Exercise 4.3.4

April 30, 2020 Issa Rice 6 Comments

Exercise statement

Prove Proposition 4.3.12.

Proposition 4.3.12 (Properties of exponentiation, II). Let $x,y$ be non-zero rational numbers, and let $n,m$ be integers.

(a) We have $x^n x^m = x^{n+m}$ , $(x^n)^m = x^{nm}$ , and $(xy)^n = x^n y^n$ .
(b) If $x \geq y > 0$ , then $x^n \geq y^n > 0$ if $n$ is positive, and $0 < x^n \leq y^n$ if $n$ is negative.
(c) If $x,y > 0$ , $n\ne 0$ , and $x^n=y^n$ , then $x=y$ .
(d) We have $|x^n| = |x|^n$ .

Hints

Induction is not suitable here. Instead, use Proposition 4.3.10.
Actually, you might want to use induction for small portions of this.

How to think about the exercise

I think this is a pretty tough exercise. The main reason is that it’s easy to accidentally use something that you aren’t supposed to know, so you have to be vigilant about making sure each step really does follow from the set of allowed facts.

Model solution

Throughout this exercise, we will use $n,m$ to denote positive or non-negative integers, and $-n,-m$ to denote negative integers. This means that instead of showing e.g. $x^n x^m = x^{n+m}$ for all integers $n,m$ , we will be showing that $x^n x^m = x^{n+m}$ for non-negative integers $n,m$ and separately $x^{-n} x^{-m} = x^{-n-m}$ for negative integers $-n, -m$ . This is consistent with the notation in Definition 4.3.11 and I find that this visual distinction helps quite a bit.

(a) Addition law for exponents: Suppose first that $n$ and $m$ are both non-negative. Then the result follows from Proposition 4.3.10(a). Next, suppose that both $-n$ and $-m$ are negative. Then we have $x^{-n} x^{-m} = \frac1{x^n} \cdot \frac1{x^m} = \frac1{x^n x^m}$ . Since $n, m$ are positive, $x^{n}x^{m} = x^{n+m}$ . Thus we have $\frac1{x^{n}x^{m}} = \frac1{x^{n+m}} = x^{-n-m}$ . Next suppose $n$ is non-negative and $-m$ is negative. We have two cases, $n-m \geq 0$ or $n-m < 0$ :

Suppose first that $n-m \geq 0$ . Then $m$ is positive, so by Proposition 4.3.10(a) we have $x^{n-m} x^{m} = x^n$ . Thus we have $x^n x^{-m} = x^{n-m} x^{m} x^{-m} = x^{n-m} x^{m} /x^{m} = x^{n-m}$ .
Suppose instead that $n-m < 0$ . Then $-n+m > 0$ and since $n \geq 0$ , we can use Proposition 4.3.10(a) to get $x^{-n+m} x^n = x^{m}$ . By definition of exponentiation, $x^{n-m} = \frac1{x^{-n+m}}$ . If we multiply both sides by $\frac1{x^n}$ we obtain $x^{n-m} \cdot \frac1{x^n} = \frac1{x^{-n+m}} \cdot \frac1{x^n}$ . Simplifying the right side, we have $\frac1{x^{-n+m}} \cdot \frac1{x^n} = \frac1{x^{-n+m} x^n} = \frac1{x^{m}}$ . But $x^{-m} = 1/x^{m}$ so we have $x^{n-m} \cdot \frac1{x^n} = x^{-m}$ . So multiplying through by $x^n$ we thus get $x^{n-m} = x^n x^{-m}$ .

The final case when $-n$ is negative and $m$ is non-negative is analogous to the previous case, but we could also note that $x^{-n} x^m = x^m x^{-n}$ and $x^{-n+m} = x^{m-n}$ by the commutativity of multiplication and addition. By the previous case, we know that $x^m x^{-n} = x^{m-n}$ so the result follows.

Multiplication law for exponents: First we show using induction that $1/y^n = (1/y)^n$ for any natural number $n$ . We already know that $y^{-n} = 1/y^n = (y^n)^{-1}$ and $(1/y)^n = (y^{-1})^n$ , so this will show that all of these expressions are equal. For the base case, we have $1/y^0 = 1/1 = 1 = (1/y)^0$ . Now suppose inductively that $1/y^n = (1/y)^n$ . We must show that $1/y^{n+1} = (1/y)^{n+1}$ . We have $1/y^{n+1} = 1/(y^n y) = (1/y^n)(1/y) = (1/y)^n (1/y) = (1/y)^{n+1}$ . This closes the induction.

First, suppose $n \geq 0$ and $m \geq 0$ . Then the result follows from Proposition 4.3.10(a). Next suppose $-n < 0$ and $-m < 0$ . We must show $(x^{-n})^{-m} = x^{(-n)(-m)}$ . We have

$\displaystyle (x^{-n})^{-m} = \frac{1}{(\frac{1}{x^n})^m} = \frac{1}{\frac{1}{(x^n)^m}} = \frac{1}{\frac{1}{x^{nm}}} = x^{nm} = x^{(-n)(-m)}$

Next suppose $n \geq 0$ and $-m < 0$ . We have $(x^n)^{-m} = \frac{1}{(x^n)^m} = \frac{1}{x^{nm}} = x^{-nm} = x^{n(-m)}$ .

Finally suppose $-n < 0$ and $m \geq 0$ . We have $(x^{-n})^m = (\frac1{x^n})^m = \frac{1}{(x^{n})^m} = x^{-nm} = x^{(-n)m}$ .

Distributing exponents: If $n \geq 0$ , the result follows from Proposition 4.3.10(a). If $-n < 0$ , then $(xy)^{-n} = \frac1{(xy)^n} = \frac1{x^n y^n} = \frac1{x^n}\cdot \frac1{y^n} = x^{-n} y^{-n}$ , which proves the result.

