$$\begin{split}
  -(a - b) & = -1(a - b) & \text{ Axiom 4} \\
  & = -1a + (-1)(-b) & \text{ Axiom 3} \\
  & = -1a + 1b & \text{ Theorem I.12} \\
  & = -a + b & \text{ Axiom 4}
  \end{split}$$

**7** $(a - b) + (b - c) = a + c$

$$\begin{split}
  (a - b) + (b - c) & = (a + (-b)) + (b + (-c)) & \text{ Theorem I.3} \\
  & = (a + ((-b) + b)) + (-c) & \text{ Axiom 2} \\
  & = (a + 0) + (-c) & \text{ Axiom 1, Axiom 4} \\
  & = a - c & \text{ Axiom 4, Theorem I.3} \\
  \end{split}$$

**8** If $a \neq 0$ and $b \neq 0$, then $(ab)^{-1} = (a^{-1}b^{-1})$

Check out the proof of **Theorem I.13** where we prove that $(b^{-1}d^{-1}) = (bd)^{-1}$

**9** $-(a / b) = (-a / b) = a / (-b)$ if $b \neq 0$

First we prove that $-1 \neq 0$. If $-1 = 0$ then $1 + (-1) = 0$ and $0 + 0 = 0$ will imply that $1 = 0$ which is not true by **Axiom 4**. Therefore by *reductio ad absurdum* (please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]) $-1 \neq 0$ must be true. 

Second we prove that $-b^{-1} = (-b)^{-1}$. Because $(-b)(-b)^1 = 1$ by **Theorem I.8**, $(-b)(-b^{-1}) = bb^{-1} = 1$ by **Theorem I.12, Theorem I.8** and $-b = -1b \neq 0$ by **Theorem I.11** we can conclude that $-b^{-1} = (-b)^{-1}$ using **Theorem I.7**. 

$$\begin{split}
  -(a / b) & = -(ab^{-1}) & \text{ Theorem I.9} \\
  & = (-ab^{-1}) & \text{ Theorem I.3, Axiom 4} \\
  & = (-a / b) & \text{ Theorem I.9} \\
  & = (-ab^{-1}) & \text{ Theorem I.9} \\
  & = (-b^{-1}a) & \text{ Axiom 1} \\
  & = -(b^{-1}a) & \text{ Theorem I.3, Axiom 4} \\
  & = (-b^{-1})a & \text{ Theorem I.12} \\
  & = (-b)^{-1}a \\
  & = a(-b)^{-1} & \text{ Axiom 1} \\
  & = a / (-b) & \text{ Theorem I.9}
  \end{split}$$
  
  **10** $(a / b) - (c / d) = (ad - bc) / (bd)$ if $b \neq 0$ and $d \neq 0$
  
  In this part remember exercise **8**

$$\begin{split}
  (a / b) - (c / d) & = (ab^{-1}) - (cd^{-1}) & \text{ Theorem I.9} \\
  & = 1(ab^{-1}) - 1(cd^{-1}) & \text{ Axiom 4} \\
  & = dd^{-1}(ab^{-1}) - bb^{-1}(cd^{-1}) & \text{ Theorem I.8, Theorem I.9} \\
  & = (b^{-1}d^{-1})ad - (b^{-1}d^{-1})bc & \text{ Axiom 2, Axiom 1} \\
  & = (ad - bc)(b^{-1}d^{-1}) & \text{ Axiom 3, Theorem I.5, Axiom 1} \\
  & = (ad - bc)(bd)^{-1} \\
  & = (ad - bc) / (bd) & \text{  Theorem I.9}
  \end{split}$$

### Exercises I 3.5 {#exrI3.5}

**1** Prove Theorems I.22 through I.25, using the earlier theorems and Axioms 1 through 9

**Theorem I.22** If $a < b$ and $c < 0$, then $ac > bc$

$$\begin{split}
  0 > c \land b > a & \Longrightarrow 0 - c \in \mathbb{R}^+ \land b - a \in \mathbb{R}^+ & \text{ By definition} \\
  & \Longrightarrow -c \in \mathbb{R}^+ \land b - a \in \mathbb{R}^+ & \text{ Theorem I.2} \\
  & \Longrightarrow -c(b - a) \in \mathbb{R}^+ & \text{ Axiom 7} \\
  & \Longrightarrow -c(b + (-a)) \in \mathbb{R}^+ & \text{ Theorem I.3} \\
  & \Longrightarrow -cb + -c(-a) \in \mathbb{R}^+ & \text{ Axiom 3} \\
  & \Longrightarrow -cb + ca \in \mathbb{R}^+ & \text{ Theorem I.5} \\
  & \Longrightarrow ca + (-cb)  \in \mathbb{R}^+ & \text{ Axiom 1} \\
  & \Longrightarrow ca - cb  \in \mathbb{R}^+ & \text{ Theorem I.3} \\
  & \Longrightarrow ca > cb & \text{ By definition}
  \end{split}$$

**Theorem I.23** If $a < b$, then $-a > -b$. In particular, if $a < 0$, then $-a > 0$.

$$\begin{split}
  b > a & \Longrightarrow & b - a \in \mathbb{R}^+ & \text{ By definition} \\
  & \Longrightarrow & b + (- a) \in \mathbb{R}^+ & \text{ Theorem I.3} \\
  & \Longrightarrow & (- a) + -(-b) \in \mathbb{R}^+ & \text{ Axiom 1, Theorem I.4} \\
  & \Longrightarrow & (- a) + (-(-b)) \in \mathbb{R}^+ & \text{ Theorem I.3, Theorem I.2, Axiom 4} \\
  & \Longrightarrow & (- a) - (-b) \in \mathbb{R}^+ & \text{ Theorem I.3} \\
  & \Longrightarrow & (- a) > (-b) & \text{ By definition} \\
  & \Longrightarrow & -a > -b & \text{ Theorem I.3, Theorem I.2, Axiom 4}
  \end{split}$$

$$\begin{split}
  a < 0 & \Longrightarrow -a > -0 & \text{ Theorem I.23} \\
  & \Longrightarrow -a > 0 & \text{ Exercises I 3.3 2}
  \end{split}$$

**Theorem I.24** If $ab > 0$, then both $a$ and $b$ are positive or both are negative

To prove this statement we are going to use the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First, we assume that $\neg((a > 0 \land b > 0) \lor (a < 0 \land b < 0))$ is true and find a contradiction by inspecting each case.