(b) Suppose $x \geq y > 0$ . First suppose $n$ is positive. By Proposition 4.3.10(c) we have $x^n \geq y^n \geq 0$ , so we just have to show that $y^n \ne 0$ . By Proposition 4.3.10(b), $y^n =0$ if and only if $y =0$ . Since $y > 0$ , we conclude that $y^n \ne 0$ .

Now suppose that $-n$ is negative. We must show that $0 < x^{-n} \leq y^{-n}$ . Since $x,y$ are positive, $\frac1{xy}$ is positive. We can thus multiply on both sides of $x \geq y$ to obtain $1/y = x \cdot \frac1{xy} \geq y \cdot \frac1{xy} = 1/x > 0$ . Applying the previous case, we thus obtain $(1/y)^n \geq (1/x)^n > 0$ . But $(1/y)^n = y^{-n}$ and $(1/x)^n = x^{-n}$ (see part (a) for proof) so the result follows.

(c) Let $x,y > 0$ be given. First suppose that $n > 0$ , and suppose $x^n = y^n$ . Suppose for sake of contradiction that $x\ne y$ . Then either $x < y$ or $x > y$ . In the former case, $x^n < y^n$ by Proposition 4.3.10(c), which contradicts the fact that $x^n=y^n$ . Similarly, if $x > y$ we obtain a contradiction. Thus the only option left is $x=y$ . Now suppose $-n < 0$ , and suppose that $x^{-n} = y^{-n}$ . By multiplying on both sides by $x^n y^n$ we obtain $x^n=y^n$ . Suppose for sake of contradiction that $x \ne y$ . Then either $x < y$ or $x > y$ . In either case, we see that $x^n \ne y^n$ , a contradiction. Thus we conclude that $x=y$ .

(d) If $n \geq 0$ , this follows from Proposition 4.3.10(d). So suppose $-n < 0$ . Then $|x^{-n}| = |(1/x)^n| = |1/x|^n = (1/|x|)^n = 1/|x|^n = |x|^{-n}$ . The first equality follows from part (a), the second equality follows from Proposition 4.3.10(d), the third equality follows from an easy proof by cases (just show that $-(1/x) = 1/(-x)$ ), the fourth equality follows from part (a) again, and the fifth equality follows from the definition of exponentiation.

Solutions

Exercise 4.3.3

April 29, 2020 Issa Rice 7 Comments

Exercise statement

Prove Proposition 4.3.10.

Proposition 4.3.10 (Properties of exponentiation, I). Let $x,y$ be rational numbers, and let $n,m$ be natural numbers.

(a) We have $x^n x^m = x^{n+m}$ , $(x^n)^m = x^{nm}$ , and $(xy)^n = x^n y^n$ .
(b) Suppose $n > 0$ . Then we have $x^n = 0$ if and only if $x = 0$ .
(c) If $x \geq y \geq 0$ , then $x^n \geq y^n \geq 0$ . If $x > y \geq 0$ and $n > 0$ , then $x^n > y^n \geq 0$ .
(d) We have $|x^n| = |x|^n$ .

Hints

Use induction.

How to think about the exercise

This is a pretty straightforward exercise.

Model solution

(a) First we show $x^n x^m = x^{n+m}$ . We fix $m$ and induct on $n$ . For the case $n=0$ , we must show $x^0 x^m = x^{0+m}$ . But $x^0 x^m = 1 \times x^m = x^m = x^{0+m}$ . Now suppose inductively that $x^n x^m = x^{n+m}$ . We must show that $x^{n+1} x^m = x^{(n+1)+m}$ . We have $x^{n+1} x^m = x^n \times x \times x^m =(x^n \times x^m) \times x$ . By the inductive hypothesis, $x^n x^m = x^{n+m}$ , so this is equal to $x^{n+m} \times x = x^{n+m+1} = x^{(n+1)+m}$ . This closes the induction.

Next we show that $(x^n)^m = x^{nm}$ . To do this, we first show that $1^m = 1$ for every natural number $m$ . Indeed we can induct on $m$ . For the base case, $1^0=1$ by definition of exponentiation, and if we suppose inductively that $1^m=1$ then we have $1^{m+1} = 1^m\times 1 = 1\times 1 = 1$ , which closes the induction. Now we actually prove the result. We fix $m$ and induct on $n$ . For the case $n=0$ , we must show $(x^0)^m = x^{0m}$ . But $(x^0)^m = 1^m=1$ and $x^{0m} = x^0 = 1$ . For the inductive case, suppose $(x^n)^m = x^{nm}$ . We must show $(x^{n+1})^m = x^{(n+1)m}$ . We have

$\begin{aligned} (x^{n+1})^m &= (x^n\times x)^m \\ &= (x^n)^m x^m \\ &= x^{nm}x^m \\ &= x^{nm+m} \\ &= x^{(n+1)m}\end{aligned}$

Here the second equality follows from the third result in part (a), which we prove next (there is no circularity since that result does not rely on this one), the third equality uses the inductive hypothesis, and the fourth equality follows from the first result in part (a).

Finally, we show $(xy)^n = x^n y^n$ . We induct on $n$ . For the base case, we have $(xy)^0 = 1 = 1\times 1 = x^0 y^0$ . Now suppose inductively that $(xy)^n = x^n y^n$ . We must show that $(xy)^{n+1} = x^{n+1} y^{n+1}$ . We have $(xy)^{n+1} = (xy)^n \times xy = x^n y^n\times xy = x^n\times x \times y^n \times y = x^{n+1} y^{n+1}$ , where the second equality uses the induction hypothesis.

(b) We can induct on $n$ , starting at $n=1$ . For the base case, we must show that $x^1 = 0$ if and only if $x=0$ . But this is true since $x^1 = x$ . Now suppose inductively that $x^n=0$ if and only if $x=0$ . We must show that $x^{n+1}=0$ if and only if $x=0$ . Suppose first that $x^{n+1}=0$ . Then $x^n \times x = 0$ so either $x^n=0$ or $x=0$ (for if both $x^n \ne 0$ and $x\ne0$ then by Proposition 4.2.4 we have $x^n(x^n)^{-1} = 1$ and $xx^{-1}=1$ so $1 = x^n(x^n)^{-1} xx^{-1} = (x^n\times x) (x^n)^{-1} x^{-1} = 0$ , a contradiction). In the latter case we’re done. In the former case, by the inductive hypothesis we see that $x=0$ . Conversely, suppose that $x=0$ . Then $x^{n+1} = 0^n \times 0 = 0$ . This closes the induction.

(If starting the induction at $n=1$ makes you nervous, you can start at $n=0$ using the statement “if $n > 0$ , then $x^n = 0$ if and only if $x = 0$ ”. The base case is then vacuous, and you will need to split into cases depending on $n=0$ or $n>0$ in the inductive case.)

(c) Suppose $x\geq y \geq 0$ . We want to show $x^n \geq y^n \geq 0$ . We will induct on $n$ . For the base case, we must show $x^0 \geq y^0 \geq 0$ . But this is true, since $x^0 = y^0 = 1 \geq 0$ . Now suppose inductively that $x^n \geq y^n \geq 0$ . We can multiply $x \geq y \geq 0$ by $x^n$ (which is non-negative by induction hypothesis) to get $x^n \times x \geq x^n \times y \geq 0$ . Similarly, we can multiply $x^n \geq y^n \geq 0$ by $y$ to get $x^n \times y \geq y^n \times y \geq 0$ . Thus we get $x^n \times x \geq x^n \times y\geq y^n \times y \geq 0$ . Now use $x^n \times x = x^{n+1}$ and $y\geq y^n = y^{n+1}$ to complete the proof.

The version for strict inequality is proved in the same way, except that the induction will start at $n=1$ .

(d) We will induct on $n$ . For the case $n=0$ , we want to show $|x^0|=|x|^0$ . But $|x^0| = |1| = 1 = |x|^0$ . Now suppose inductively that $|x^n| = |x|^n$ . We want to show that $|x^{n+1}| = |x|^{n+1}$ . We have $|x^{n+1}| = |x^n \times x| = |x^n|\times |x| = |x|^n \times |x| = |x|^{n+1}$ . Here, the second equality follows from Proposition 4.3.3(d) and the third equality follows from the inductive hypothesis.

Solutions

Exercise 4.3.2

April 27, 2020 Issa Rice 1 Comment

Exercise statement

Prove the remaining claims in Proposition 4.3.7.

Proposition 4.3.7. Let $x,y,z,w$ be rational numbers.

(a) If $x=y$ , then $x$ is $\varepsilon$ -close to $y$ for every $\varepsilon > 0$ . Conversely, if $x$ is $\varepsilon$ -close to $y$ for every $\varepsilon > 0$ , then we have $x=y$ .
(b) Let $\varepsilon > 0$ . If $x$ is $\varepsilon$ -close to $y$ , then $y$ is $\varepsilon$ -close to $x$ .
(c) Let $\varepsilon, \delta > 0$ . If $x$ is $\varepsilon$ -close to $y$ , and $y$ is $\delta$ -close to $z$ , then $x$ and $z$ are $(\varepsilon+\delta)$ -close.
(d) Let $\varepsilon,\delta > 0$ . If $x$ and $y$ are $\varepsilon$ -close, and $z$ and $w$ are $\delta$ -close, then $x+z$ and $y+w$ are $(\varepsilon+\delta)$ -close, and $x-z$ and $y-w$ are also $(\varepsilon+\delta)$ -close.
(e) Let $\varepsilon > 0$ . If $x$ and $y$ are $\varepsilon$ -close, they are also $\varepsilon'$ -close for every $\varepsilon' > \varepsilon$ .
(f) Let $\varepsilon > 0$ . If $y$ and $z$ are both $\varepsilon$ -close to $x$ , and $w$ is between $y$ and $z$ (i.e., $y \leq w \leq z$ or $z \leq w \leq y$ ), then $w$ is also $\varepsilon$ -close to $x$ .
(g) Let $\varepsilon > 0$ . If $x$ and $y$ are $\varepsilon$ -close, and $z$ is non-zero, then $xz$ and $yz$ are $\varepsilon|z|$ -close.
(h) Let $\varepsilon, \delta > 0$ . If $x$ and $y$ are $\varepsilon$ -close, and $z$ and $w$ are $\delta$ -close, then $xz$ and $yw$ are $(\varepsilon|z| + \delta|x| + \varepsilon\delta)$ -close.

Hints

You may want to use Proposition 4.3.3.

How to think about the exercise

Part (e) of this proposition justifies why we should care about small values of $\varepsilon$ in analysis. Since the property is satisfied automatically for all large values of $\varepsilon$ , it is only with the small values that we are in danger of not satisfying the property.

I don’t have much to say about the exercise itself. There are a lot of parts, but each part is straightforward.

Model solution

(a) If $x=y$ , then $x-y=0$ so $d(x,y) = |x-y| = 0 \leq \varepsilon$ .

Conversely, suppose $x$ and $y$ are $\varepsilon$ -close for every $\varepsilon > 0$ . Suppose for sake of contradiction that $x \ne y$ . Then $x-y \ne 0$ so $|x-y|$ is a positive number. Since $x$ and $y$ are $\varepsilon$ -close for every $\varepsilon > 0$ , in particular for $\varepsilon := |x-y|/2$ we must have $|x-y| \leq |x-y|/2$ . We can add $-|x-y|/2$ to both sides to get $|x-y|/2 \leq 0$ , a contradiction.

(b) This follows from Proposition 4.3.3(f), the symmetry of distance.

(c) We have $d(x,y) \leq \varepsilon$ and $d(y,z) \leq \delta$ . By the triangle inequality, we have $d(x,z) \leq d(x,y) + d(y,z) \leq \varepsilon + \delta$ .