Second, if $a = 0$ or $b = 0$ then $ab = 0$ by **Theorem I.6**. But we would have that $ab > 0$ and $ab = 0$ which is a contradiction by **Theorem I.16**

Third, if $a < 0$ and $b > 0$ then $ab < a0 = 0$ by **Theorem I.19** and **Theorem I.6**. But we would have that $ab < 0$ and $ab = 0$ which is a contradiction by **Theorem I.16**

Fourth, if $b < 0$ and $a > 0$ then $ab = ba < b0 = 0$ by **Theorem I.19**, **Theorem I.6** and **Axiom 1**. But we would have that $ab < 0$ and $ab = 0$ which is a contradiction by **Theorem I.16**

Fifth, we conclude that the original statement is true: $(a > 0 \land b > 0) \lor (a < 0 \land b < 0)$.

**Theorem I.25** If $a < c$ and $b < d$ then $a + b < c + d$

$$\begin{split}
  c > a \land d > b & \Longrightarrow c - a \in \mathbb{R}^+ \land d - b \in \mathbb{R}^+ & \text{ By definition} \\
  & \Longrightarrow (c - a) + (d - b) \in \mathbb{R}^+ & \text{ Axiom 7} \\
  & \Longrightarrow (c + (-a)) + (d + (-b)) \in \mathbb{R}^+ & \text{ Theorem I.3} \\
  & \Longrightarrow (c + ((-a) + d)) + (-b) \in \mathbb{R}^+ & \text{ Axiom 2} \\
  & \Longrightarrow (c + (d + (-a))) + (-b) \in \mathbb{R}^+ & \text{ Axiom 1} \\
  & \Longrightarrow (c + d) + ((-a) + (-b)) \in \mathbb{R}^+ & \text{ Axiom 2} \\
  & \Longrightarrow (c + d) + (-a + (-b)) \in \mathbb{R}^+ & \text{ Theorem I.3, Theorem I.2, Axiom 4} \\
  & \Longrightarrow (c + d) + (-a - b) \in \mathbb{R}^+ & \text{ Theorem I.3} \\
  & \Longrightarrow (c + d) + (-(a + b)) \in \mathbb{R}^+ & \text{ Exercises I 3.3 5} \\
  & \Longrightarrow (c + d) - (a + b) \in \mathbb{R}^+ & \text{ Theorem I.3} \\
  & \Longrightarrow c + d > a + b & \text{ By definition}
  \end{split}$$

**2** There is no real number $x$ such that $x^2 + 1 = 0$

To prove this statement we are going to use the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First, we assume that $x^2 + 1 = 0$ is true and find a contradiction

Second, if $x = 0$ then $x^2 = 0$ by **Theorem I.6**. But we would have that $0 + 1 = 1 = 0$ by **Axiom 4** and **Axiom 1**. But $1 > 0$ by **Theorem I.21** where this is a contradiction by **Theorem I.16**      

Third, if $x \neq 0$ then $x^2 + 1 = 1 + x^2 = 0$ and therefore $x^2 = -1$ by **Axiom 1** and **Theorem I.2**

Fourth, $-1 < 0$ by **Theorem I.21**, **Theorem I.23** and **Exercises I 3.3 2** so $-1 = x^2 < 0$. But $x^2 > 0$ by **Theorem I.20** where this is a contradiction by **Theorem I.16** 

Fifth, we conclude that $x^2 + 1 = 0$ is false so $x^2 + 1 \neq 0$ must be true.

**3** The sum of two negative numbers is negative

$$\begin{split}
  a < 0 \land b < 0 & \Longrightarrow -a > 0 \land -b > 0 & \text{ Theorem I.23} \\
  & \Longrightarrow -a \in \mathbb{R} \land -b \in \mathbb{R} & \text{ By definition} \\
  & \Longrightarrow -a + (-b) \in \mathbb{R} & \text{ Axiom 7} \\
  & \Longrightarrow -a - b \in \mathbb{R} & \text{ Theorem I.3} \\
  & \Longrightarrow -(a + b) \in \mathbb{R} & \text{ Exercises I 3.3 5} \\
  & \Longrightarrow -(a + b) > 0 & \text{ By definition} \\
  & \Longrightarrow -(-(a + b)) < 0 & \text{ Theorem I.23, Exercises I 3.3 2} \\
  & \Longrightarrow a + b < 0 & \text{ Theorem I.4} \\
  \end{split}$$

**4** If $a > 0$, then $1/a > 0$; if $a < 0$, then $1/a < 0$

$$\begin{split}
  a > 0 & \Longrightarrow a > 0 \land aa^{-1} = 1 \land 1 > 0 & \text{ Theorem I.8, Theorem I.21} \\
  & \Longrightarrow a^{-1} > 0 & \text{ Theorem I.24} \\
  & \Longrightarrow 1/a > 0 & \text{ By definition}
  \end{split}$$

$$\begin{split}
  a < 0 & \Longrightarrow a < 0 \land aa^{-1} = 1 \land 1 > 0 & \text{ Theorem I.8, Theorem I.21} \\
  & \Longrightarrow a^{-1} < 0 & \text{ Theorem I.24} \\
  & \Longrightarrow 1/a < 0 & \text{ By definition}
  \end{split}$$

**5** If $0 < a < b$, then $0 < b^{-1} < a^{-1}$

$$\begin{split}
  0 < a < b & \Longrightarrow 0 < b \land aa^{-1} < ba^{-1} & \text{ Theorem I.17, Theorem I.19} \\
  & \Longrightarrow 0 < b \land 1 < a^{-1}b & \text{ Theorem I.8, Axiom 1} \\
  & \Longrightarrow 0 < b \land 1b^{-1} < (a^{-1}b)b^{-1} & \text{ Theorem I.19} \\
  & \Longrightarrow 0 < b \land b^{-1} < a^{-1}1 & \text{ Axiom 2, Theorem I.8} \\
  & \Longrightarrow 0 < b \land b^{-1} < a^{-1} & \text{ Axiom 2, Axiom 4} \\
  & \Longrightarrow 0 < b^{-1} < a^{-1} & \text{ Exercises I 3.5 4} \\
  \end{split}$$

**6** If $a \leq b$ and $b \leq c$, then $a \leq c$

$$\begin{split}
  a \leq b \land b \leq c & \Longrightarrow (a < b \lor a = b) \land (b < c \lor b = c) & \text{ By definition} \\
  & \Longrightarrow (a < b \land b < c) \lor (a < c) \lor (a = c) \\
  & \Longrightarrow (a < c) \lor (a = c) & \text{ Theorem I.17} \\
  & \Longrightarrow a \leq c & \text{ By definition} \\
  \end{split}$$

**7** If $a \leq b$ and $b \leq c$, and $a = c$, then $b = c$

$$\begin{split}
  a \leq b \land b \leq c \land a = c & \Longrightarrow  c \leq b \land b \leq c \\
  & \Longrightarrow (c < b \land b < c) \lor (b < b) \lor (c = b) \\
  & \Longrightarrow c = b & \text{ Theorem I.16}
  \end{split}$$

**8** For all real $a$ and $b$ we have $a^2 + b^2 \geq 0$. If $a$ and $b$ not both $0$, then $a^2 + b^2 > 0$

$$\begin{split}
  a \in \mathbb{R} \land b \in \mathbb{R} & \Longrightarrow a^2 \geq 0 \land b^2 \geq 0 & \text{ Theorem I.20, Theorem I.6} \\
  & \Longrightarrow \left[ \begin{split}
                     (a^2 > 0 \land b^2 > 0) & \lor \\
                     (a^2 = 0 \land b^2 > 0) & \lor \\
                     (a^2 > 0 \land b^2 = 0) & \lor \\
                     (a^2 = 0 \land b^2 = 0)
                     \end{split} \right] \\
  & \Longrightarrow \left[ \begin{split}
                     (a^2 + b^2 > 0 + 0) & \lor \\
                     (0 + b^2 > 0 + 0) & \lor \\
                     (a^2 + 0 > 0 + 0) & \lor \\
                     (a^2 + b^2 = 0 + 0)
                     \end{split} \right] & \text{ Theorem I.25, Theorem I.18} \\
  & \Longrightarrow \left[ \begin{split}
                     (a^2 + b^2 > 0) & \lor \\
                     (0 + b^2 > 0) & \lor \\
                     (a^2 + 0 > 0) & \lor \\
                     (a^2 + b^2 = 0)
                     \end{split} \right] & \text{ Axiom 4, Theorem I.18} \\
  & \Longrightarrow \left[ \begin{split}
                     (a^2 + b^2 > 0) & \lor \\
                     (a^2 + b^2 = 0)
                     \end{split} \right] \\
  & \Longrightarrow a^2 + b^2 \geq 0 & \text{ By definition} 
                     \end{split}$$

**9** There is no real number $a$ such that $x \leq a$ for all real $x$

To prove this statement we are going to use the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First, let's assume that there is a real number $a$ such that $x \leq a$ for all real $x$ and find a contradiction.

Second, $1 > 0$ by **Theorem 1.21** therefore:

$$\begin{split}
  1 > 0 & \Longrightarrow & 1 + 0 > 0 & \text{ Theorem I.4} \\
  & \Longrightarrow & 1 + (a + (-a)) > 0 & \text{ Theorem I.2} \\
  & \Longrightarrow & (1 + a) + (-a) > 0 & \text{ Axiom 2} \\
  & \Longrightarrow & (1 + a) - a > 0 & \text{ Theorem I.2} \\
  & \Longrightarrow & (1 + a) - a \in \mathbb{R}^+ & \text{ By definition} \\
  & \Longrightarrow & 1 + a > a & \text{ By definition} \\
  \end{split}$$

Third, there is a contradiction because it exists a real number $1 + a$ such that $1 + a > a$. Therefore the original statement is true: $\neg (\exists a: a \in \mathbb{R} \land (\forall x:x \in \mathbb{R} \Longrightarrow x \leq a))$

**10** If $x$ has the property that $0 \leq x < h$ for every positive real number $h$, then $x = 0$

To prove this statement we are going to use the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First, let's assume that $x > 0$ then $x \in \mathbb{R}^+$. Therefore it must be the case that $0 \leq x < x$ because $x$ is a positive number. But $x = x$ so there is a contradiction by **Theorem I.16**. We can conclude that $x > 0$ is false and $x \leq 0$ must be true.

Second, apart from $x \leq 0$ we also have that $x \geq 0$. Therefore because $(x < 0 \lor x = 0) \land (x > 0 \lor x = 0)$ by **Theorem I.16** it must be the case that $x = 0$. 

### Exercises I 3.12 {#exrI3.12}

**1** If $x$ and $y$ are arbitrary real numbers with $x < y$, prove that there is at least one real $z$ satisfying $x < z < y$

Let choose an arbitary $w \in \mathbb{R}^+$ and use $\mathbb{Z}^+$ to denote the set of positive integers:

$$\begin{split}
  w \in \mathbb{R}^+ \land y - x \in \mathbb{R}^+ & \Longrightarrow n(y - x) > w \land n \in \mathbb{Z}^+ \text{ Theorem I.30} \\
  & \Longrightarrow n(y - x) > w \land n^{-1} \in \mathbb{R}^+ \text{ Exercises I 3.5 4, Theorem I.8} \\
  & \Longrightarrow 1(y - x) > wn^{-1} \text{ Theorem I.19, Axiom 1, Theorem I.18} \\
  & \Longrightarrow y - x > w/n \text{ Axiom 4, Theorem I.9} \\
  & \Longrightarrow y > w/n + x \text{ Theorem I.18, Axiom 4, Axiom 5} \\
  & \Longrightarrow y > w/n + x \land wn^{-1} \in \mathbb{R}^+ \text{ Axiom 7} \\
  & \Longrightarrow y > w/n + x > x \text{ Theorem 1.18, Axiom 4, Theorem 1.9} \\
  & \Longrightarrow z = w/n + x
  \end{split}$$
  
Some examples of $z$ are $|y - x| / n + x$ for $n = 2, 3, \ldots$ where $|y - x|$ is the absolute value of $y - x$

**2** If $x$ is an arbitrary real number, prove that there are integers $m$ and $n$ such that $m < x < n$

First, by **Theorem I.29** $n > x$ were $n$ is a positive integer and therefore an integer.