(d) We have $d(x,y) \leq \varepsilon$ and $d(z,w) \leq \delta$ . Using the triangle inequality, we have

$\begin{aligned} d(x+z, y+w) &= |(x+z) - (y+w)| \\ &= |(x-y) + (z-w)| \\ &\leq |x-y| + |z-w| \\ &= d(x,y) + d(z,w) \\ &\leq \varepsilon + \delta\end{aligned}$

Similarly we have

$\begin{aligned} d(x-z, y-w) &= |(x-z) - (y-w)| \\ &= |(x-y) + (w-z)| \\ &\leq |x-y| + |w-z| \\ &= d(x,y) + d(w,z) \\ &\leq \varepsilon + \delta\end{aligned}$

The only extra step here is to use the symmetry of distance, $d(w,z)=d(z,w)$ .

(e) We have $d(x,y) \leq \varepsilon < \varepsilon'$ so the result follows from the transitivity of order.

(f) Suppose first that $y \leq w \leq z$ . Subtract $x$ from each expression to get $y-x \leq w-x \leq z-x$ . Since $|y-x| \leq \varepsilon$ and $|z-x| \leq \varepsilon$ , we can use Proposition 4.3.3(c) to get $-\varepsilon \leq y-x \leq \varepsilon$ and $-\varepsilon \leq z-x \leq \varepsilon$ . So summarizing all the inequalities, we have $-\varepsilon \leq y-x \leq w-x \leq z-x \leq \varepsilon$ which means that $-\varepsilon \leq w-x \leq \varepsilon$ , so $|w-x| \leq \varepsilon$ by Proposition 4.3.3(c) again.

The case for $z \leq w \leq y$ is similar.

(g) We have $|x-y| \leq \varepsilon$ . Since $z$ is non-zero, we can multiply by $|z| > 0$ on both sides to get $|x-y||z| \leq \varepsilon |z|$ . But $|x-y||z| = |(x-y)z| = |xz-yz|$ by Proposition 4.3.3(d). Thus $|xz-yz| \leq \varepsilon|z|$ , which means that $xz$ and $yz$ are $\varepsilon|z|$ -close.

(h) This was already shown in the book.

Solutions

Exercise 4.3.1

April 26, 2020 Issa Rice 8 Comments

Exercise statement

Prove Proposition 4.3.3.

Proposition 4.3.3 (Basic properties of absolute value and distance). Let $x,y,z$ be rational numbers.

(a) (Non-degeneracy of absolute value) We have $|x| \geq 0$ . Also, $|x| = 0$ if and only if $x$ is $0$ .
(b) (Triangle inequality for absolute value) We have $|x+y| \leq |x| + |y|$ .
(c) We have the inequalities $-y \leq x \leq y$ if and only if $y \geq |x|$ . In particular, we have $-|x| \leq x \leq |x|$ .
(d) (Multiplicativity of absolute value) We have $|xy| = |x| |y|$ . In particular, $|{-x}| = |x|$ .
(e) (Non-degeneracy of distance) We have $d(x,y) \geq 0$ . Also, $d(x,y) = 0$ if and only if $x=y$ .
(f) (Symmetry of distance) $d(x,y) = d(y,x)$ .
(g) (Triangle inequality for distance) $d(x,z) \leq d(x,y) + d(y,z)$ .

Hints

While all of these claims can be proven by dividing into cases, such as when $x$ is positive, negative, or zero, several parts of the proposition can be proven without such a tedious division into cases. For instance one can use earlier parts of the proposition to prove later ones.
You can also prove some of the later parts first, to help with proving the earlier parts. This is fine to do as long as you don’t then use the earlier part to prove the later part (which would be circular).

How to think about the exercise

I want to prove a few supplementary facts here that will be useful in the proofs below. I think proving these will make some of the proofs below conceptually cleaner. It’s totally possible to do this exercise without proving any of these supplementary facts.

In two of the proofs for the exercise (parts (b) and (d)) I prove a square version first, and use it to conclude the non-square version. I wish I had an explanation of why this works, but I don’t. It’s a simple and standard trick, but I don’t know why it should work.

Fact 1. Let $x$ be a rational number. Then $|x|^2 = x^2$ .

Proof. If $x$ is positive or zero, this is true by definition of absolute value. If $x$ is negative, then we have $|x|^2 = (-x)(-x) = x^2$ . $\Box$

Fact 2. Let $x,y,z$ be rational numbers. If $x \leq y$ and $z$ is non-negative, then $xz \leq yz$ .

Proof. We have several cases. If $x < y$ and $z$ is positive, then this follows from Proposition 4.2.9(e). If $z=0$ , then $xz = 0 \leq 0 = yz$ . If $x = y$ , then $xz = yz$ so $xz \leq yz$ . $\Box$

Fact 3. Let $x,y$ be non-negative rational numbers. If $x^2 = y^2$ , then $x = y$ .

Proof. Suppose first that $x=0$ or $y=0$ . If $x=0$ then $0=x^2 =y^2$ . Thus $y^2=0$ so $y$ cannot be positive or negative. (For if $y > 0$ then by Proposition 4.2.9(e) we could multiply by $y$ on both sides to obtain $y^2 > 0$ , a contradiction. Similarly for $y < 0$ .) So by trichotomy we see that $y=0$ . Next, if $y=0$ , then $x^2 = y^2 = 0$ so we have $x=0$ . In either case, $x=y$ .

Now suppose both $x$ and $y$ are positive. Suppose for sake of contradiction that $x\ne y$ . Then by order trichotomy we have two cases, $x<y$ or $x>y$ . If $x < y$ then $x^2 < y^2$ , a contradiction. And if $x>y$ then $x^2 > y^2$ , which again is a contradiction. Thus we are left to conclude that $x=y$ . $\Box$

Fact 4. Let $x,y$ be non-negative rational numbers. If $x^2 \leq y^2$ , then $x \leq y$ .