Second, let's use $\mathbb{Z}^-$ to denote the set of negative integers, and prove analogous results similar to **Theorem I.28** and **Theorem I.29** for $\mathbb{Z}^-$.

Third, we prove the set $\mathbb{Z}^-$ of negative integers $-1, -2, -3, \ldots$ is unbounded below using the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]. Assume to the contrary that $\mathbb{Z}^-$ is bounded below so by  **Theorem I.27** $\mathbb{Z}^-$ has a greatest lower bound. Therefore exists $c \in \mathbb{R}$ such that $c \leq x$ for all $x \in \mathbb{Z}^-$. Then $c + 1$ can not be a lower bound for $\mathbb{Z}^-$ because $c < c + 1$ so there is some $m \in \mathbb{Z}^-$ such that $m < c + 1$. So it must be the case that  $-(-m + 1) < c$ where $-(-m + 1) \in \mathbb{Z}^-$ but this is a contradiction because $c$ is the greatest lower bound of $\mathbb{Z}^-$. Therefore we must conclude that $\mathbb{Z}^-$ is unbounded below.

Fourth, we prove that for every real $x$ there exists a negative integer $m$ such that $x > m$. We prove this statement using the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]. If this is not the case $x$ would be a lower bound for $\mathbb{Z}^-$ but this is not true. So we must conclude that for every real $x$ there exists a negative integer $m$ such that $x > m$.

Fifth, we have that if $x$ is an arbitrary real number there are integers $m$ and $n$ such that $n > x > m$ where in this case $n$ is a positive integer and $m$ is a negative integer.

**3** If $x > 0$, prove that there is a positive integer $n$ such that $1 / n < x$

$$\begin{split}
  x > 0 & \Longrightarrow nx > 1 & \text{ Theorem I.30} \\
  & \Longrightarrow x > n^{-1}1 & \text{ Theorem I.19, Axiom 1, Theorem I.8, Axiom 4} \\
  & \Longrightarrow x > 1 / n & \text{ Axiom 1, Theorem I.9}
  \end{split}$$

**4** If $x$ is an arbitrary real number, prove that there is exactly one integer $n$ which satisfies the inequalities $n \leq x < n + 1$. This $n$ is called the greatest integer in $x$ and is denoted by $[x]$. For example $[5] = 5$, $\left[ \frac{5}{2} \right] = 2$, $\left[ -\frac{8}{3} \right] = -3$

First, we prove that if $S \subseteq \mathbb{Z}$, $S \neq \emptyset$ and $S$ is bounded above then $sup \; S$ exists and $sup \; S = max \; S$. We have that $sup \; S$ exists by **Axiom 10**. Now, we prove that $sup \; S = max \; S$ using the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]. Let's assume that $sup \; S \notin S$ then for all $s \in S$ we have that $sup \; S > s$. Because $sup \; S > Sup \; S - 1$ then $sup \; S - 1$ is not an upper bound so there exists some $s^* \in S$ such that $s^* > sup \; S - 1$. Also because $s^*$ is not an upper bound there exists some $s^{**} \in S$ such that $s^{**} > s^*$. Therefore we have that $sup \; S > s^{**} > s^* > sup \; S - 1$ so we have that $1 = sup \; S - (sup \; S - 1) > s^{**} - (sup \; S - 1) > s^{**} - s^*$ where this means that $1 > s^{**} - s^* > 0$. But this is a contradiction because if $s^{**} \in S$ and $s^* \in S$ then $s^{**} \in \mathbb{Z}$ and $s^* \in \mathbb{Z}$ so if $s^{**} > s^*$ it must be that $s^{**} - s^* \geq 1$. In that case $sup \; S \in S$ so $sup \; S = max \; S$. 

Second, let $S = \{y: y \leq x \land y \in \mathbb{Z} \text{ where } x \in \mathbb{R} \}$. We have that $S \subseteq \mathbb{Z}$, $S$ is bounded above by $x$ and also by **Exercises I 3.12 2** $S \neq \emptyset$. Therefore by the first part $sup \; S$ exists and $sup \; S = max \; S$. Because $sup \; S \in S$ then $sup \; S \leq x$ and $sup \; S \in \mathbb{Z}$. Also $sup \; S = max \; S$, $sup \; S < sup \; S + 1$ and $sup \; S + 1 \in \mathbb{Z}$ then $sup \; S + 1 \notin S$ so $x < sup \; S + 1$. Finally, we have that $sup \; S < x < sup \; S + 1$, $sup \; S \in \mathbb{Z}$ and $sup \; S$ is unique by **Theorem I.26**.

**Observation 1**: Also we have that:

$$\begin{split}
  n \leq x < n + 1 & \Longrightarrow \left[ x \right] \leq x \land x < \left[ x \right] + 1 & \text{ By definition} \\
  & \Longrightarrow \left[ x \right] + 1 \leq x + 1 \land x < \left[ x \right] + 1 & \text{ Theorem I.18} \\
  & \Longrightarrow x < \left[ x \right] + 1 \leq x + 1 & \text{ Theorem I.17} \\
  \end{split}$$

**Observation 2**: A better notation would be $n = \lfloor x \rfloor$ for $n \leq x < n + 1$ to differentiate it from $m = \lceil x \rceil$ for $m - 1 < x \leq m$. For example $\lfloor 2.5 \rfloor = 2$, $\lceil 2.5 \rceil = 3$ and $\lfloor 2 \rfloor = 2 = \lceil 2 \rceil$

**5** If $x$ is an arbitrary real number, prove that there is exactly one integer $n$ which satisfies $x \leq n < x + 1$.