Proof. We prove the contrapositive. Suppose $x > y$ . We want to show that $x^2 > y^2$ . Since $x > y$ , we must have $x \geq y$ and $x\ne y$ . By Fact 2, we have $x^2 \geq y^2$ . If $x^2=y^2$ then by Fact 3 we would have $x=y$ , which contradicts the fact that $x\ne y$ . Thus we must have $x^2 \geq y^2$ and $x^2\ne y^2$ , i.e. $x^2 > y^2$ . $\Box$

Finally, a psychological note: I found this to be the hardest post to write so far. I wrote parts of it one day, then gave up, then wrote some other parts on a second day, then put it down again, and came back to it on a third day and didn’t do much, put it down, and finally came back to finish it on a fourth day. For some reason, it felt very annoying to type everything up. I don’t think I got this feeling when working through the book myself, originally.

Model solution

(a) By order trichotomy, we have three cases: $x > 0$ , $x=0$ , and $x < 0$ . If $x>0$ then $x-0 = x$ is a positive number, so $|x| = x \geq 0$ . If $x=0$ then $|x| = 0 \geq 0$ . If $x < 0$ then $x-0 = x$ is negative, so $|x| = -x$ is positive, hence $-x-0$ is positive which means we have $-x > 0$ . Therefore, $|x| = -x \geq 0$ .

Now suppose $|x|=0$ . If $x$ were positive or negative, the absolute value would be a positive number. So by trichotomy, we see that $x$ must be zero. Conversely, if $x=0$ then by definition of absolute value we have $|x|=0$ .

(b) There are several ways to prove this. Here are two:

We can split into cases depending on the value of $x+y$ . If $x+y=0$ then $|x+y| = 0 \leq |x|+|y|$ since by part (a) we know that both $|x| \geq 0$ and $|y| \geq 0$ . If $x+y$ is positive then $|x+y| = x+y \leq |x|+|y|$ by part (c) (there is no risk of circularity since the proof of part (c) does not require part (b)). If $x+y$ is negative then $|x+y| = (-x) + (-y) \leq |{-x}| + |{-y}| = |x|+|y|$ by parts (c) and (d) (again there is no circularity since the proof of (d) does not require (b)).
By Fact 4, it suffices to show that $|x+y|^2 \leq (|x|+|y|)^2$ . We have $(|x|+|y|)^2 = |x|^2 + 2|x||y| + |y|^2$ by Exercise 2.3.4 (more precisely, the version of this result for rational numbers, which can be proved in exactly the same way). By parts (c) and (d) we have $2|x||y| =2|xy| \geq 2xy$ . Thus we have $(|x|+|y|)^2 \geq x^2 + 2xy + y^2 = (x+y)^2 = |x+y|^2$ , where we have used Fact 1.

(c) Suppose $-y \leq x \leq y$ . If $x \geq 0$ then $|x| = x$ so we have $|x| = x \leq y$ . If $x < 0$ then we have $|x| = -x$ . Using $-y \leq x$ and a slight modification of Exercise 4.2.6, we have $y \geq -x = |x|$ .

Now suppose $y \geq |x|$ . If $x \geq 0$ then $y \geq |x| = x \geq 0$ . This means $y\geq 0$ , so $-y \leq 0$ . Thus we have $y \geq x \geq 0 \geq -y$ . If $x < 0$ then $|x| = -x$ . Thus we have $y \geq |x| = -x > 0$ . This means $y$ is positive, so $-y$ is negative. So we have $y \geq -x \geq -y$ , which means $-y \leq x \leq y$ .

To show that $-|x| \leq x \leq |x|$ , take $y:=|x|$ . Then we have $|x|\geq |x|$ so by the above result we must have $-|x| \leq x \leq |x|$ .

(d) Here are two ways to prove this:

We can prove by cases on the values of $x$ and $y$ . For instance, if $x$ is negative and $y$ is positive, then $xy$ is negative so $|xy| = -xy = |x||y|$ . The other cases can be shown in a similar way, and are omitted.
Both $|xy|$ and $|x||y|$ are non-negative by part (a), so by Fact 3 it suffices to show that $|xy|^2 = (|x||y|)^2$ . We have $|xy|^2 = (xy)^2 = x^2y^2 = |x|^2 |y|^2 = (|x||y|)^2$ , where we have used Fact 1.

To show $|{-x}| = |x|$ , take $y:=-1$ .

(e) By part (a), $d(x,y) = |x-y| \geq 0$ .

By part (a), we have $|x-y| = 0$ if and only if $x-y = 0$ . But we know that $|x-y| = d(x,y)$ and that $x-y=0$ iff $x=y$ . Thus $d(x,y) = 0$ if and only if $x=y$ .

(f) By part (d), $d(x,y) = |x-y| = |{-(x-y)}| = |y-x| = d(y,x)$ .

(g) By part (b) using $x-y$ and $y-z$ as inputs, we have $|(x-y) + (y-z)| \leq |x-y| + |y-z|$ . But $|(x-y) + (y-z)| = |x-z| = d(x,z)$ and $|x-y| + |y-z| = d(x,y) + d(y,z)$ so the result follows.

Solutions

Exercise 4.2.6

April 19, 2020 Issa Rice 1 Comment

Exercise statement

Show that if $x,y,z$ are rational numbers such that $x < y$ and $z$ is negative, then $xz > yz$ .

Hints

None.

How to think about the exercise

This is a simple exercise, so I don’t have anything to say.

Model solution 1

Suppose $x < y$ . By definition of order, this means that $x-y$ is a negative number. Since $z$ is also negative, we have $x-y = (-a)/b$ and $z=(-c)/d$ for positive rational numbers $a,b,c,d$ . Thus $xz - yz = (x-y)z = (ac)/(bd)$ is a positive rational, so $xz > yz$ as required.

Model solution 2

Suppose $x < y$ . By Proposition 4.2.9(d) we can add $-y$ to both sides to obtain $x-y < 0$ . Again by Proposition 4.2.9(d) we can add $-x$ to both sides to obtain $-y < -x$ . Since $z$ is a negative rational, $-z$ is a positive rational. Thus by Proposition 4.2.9(e) we have $(-y)(-z) < (-x)(-z)$ , i.e. $yz < xz$ . By Proposition 4.2.9(b) we have $xz > yz$ as required.