First, we prove that if $S \subseteq \mathbb{Z}$, $S \neq \emptyset$ and $S$ is bounded below then $inf \; S$ exists and $inf \; S = min \; S$. We have that $inf \; S$ exists by **Theorem I.27**. Now let $-S = \{-y: y \in S \}$ then $-S \subseteq \mathbb{Z}$, $-S \neq \emptyset$ and $-S$ is bounded above. By the first part in the proof of **Exercises I 3.12 4** $-S$ we have that $sup -S = max -S$ so $-(-sup -S) = -(-max -S)$ by **Theorem I.4**. Because $-(-sup -S) \in -S$ then $-sup -S \in S$ and also we have that $-sup -S = inf \; S$. Therefore $inf \; S = min \; S$.

Second, let $S = \{y: x \leq y \land y \in \mathbb{Z} \text{ where } x \in \mathbb{R} \}$. We have that $S \subseteq \mathbb{Z}$, $S$ is bounded below by $x$ and also by **Exercises I 3.12 2** $S \neq \emptyset$. Therefore by the first part $inf \; S$ exists and $inf \; S = min \; S$. Because $inf \; S \in S$ then $x \leq inf \; S$ and $inf \; S \in \mathbb{Z}$.

Third, if $inf^* \; S$ is another greater lower bound then $inf \; S \geq inf^* \; S$. But $inf \; S$ is also a greater bound so $inf^* \; S \geq inf \; S$. Therefore $inf \; S = inf^* \; S$ where this means that $inf \; S$ is unique.

Fourth, if $inf \; S = x$ then by **Exercises I 3.12 4** $inf \; S = x < inf \; S + 1 = x + 1$. Also if $x < inf \; S = (inf \; S - 1) + 1$ by **Exercises I 3.12 4** $inf \; S - 1 < x$ so $inf \; S < x + 1$.

Fifth, we have that $x \leq inf \; S < x + 1$, $inf \; S \in \mathbb{Z}$ and $sup \; S$ is unique.

**Observation**: We can denote this $inf \; S = n$ as $n = \lceil x \rceil$ for $x \leq \lceil x \rceil < x + 1$. For example $\lceil 2.5 \rceil = 3$ and $\lceil 2 \rceil = 2$

**6** If $x$ and $y$ are arbitrary real numbers, $x < y$, prove that there exists at least one rational number $r$ satisfying $x < r < y$, and hence infinitely many. This property is often described by saying that the rational numbers are dense in the real-number system.

First, we have that:

$$\begin{split}
  y > x & \Longrightarrow y - x > 0 \text{ By definition} \\
  & \Longrightarrow (y - x)^{-1} \text{ Theorem I.8} \\
  & \Longrightarrow  (y - x)^{-1} + 1 \geq \left[ (y - x)^{-1} \right]  + 1 > (y - x)^{-1} \text{ Exercises I 3.12 4 Observation 1} \\
  & \Longrightarrow  (y - x)(\left[ (y - x)^{-1} \right] + 1) > 1 \text{ Theorem I.19, Theorem I.8} \\
  & \Longrightarrow  (y - x)n > 1 \text{ where } n = \left[ (y - x)^{-1} \right] + 1 \land n \in \mathbb{Z}^+ \\
  \end{split}$$

Second, we have that:

$$\begin{split}
  y > x & \Longrightarrow y - x > 0 & \text{ By definition} \\
  & \Longrightarrow n(y - x) > 1 \text{ where } n \geq \left[ (y - x)^{-1} \right] + 1 & \text{ Theorem I.30} \\
  & \Longrightarrow ny > 1 + nx \\
  & \Longrightarrow ny > 1 + nx \geq 1 + \left[ nx \right] > nx & \text{ Exercises I 3.12 4 Observation 1} \\
  & \Longrightarrow ny > 1 + \left[ nx \right] > nx \\
  & \Longrightarrow y > \frac{1 + \left[ nx \right]}{n} > x \\
  & \Longrightarrow y > r > x
  \end{split}$$
  
**Observation 1**: For example, $2\pi > \pi$ and $2\pi - \pi > 1 > 0$. Therefore we can fix $n = \left[ \pi^{-1} \right] + 1 = 1$ where $1*(2\pi - \pi) > 1$ and $r = \frac{1 + [1\pi]}{1} = \frac{1 + 3}{1} = \frac{4}{1}$.  
  
**Observation 2**: For example, $\pi > e$ and $1 \geq \pi - e > 0$. Therefore we can fix $n = \left[ (\pi - e)^{-1} \right] + 1 = 2 + 1 = 3$ where $3*(\pi - e) > 1$ and $r = \frac{1 + [3e]}{3} = \frac{1 + 8}{3} = \frac{3}{1}$.

Third, we have that $y > r > x$ so we can choose $n = \left[ (y - x)^{-1} \right] + 1, \left[ (y - x)^{-1} \right] + 2, \ldots$ to fix $r$. Therefore there are countable infinite $r$'s between $y$ and $x$ that we can build in this way.

**7** If $x$ is rational, $x \neq 0$ and $y$ irrational, prove that $x + y$, $x - y$, $xy$, $x/y$ and $y/x$ are all irrational.

To prove these statements we are going to use the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First, assume that $x + y$ is rational: 

$$\begin{split}
  x + y = \frac{a}{b} & \Longrightarrow \frac{c}{d} + y = \frac{a}{b} & \text{ By definition} \\
  & \Longrightarrow y = \frac{a}{b} - \frac{c}{d} \\
  & \Longrightarrow y = \frac{ad - bc}{bd} & \text{ Exercises I 3.3 10} \\
  & \Longrightarrow y \in \mathbb{Q} & \text{ This is a contradiction}
  \end{split}$$

Therefore $x + y \notin \mathbb{Q}$ but $x + y \notin \mathbb{R}$ so $x + y$ is irrational. 

Second, for $x - y$ we prove that if $y$ is irrational then $-y$ is irrational. Assume that $y$ is irrational and $-y$ rational so $- y = \frac{a}{b}$ for $a \in \mathbb{Z}$, $b \in \mathbb{Z}$ and $b \neq 0$ but $y = -\frac{a}{b}$ would be rational where this is a contradiction. Therefore if $y$ is irrational then $-y$ is irrational is a true statement. Now we can apply the first part to $x + (-y)$ and we have that $x + (-y)$ is irrational.

Third, assume that $xy$ is rational:

$$\begin{split}
  xy = \frac{a}{b} & \Longrightarrow \frac{c}{d}y = \frac{a}{b} & \text{ By definition} \\
  & \Longrightarrow y = \frac{d}{c}\frac{a}{b} & \text{ Theorem I.9, Axiom 2, Theorem 1.8, Axiom 4} \\
  & \Longrightarrow y = \frac{da}{cb} & \text{ Theorem I.14} \\
  & \Longrightarrow y \in \mathbb{Q} & \text{ This is a contradiction}
  \end{split}$$

Therefore $xy \notin \mathbb{Q}$ but $xy \notin \mathbb{R}$ so $xy$ is irrational.

Fourth, for $x / y = xy^{-1}$ we prove that if $y$ is irrational then $y^{-1}$ is irrational. Assume that $y$ is irrational and $y^{-1}$ rational so $y^{-1} = \frac{a}{b}$ for $a \in \mathbb{Z}$, $b \in \mathbb{Z}$ and $b \neq 0$ but $y = (y^{-1})^{-1} = (a / b)^{-1} = \frac{1}{a / b} = \frac{1 \cdot 1^{-1}}{a / b} = \frac{1 / 1}{a / b} = \frac{b}{a}$ would be rational where this is a contradiction. Therefore if $y$ is irrational then $y^{-1}$ is irrational is a true statement. Now we can apply the first part to $x / y = xy^{-1}$ and we have that $x / y$ is irrational.

Fifth, for $y / x = yx^{-1} = x^{-1}y$ we prove that if $x$ is rational then $x^{-1}$ is rational. If $x$ is rational then $x^{-1} = \left( \frac{a}{b} \right)^{-1} = \frac{1}{a / b} = \frac{1 \cdot 1^{-1}}{a / b} = \frac{1 / 1}{a / b} = \frac{b}{a}$ so $x^{-1}$ is rational. Now we can apply the third part to $y / x = yx^{-1} = x^{-1}y$ and we have that $y / x$ is irrational.

**8** Is the sum or product of two irrational numbers always irrational?

The answer is no. If $x$ is rational, $x \neq 0$ and $y$ is irrational then $x + y$, $-y$ and $y^{-1}$ are irrationals by **Exercises I 3.12 7**. However $(x + y) + (-y) = x$ and $(xy)y^{-1} = x$ are rationals.

**9** If $x$ and $y$ are arbitrary real numbers, $x < y$, prove that there exists at least one irrational number $z$ satisfying $x < z < y$, and hence infinitely many.

First, let $w > 0$ and $w \in \mathbb{R} - \mathbb{Q}$. The condition $w > 0$ is not restrictive because if $w < 0$ and $w \in \mathbb{R} - \mathbb{Q}$ then $-w > 0$ and $-w \in \mathbb{R} - \mathbb{Q}$. What we need is at least one $w \in \mathbb{R} - \mathbb{Q}$.

Second, if $y > x$, $w > 0$ and $w \in \mathbb{R} - \mathbb{Q}$

$$\begin{split}
  y > x  & \Longrightarrow y > r_1 > x \text{ where } r_1 \in \mathbb{Q} & \text{ Exercises I 3.12 6} \\
  & \Longrightarrow y > r_1 > r_2 > x & \text{ Exercises I 3.12 6} \\
  & \Longrightarrow n(r_1 - r_2) > w & \text{ Theorem I.30} \\
  & \Longrightarrow r_1 - r_2 > \frac{w}{n} \text{ where } \frac{w}{n} \in  \mathbb{R} - \mathbb{Q} & \text{ Exercises I 3.12 7} \\
  & \Longrightarrow r_1 > \frac{w}{n} + r_2 \text{ where } \frac{w}{n} + r_2 \in  \mathbb{R} - \mathbb{Q} & \text{ Exercises I 3.12 7} \\
  & \Longrightarrow r_1 > \frac{w}{n} + r_2 > r_2 \\
  & \Longrightarrow y > r_1 > \frac{w}{n} + r_2 > r_2 > x \\
  & \Longrightarrow y > \frac{w}{n} + r_2 > x
  \end{split}$$

Third, we have that $w(r_1 - r_2)^{-1} + 1 \geq \left[ w(r_1 - r_2)^{-1} \right] + 1 > w(r_1 - r_2)^{-1}$ by **Exercises I 3.12 4 Observation 1**. Therefore $(\left[ w(r_1 - r_2)^{-1} \right] + 1)(r_1 - r_2) > w$ by **Theorem I.19**, **Exercises I 3.5 4 Observation 1**, **Theorem I.8** and **Axiom 4**. Also by **Exercises I 3.5 4** and  because $\left[ w(r_1 - r_2)^{-1} \right] \geq 0 \land 1 > 0$ we have that $\left[ w(r_1 - r_2)^{-1} \right] + 1) \in \mathbb{Z^+}$. In that sense we can fix $n = \left[ w(r_1 - r_2)^{-1} \right] + 1), \left[ w(r_1 - r_2)^{-1} \right] + 2), \ldots$

Fourth, we have that $y > \frac{w}{n} + r_2 > x$ where $n \geq \left[ w(r_1 - r_2)^{-1} \right] + 1$, $n \in \mathbb{Z}^+$, $y > r_1 > r_2 > x$, $r_1 \in \mathbb{Q}$, $r_2 \in \mathbb{Q}$, $w \in \mathbb{R} - \mathbb{Q}$ and $\frac{w}{n} + r_2 \in \mathbb{R} - \mathbb{Q}$

**Observation 1**: In **Exercises I 3.12 11** we are going to prove that $\sqrt{2} \in \mathbb{R} - \mathbb{Q}$ and also we have that $\sqrt{2} > 0$ because $\sqrt{2}$ is the positive root of $2$. Therefore we can fix $w = \sqrt{2}$