Solutions

Exercise 4.2.5

April 19, 2020 Issa Rice 3 Comments

Exercise statement

Prove Proposition 4.2.9.

Proposition 4.2.9 (Basic properties of order on the rationals). Let $x,y,z$ be rational numbers. Then the following properties hold.

(a) (Order trichotomy) Exactly one of the three statements $x=y$ , $x < y$ , or $x > y$ is true.
(b) (Order is anti-symmetric) One has $x < y$ if and only if $y > x$ .
(c) (Order is transitive) If $x < y$ and $y < z$ , then $x < z$ .
(d) (Addition preserves order) If $x < y$ , then $x+z < y+z$ .
(e) (Positive multiplication preserves order) If $x < y$ and $z$ is positive, then $xz < yz$ .

Hints

None.

How to think about the exercise

I want to say a few things about the definition of order on the rationals, and compare this definition to the definitions for the natural numbers and integers. None of what I say will be directly relevant to this exercise, but I think this is probably the best place to say these things (as opposed to saying these things in a post about a different exercise).

You might be wondering why the definition of order for the rationals is so different from the definitions for naturals and integers. A few things you might notice:

For rationals, the strict inequality $<$ is taken to be fundamental, whereas in the naturals and integers, the non-strict inequality $\leq$ is fundamental. By “fundamental”, I just mean that it’s the thing that’s defined in a “novel” way, rather than the one that’s defined by including or excluding the case of equality.
For rationals, order is defined using subtraction and with respect to positive and negative numbers, whereas for naturals and integers, it’s defined via addition by saying $a \geq b$ iff $a = b+n$ for some natural number $n$ .

For (1), mathematically this turns out to not be a very important distinction. We could have defined order on naturals and integers as follows: $a > b$ iff $a = b+n$ for a positive natural number $n$ ; indeed, this is exactly what Proposition 2.2.12(f) is about. Then we could have defined $a\geq b$ to mean $a>b$ or $a=b$ . Similarly, we could have defined order on rationals by saying that $x \leq y$ iff $x-y$ is either zero or negative. Then we could have defined $x < y$ to mean $x\leq y$ and $x\neq y$ .

So if there is an important distinction, I think it would have to be psychological (i.e. pedagogical). And here I’m not sure why Tao chose these particular definitions. The definition for rationals is slightly cleaner if one uses strict inequality, because otherwise one must include the “or zero” condition. And I think (but I’m not sure) that proving some of the results for natural numbers is easier if order is defined using non-strict inequality.

Moving onto (2), we have the “positive” concept for natural numbers, but we don’t have subtraction. So we can’t say $a > b$ iff $a-b$ is positive. But saying $a-b = n$ for some positive number $n$ is like saying $a = b+n$ for some positive number $n$ , so we are doing something similar even in the natural numbers case. For integers, we do have subtraction, so we indeed could have said that $a < b$ iff $a-b$ is negative, and $a > b$ iff $a-b$ is positive. This is actually precisely what Lemma 4.1.11(a) is about.

What about the other way around? For rationals, could we have defined $x > y$ to mean that $x = y + z$ for some positive $z$ ? As long as we allow $z$ to be rational (rather than just a positive integer), then I think this would work.

So to summarize, for (2), we could have defined order for the integers and rationals in whatever way we like, but for natural numbers we are restricted to not being able to use subtraction. I’m not sure why Tao decided to define order for the integers one way and for the rationals a different way.

Another question you might have is, how does the list of properties of order compare to the ones given in Lemma 4.1.11 and Proposition 2.2.12? Looking at Lemma 4.1.11, we see that its (a) is part of the definition of order for rationals (so we don’t need to prove it), its (b) we have, its (c) we have, its (d) appears in Exercise 4.2.6, its (e) we have, and its (f) we have. Lemma 4.1.11 doesn’t have anti-symmetry, but that’s actually part of the definition of order for integers.

Looking at Proposition 2.2.12, we see that its (a) follows trivially from the reflexivity of equality, its (b) we have (after some rewriting to deal with cases), its (c) we have (again after some rewriting), its (d) we have, its (e) we don’t have (but that doesn’t matter, since the notion of successor does not really matter for rationals), and its (f) we have (as long as we allow $d$ to be rational). Proposition 2.2.12 doesn’t have trichotomy (because there are no negative natural numbers), and for some reason Tao decided not to include the positive multiplication result (I’m not sure why).

Model solution

(a) Consider the number $x-y$ . By Lemma 4.2.7, this number is exactly one of zero, positive, or negative. But $x-y=0$ iff $x=y$ ; $x-y$ is positive iff $x > y$ ; and $x - y$ is negative iff $x < y$ . So each possibility in the order trichotomy corresponds to some possibility in the trichotomy of rationals. (If this reasoning makes you uncomfortable, you can also do this in the usual way, by first showing that at least one of the statements is true, and then showing that at most one of the statements is true. This would again make use of Lemma 4.2.7 extensively.)

(b) Suppose $x < y$ . By definition of order, this means that $x - y$ is negative, i.e. $x-y = (-a)/b$ for some positive integers $a$ and $b$ . Thus $y-x = -(x-y) = a/b$ is a positive number. By definition of order, this means $y > x$ as required.

The other direction is proved in the same way.

(c) Suppose $x < y$ and $y < z$ . Thus $x-y$ and $y-z$ are negative, which means that $x-y = (-a)/b$ and $y-z = (-c)/d$ for some positive integers $a,b,c,d$ . But now

$\begin{aligned}x-z &= (x-y)+(y-z) \\ &= (-a)/b + (-c)/d \\ &= (-ad-bc)/(bd) \\ &= (-(ad+bc))/(bd)\end{aligned}$

By Lemma 2.3.3 $ad$ , $bc$ , and $bd$ are all positive, and by Proposition 2.2.8 $ad+bc$ is positive. This means that $x-z$ is a negative rational, so $x < z$ as required.

(d) Suppose $x < y$ , so that $x-y$ is a negative number. But $(x+z) - (y+z) = x-y$ , so $x+z < y+z$ .