**Observation 2**: For example, $\pi > e$ and $1 \geq \pi - e > 0$. Therefore we can fix $r_2 = \frac{1 + [(\left[ (\pi - e)^{-1} \right] + 1)e]}{\left[ (\pi - e)^{-1} \right] + 1} = \frac{1 + [3e]}{3} = \frac{1 + 8}{3} = 3$ and $r_2 = \frac{1 + [(\left[ (\pi - e)^{-1} \right] + 2)e]}{\left[ (\pi - e)^{-1} \right] + 2} = \frac{1 + [4e]}{4} = \frac{1 + 10}{4} = \frac{11}{4}$. Furthermore we can fix $\frac{w}{n} + r_2 = \frac{w}{\left[ w(r_1 - r_2)^{-1} \right] + 1} + r_2 =  \frac{\sqrt{2}}{\left[ \sqrt{2}(3 - \frac{11}{4})^{-1} \right] + 1} + \frac{11}{4} = \frac{\sqrt{2}}{5 + 1} + \frac{11}{4} = \frac{2\sqrt{2} + 33}{12}$. So $\pi > \frac{2\sqrt{2} + 33}{12} > e$ where $\frac{2\sqrt{2} + 33}{12} \in \mathbb{R} - \mathbb{Q}$

**10** An integer $n$ is called even if $n = 2m$ for some integer $m$, and odd if $n + 1$ is even. Proof the following statements.

**10a** An integer cannot be both even and odd

To prove this statement we are going to use the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First, if $m \in \mathbb{Z}$ and $s \in \mathbb{Z}$ then $m - s \in \mathbb{Z}$

Second, assume that for some $n \in \mathbb{Z}$ we have that $n$ is even and odd:

$$\begin{split}
  n = 2m \land n + 1 = 2s & \Longrightarrow n = 2m \land n = 2s - 1 \\
  & \Longrightarrow 2m = 2s - 1 \\
  & \Longrightarrow 2(m - s) = - 1 \\
  & \Longrightarrow m - s = -\frac{1}{2} \\
  & \Longrightarrow m - s \notin \mathbb{Z} & \text{ This is a contradiction}
  \end{split}$$

Therefore, there is a contradiction so $n$ can not be even and odd at the same time.

**10b** Every integer is either even or odd

First, we prove that if $n \in \mathbb{Z}$ and if $s < s + 2$ then there is a unique $t \in \mathbb{Z}$ and is $t = s + 1$.

$$\begin{split}
  s < s + 2 & \Longrightarrow s = s < s + 1 \land s + 1 = s + 1 < s + 2 & \text{ where } s \text{ is unique } \text{ Exercises I 3.12 5} \\
  & \Longrightarrow s < s + 1 < s + 2
  \end{split}$$

Second

$$\begin{split}
  n \in \mathbb{Z} & \Longrightarrow \frac{n}{2} \in \mathbb{Q} \subset \mathbb{R} \\
  & \Longrightarrow m \leq \frac{n}{2} < m + 1 & \text{ Exercises I 3.12 5} \\
  & \Longrightarrow 2m \leq n < 2(m + 1) \\
  & \Longrightarrow 2m + 1 \leq n  + 1 < 2m + 3 \\
  \end{split}$$

Third, if $2m = n$ then $n$ is even. But if $2m < n$ then $2m + 1 < n + 1$ so $2m + 1 < n + 1 < 2m + 3$. Because $2m + 1 < 2m + 2 < 2m + 3$ by the first part $n + 1 = 2m + 2 = 2(m + 1)$ so $n + 1$ is even and $n$ is odd. Therefore $n$ is even or odd but no both because of **Exercises I 3.12 10a**

**10c** The sum or product of two even integers is even. What can you say about the sum or product of two odd integers?

First, the sum of two even integers

$$\begin{split}
  2m = n \in \mathbb{Z} \land 2s = p \in \mathbb{Z} & \Longrightarrow n + p = 2m + 2s \\
  & \Longrightarrow n + p = 2(m + s) \text{ where } m + s  \in \mathbb{Z} \\
  & \Longrightarrow n + p \text{ is even}
  \end{split}$$

Second, the product of two even integers

$$\begin{split}
  2m = n \in \mathbb{Z} \land 2s = p \in \mathbb{Z} & \Longrightarrow np = (2m)(2s) \\
  & \Longrightarrow np = 2(m(2s)) \text{ where } m(2s) \in \mathbb{Z} \\
  & \Longrightarrow np \text{ is even}
  \end{split}$$

Third, the sum of two odd integers

$$\begin{split}
  n \text{ and } p \text{ are odd} & \Longrightarrow n + 1 = 2m \land p + 1 = 2s \\
  & \Longrightarrow (n + 1) + (p + 1) = 2m + 2s \\
  & \Longrightarrow (n + p) + 2 = 2(m + s) \\
  & \Longrightarrow n + p = 2(m + s) - 2 \\
  & \Longrightarrow n + p = 2((m + s) - 1) \text{ where } 2((m + s) - 1) \in \mathbb{Z} \\
  & \Longrightarrow n + p \text{ is even} \\
  \end{split}$$

Fourth, the product of two odd integers

$$\begin{split}
  n \text{ and } p \text{ are odd} & \Longrightarrow n + 1 = 2m \land p + 1 = 2s \\
  & \Longrightarrow (n + 1)(p + 1) = 2m(2s) \\
  & \Longrightarrow (np + 1) + (n + p) = 2m(2s) \\
  & \Longrightarrow np + 1 = 2m(2s) - (n + p) \\
  & \Longrightarrow np + 1 = 2m(2s) - 2t \text{ where } t \in \mathbb{Z} & \text{ By the third part} \\
  & \Longrightarrow np + 1 = 2(m(2s) - t)  \text{ where } 2(m(2s) - t) \in \mathbb{Z} \\
  & \Longrightarrow np + 1 \text{ is even} \\
  & \Longrightarrow np \text{ is odd}
  \end{split}$$

**10d** If $n^2$ is even, so is $n$. If $a^2 = 2b^2$, where $a$ and $b$ are integers, then both $a$ and $b$ are even.

First, assume $n^2$ is even. Also by  **Exercises I 3.12 10a** and **10b** $n$ is even or odd but not both. If $n$ is odd by **Exercises I 3.12 10c third part** $n^2$ would be odd so $n$ must be even.