(e) Suppose $x < y$ and $z$ is positive. Then $x-y$ is negative. We thus have $x-y = (-a)/b$ and $z=c/d$ for positive integers $a,b,c,d$ . But now $xz-yz = (x-y)z = (-a)/b \times c/d = (-ac)/(bd)$ . Since $ac$ and $bd$ are positive by Lemma 2.3.3, we see that $(-ac)/(bd)$ is negative. Thus $xz < yz$ as required.

Solutions

Exercise 4.2.4

April 17, 2020 Issa Rice 1 Comment

Exercise statement

Prove Lemma 4.2.7.

Lemma 4.2.7 (Trichotomy of rationals). Let $x$ be a rational number. Then exactly one of the following three statements is true: (a) $x$ is equal to $0$ . (b) $x$ is a positive rational number. (c) $x$ is a negative rational number.

Hints

Note that, as in Proposition 2.2.13, you have to prove two different things: firstly, that at least one of (a), (b), (c) is true; and secondly, that at most one of (a), (b), (c) is true.

How to think about the exercise

This is a pretty straightforward exercise, so I don’t have anything to say.

Model solution

Let $x = a//b$ be a rational number, where $a$ is an integer and $b$ is a non-zero integer. First we show that at least one of (a), (b), (c) is true. If $a=0$ , then $x=0$ . If $a \ne 0$ , then by trichotomy of integers $a$ is positive or negative. Similarly, $b$ is positive or negative. So we have four cases:

Both $a,b$ are positive: We see that $x = a//b = a/b$ , so $x$ is positive.
$a$ is negative and $b$ is positive: We see that $-a$ is positive, so $(-a)//b$ is positive. But $x = -((-a)//b)$ , so $x$ is negative.
$a$ is positive and $b$ is negative: Then $-b$ is positive so $a//(-b)$ is positive. We have $x = -(a//(-b))$ so $x$ is negative.
Both $a,b$ are negative: Then $-a$ and $-b$ are both positive, so $(-a)//(-b)$ is positive. But $(-a)//(-b) = a//b =x$ so $x$ is positive.

Thus, in each case $x$ is zero, positive, or negative.

Now we show that at most one of (a), (b), (c) is true. Suppose for sake of contradiction that $x$ is equal to zero and is also positive. Thus $x = c//d = 0//1$ for positive integers $c,d$ . But this means $c1 = 0d$ , i.e. $c = 0$ . This contradicts the trichotomy for integers, since $c$ cannot be both positive and equal to zero. Thus we conclude that $x$ cannot be both positive and equal to zero.

Similarly, $x$ cannot be equal to zero and also negative, since this would imply that $x = (-c)//d = 0//1$ for positive integers $c,d$ , which would mean $-c = 0$ , i.e. $c=0$ .

Finally we show that $x$ cannot be both positive and negative. Suppose $x = (-c)//d = e//f$ where $c,d,e,f$ are all positive integers. Thus $-cf = ed$ . By Lemma 2.3.3, $cf$ and $ed$ are positive, so $-cf$ is negative. But this means that $-cf$ is both positive and negative, which contradicts the trichotomy of integers. Thus $x$ cannot be both positive and negative after all.

Solutions

Exercise 4.2.3

April 16, 2020 Issa Rice 1 Comment

Exercise statement

Prove the remaining components of Proposition 4.2.4.

Proposition 4.2.4 (Laws of algebra for rationals). Let $x,y,z$ be rationals. Then the following laws of algebra hold:

$x+y = y+x$
$(x+y) + z = x + (y+z)$
$x+0 = 0+x = x$
$x + (-x) = (-x) + x = 0$
$xy = yx$
$(xy)z = x(yz)$
$x1 = 1x = x$
$x(y+z) = xy+xz$
$(y+z)x = yx + zx$

If $x$ is non-zero, we also have

$xx^{-1} = x^{-1}x = 1$

Hints

As with Proposition 4.1.6, you can save some work by using some identities to prove others.

How to think about the exercise

This is a straightforward exercise, so I don’t have anything to say.

Model solution

Let $x = a//b$ , $y = c//d$ , and $z=e//f$ for integers $a,c,e$ and non-zero integers $b,d,f$ .

First we show that $x+y = y+x$ . We have $x+y = (ad+bc)//(bd)$ and $y+x = (cb+da)//(db) = (ad+bc)//(bd)$ . So we see that the two are equal.

The law $(x+y) + z = x + (y+z)$ was already shown in the book.

Next we show $x+0 = 0+x = x$ . Since we already showed addition is commutative, we already know that $x+0=0+x$ . Thus we just need to show that $0+x = x$ . We have $0+x = 0//1 + a//b = (0b+1a)//(1b) = a//b = x$ as required.

Next we have to show $x + (-x) = (-x) + x = 0$ . By commutativity of addition, we get $x + (-x) = (-x) + x$ , so we can just show $(-x) + x = 0$ . We have $(-x) +x = (-a)//b + a//b = (-ab + ba)//(b^2) = 0//(b^2)$ . We have $0//(b^2) = 0//1$ since $0\times 1 = 0b^2$ . Thus $(-x)+x = 0$ as desired.

Next we show that $xy = yx$ . We have $xy = (ac)//(bd)$ and $yx = (ca)//(db) = (ac)//(bd)$ . Thus the two are equal.

Next we show that $(xy)z = x(yz)$ . We have $(xy)z = (ace)//(bdf) = x(yz)$ .

Next we show that $x1 = 1x = x$ . By commutativity of multiplication, we have $x1 = 1x$ , so we just need to show that $1x = x$ . We have $1x = (1a)//(1b) = a//b = x$ .

Next we show that $x(y+z) = xy+xz$ . We have

$\begin{aligned}x(y+z) &= (a//b)\times (c//d + e//f) \\ &= (a//b) \times (cf +de)//(df) \\ &= (acf + ade)//(bdf)\end{aligned}$

$\begin{aligned}xy + xz &= (a//b)(c//d) + (a//b)(e//f) \\ &= (ac)//(bd) + (ae)//(bf) \\ &= (acbf + bdae)//(b^2df) \\ &= (acf + dae)//(bdf)\end{aligned}$

Thus the two are equal.