Second, because $a^2 = 2b^2$ then $a^2$ is even. By the first part $a$ is even. Therefore $a^2 = (2t)^2$ where $t \in \mathbb{Z}$ so:

$$\begin{split}
  a^2 = (2t)^2 \land a^2 = 2b^2 & \Longrightarrow (2t)^2 = 2b^2 \\
  & \Longrightarrow 4t^2 = 2b^2 \\
  & \Longrightarrow 2t^2 = b^2 \text{ where } t^2 \in \mathbb{Z} \\
  & \Longrightarrow b^2 \text{ is even} \\
  & \Longrightarrow b \text{ is even } & \text{ By the first part}
  \end{split}$$

Third, $a$ and $b$ are even.

**10e** Every rational number can be expressed in the form $a / b$, where $a$ and $b$ are integers, at least one of which is odd.

First, we assume that every rational number, $c \in \mathbb{Q}$, has a unique representation as an irreducible fraction[^1], $c = \frac{a}{b}$, where $a \in \mathbb{Z}$, $b \in \mathbb{Z}$, $b > 0$ and $a, b$ have no common divisors except $1$. For example, $\frac{3}{12}$ or $\frac{-3}{-12}$,  $\frac{-2}{10}$ or $\frac{2}{-10}$ and $\frac{0}{-1}$ or $\frac{0}{2}$ can be represented as $\frac{1}{4}$,  $\frac{-1}{5}$ and $\frac{0}{1}$ in its reduced form.

[^1]: The complete proof can be found in [Existence of Canonical Form of Rational Number](https://proofwiki.org/wiki/Existence_of_Canonical_Form_of_Rational_Number) and [Canonical Form of Rational Number is Unique](https://proofwiki.org/wiki/Canonical_Form_of_Rational_Number_is_Unique)

Second, assume $a \text{ is even } \land b \text{ is even }$ and $c = \frac{a}{b} \in \mathbb{Q}$ where $\frac{a}{b}$ is the unique representation as an irreducible fraction of $c$. Then $a = 2s$ and $b = 2t$ so $c = \frac{a}{b} = \frac{2s}{2t} = \frac{s}{t}$. But this contradicts that $\frac{a}{b}$ is the unique representation as an irreducible fraction of $c$. Therefore $a \text{ is even } \land b \text{ is even }$ is false so $a \text{ is odd } \lor b \text{ is odd }$ must be true.

**11** Prove that there is no rational number whose square is $2$. $[$Hint: Argue by contradiction. Assume $(a/b)^2 = 2$, where $a$ and $b$ are integers, at least one of which is odd. Use parts of Exercise 10 to deduce a contradiction.$]$

To prove this statement we are going to use the method of *reductio ad absurdum*. Please check out [@mendelson_number_2008, Chapter 1, p 7, Exercises 2 and 5]

First, assume $\left( \frac{a}{b} \right)^2 = 2$ where $\frac{a}{b}$ is the unique representation as an irreducible fraction of a rational number.

Second, using the results of **Exercises I 3.12 10** we have that $a \text{ is even } \land b \text{ is even }$ but this is a contradiction because $\frac{a}{b}$ is the unique representation as an irreducible fraction of a rational number. Therefore $\frac{a}{b} \notin \mathbb{Q}$ so it must be the case that $a/b = ab^{-1} \in \mathbb{R} - \mathbb{Q}$ because $a \in \mathbb{Z} \subset \mathbb{R} \land b \in \mathbb{Z}^+ \subset \mathbb{R}$

**12** The Archimedean property of the real-number system was deduced as a consequence of the least-upper-bound axiom. Prove that the set of rational numbers satisfies the Archimedean property but not the least-upper-bound property. This shows that the Archimedean property does not imply the least-upper-bound axiom.

First, the archimedean property was deduced by proving that $\mathbb{Z}^+$ was unbounded above. Also because $\mathbb{Q} \subset \mathbb{R}$ if $c \in \mathbb{Q}$ then we can apply **Theorem I.29** and **Theorem I.30** to $c$. Therefore any $c \in \mathbb{Q}$ satisfies the archimedean property.

Second, if we apply the least-upper-bound axiom to $\mathbb{Q}$ we would have that every $S \subseteq \mathbb{Q} \land S \neq \emptyset$ which is bounded above has a supremum; that is, there is a rational number $B$ such that $B = \text{sup } S$.

Third, let $S = \{s: s \in \mathbb{Q} \land s^2 < 2 \}$. We have that $S \subset \mathbb{Q}$ and $S \neq \emptyset$ because $0 \in \mathbb{Q}$ and $0^2 = 0 < 2$ so $0 \in S$.

Fourth, by **Theorem 1.16** if $s \in S \subset \mathbb{Q} \subset \mathbb{R}$ we must have that $s = 0$, $s < 0$ or $s > 0$. If $s = 0$ or $s < 0$ then $s \leq 0 < 2$. If $s > 0$ then $s \geq 1$ or $1 > s > 0$. In the case of $s \geq 1$ we have that $s^2 \geq s$ by **Theorem 1.16** so $2 > s^2 \geq s$ and in the case of $1 > s > 0$ we have that $2 > 1 > s > 0$. Therefore in any case if $s \in S$ we have that $2 > s$ so $2$ is an upper bound of $S$.

Fifth, now we assume that there is some $B \in \mathbb{Q}$ such that for all $s \in S$ we have that $s \leq B$. However by **Exercises I 3.12 6** there is some $r \in \mathbb{Q}$ such that $B < r < 2$ so $B$ is not a least upper bound of $S$. Therefore not every $S \subseteq \mathbb{Q} \land S \neq \emptyset$ which is bounded above has a supremum $B$ where $B \in \mathbb{Q}$.   

## Mathematical Induction, Summation Notation, and Related Topics

### Exercises I 4.4 {#exrI4.4}

### Exercises I 4.7 {#exrI4.7}

### Exercises I 4.9 {#exrI4.9}

### Exercises I 4.10 {#exrI4.10}