Next we show that $(y+z)x = yx + zx$ . This follows from commutativity of multiplication and the first form of the distributive law, both of which were proved above. We have $(y+z)x = x(y+z) = xy+xz = yx+zx$ .

Finally, suppose $x \ne 0$ . We want to show that $xx^{-1} = x^{-1}x = 1$ . The first equality follows from the commutativity of multiplication. For the second equality, we have $x^{-1}x = (b//a)(a//b) = (ba)//(ab)$ . And $(ba)//(ab) = 1//1$ since $ba=ab$ .

Solutions

Exercise A.7.1

April 15, 2020 Issa Rice Leave a comment

Exercise statement

Suppose you have four real numbers $a,b,c,d$ and you know that $a=b$ and $c=d$ . Use the above four axioms to deduce that $a+d=b+c$ .

Hints

None.

How to think about the exercise

So we have $a=b$ and $c=d$ . We want $a+d=b+c$ . Since we want $a+d = \text{something}$ , we can start by adding $d$ to the equation involving $a$ . This yields $a+d=b+d$ . We could also have added $a$ to both sides of $c=d$ , to get $a+c=a+d$ .

Now what? We want $\text{something} = b+c$ , so we can add $c$ to both sides of $a=b$ to get $a+c=b+c$ . Or we could have added $b$ to both sides of $c=d$ to get $b+c=b+d$ .

Now we’ve over-solved the problem. We could take $a+d=b+d$ and $b+c=b+d$ to get what we want, or we could also take $a+c=a+d$ and $a+c=b+c$ .

Model solution

Since $a=b$ , we have $a+d = b+d$ . Formally, we define a function $f : \mathbf R\to \mathbf R$ by $f(x) := x+d$ . Since $a=b$ , we must have $f(a) = f(b)$ , i.e. $a+d=b+d$ . (Strictly speaking, we would need to say that $f(a) = a+d$ , so by symmetry, $a+d=f(a)$ . Since $f(a)=f(b)$ , by transitivity we have $a+d=f(b)$ . And since $f(b)=b+d$ , by transitivity again we have $a+d=b+d$ . We shall skip this in the following.)

Analogously, since $c=d$ we have $b+c = b+d$ . By symmetry, we have $b+d=b+c$ . So by transitivity we have $a+d = b+c$ as desired.

Solutions

Exercise A.5.1

April 15, 2020 Issa Rice Leave a comment

Exercise statement

What does each of the following statements mean, and which of them are true? Can you find gaming metaphors for each of these statements?

(a) For every positive number $x$ , and every positive number $y$ , we have $y^2=x$ .
(b) There exists a positive number $x$ such that for every positive number $y$ , we have $y^2=x$ .
(c) There exists a positive number $x$ , and there exists a positive number $y$ , such that $y^2=x$ .
(d) For every positive number $y$ , there exists a positive number $x$ such that $y^2=x$ .
(e) There exists a positive number $y$ such that for every positive number $x$ , we have $y^2=x$ .

Hints

None.

How to think about the exercise

I think this is a pretty simple exercise, but that’s possibly only because I’ve gotten used to thinking about nested quantifiers. If you find this confusing, you might want to find a book on proofs, such as Velleman’s How to Prove It.

Model solution

(a) False. This is saying that every positive number is the square of every positive number. That can’t be right, since most of the time, if one picks two positive numbers, one of them isn’t the square of the other. Consider $x=1$ and $y=2$ . Then we have $y^2 = 4 \ne 1 = x$ . In terms of the gaming metaphor, your opponent hands you two positive numbers $x,y$ , and you win if $y^2=x$ . You don’t get to interact in this game; your opponent chooses two numbers, and that determines the outcome of the game. Obviously, you can’t always win this game since your opponent can choose one number, and then choose a different number which is not the square of the first number.

(b) False. This is saying that there is a single number which is the square of every positive number. But that can’t be right, since if two positive numbers are distinct, then their squares are also distinct. To prove this, suppose for contradiction that such an $x$ exists. Then in particular for $y=1$ and $y=2$ we must have $1^2 = x$ and $2^2=x$ . Thus $1=4$ , a contradiction. This contradiction shows that the statement must have been false. In terms of the gaming metaphor, you are asked to present a number $x$ . Then your opponent gets to pick a number $y$ . You win if $y^2 = x$ . Obviously you can’t always win, since your opponent gets to see your number and pick one which doesn’t square to yield your number. For instance, your opponent can pick $y:= x+1$ .

(c) True. This is saying that there are two positive numbers, one of which is the square of the other. That’s clearly true, since we can just take any positive number, then square it to get a positive number. In particular, take $x=4$ and $y=2$ . Then $y^2 = 4 = x$ . In terms of the gaming metaphor, the game begins as your move, and you get to pick two numbers. Your opponent doesn’t get to move. You can always win the game, just by picking two numbers that work.

(d) True. This is saying that every positive number can be squared to yield a positive number. So this statement is saying that every positive number has a positive square. This is true, since a positive number times a positive number is a positive number. In terms of the gaming metaphor, your opponent hands you a positive number, and you have to find a positive number which is its square; you can always win regardless of the number your opponent hands you, since you can just square your opponent’s number.

(e) False. This is saying that there is a single number which is the positive square root of every positive number. This can’t be right, since distinct positive numbers have distinct positive square roots. To prove this, suppose for sake of contradiction that such a $y$ existed. Then in particular for $x=1$ and $x=4$ we must have $y^2 = 1$ and $y^2=4$ . Thus $1=4$ , a contradiction. In terms of the gaming metaphor, you are asked to pick a number $y$ , then your opponent picks a number $x$ . You win if $y^2 = x$ . Obviously you can’t always win, since your opponent gets to see your number. For instance, your opponent can square your number and add one to get a different number.