Category Archives: Solutions

Exercise 6.1.6

May 31, 2020 Issa Rice 10 Comments

Exercise statement

Prove Proposition 6.1.15, using the following outline. Let $(a_n)_{n=m}^\infty$ be a Cauchy sequence of rationals, and write $L := \mathrm{LIM}_{n\to\infty} a_n$ . We have to show that $(a_n)_{n=m}^\infty$ converges to $L$ . Let $\varepsilon > 0$ . Assume for sake of contradiction that the sequence $(a_n)_{n=m}^\infty$ is not eventually $\varepsilon$ -close to $L$ . Use this, and the fact that $(a_n)_{n=m}^\infty$ is Cauchy, to show that there is an $N \geq m$ such that either $a_n > L + \varepsilon/2$ for all $n\geq N$ , or $a_n < L -\varepsilon/2$ for all $n \geq N$ . Then use Exercise 5.4.8.

Proposition 6.1.15 (Formal limits are genuine limits). Suppose that $(a_n)_{n=1}^\infty$ is a Cauchy sequence of rational numbers. Then $(a_n)_{n=1}^\infty$ converges to $\mathrm{LIM}_{n\to\infty} a_n$ , i.e.

$\displaystyle \mathrm{LIM}_{n\to\infty} a_n = \lim_{n\to\infty} a_n$ .

Hints

None.

How to think about the exercise

Minor note: the starting index of the sequence is different between the proposition statement and the exercise. This doesn’t really matter, as we already showed in Exercise 6.1.3.

Model solution

Let $(a_n)_{n=m}^\infty$ be a Cauchy sequence of rationals, and write $L := \mathrm{LIM}_{n\to\infty} a_n$ . We have to show that $(a_n)_{n=m}^\infty$ converges to $L$ . Assume for sake of contradiction that $(a_n)_{n=m}^\infty$ does not converge to $L$ . This means that there is some $\varepsilon > 0$ such that $(a_n)_{n=m}^\infty$ is not eventually $\varepsilon$ -close to $L$ . In other words, there is some $\varepsilon > 0$ such that for every $N \geq m$ , there is some $n \geq N$ such that $|a_n - L| > \varepsilon$ . So choose this $\varepsilon > 0$ . Since $(a_n)_{n=m}^\infty$ is Cauchy, there is some $N \geq m$ such that $|a_n - a_k| \leq \varepsilon/2$ for all $n,k \geq N$ ; fix this $N$ . Since $(a_n)_{n=m}^\infty$ is not eventually $\varepsilon$ -close to $L$ , for this $N$ in particular, there is some $n_0 \geq N$ such that $|a_{n_0} - L| > \varepsilon$ ; fix this $n_0$ . Since $n_0\geq N$ , we can fix $k$ above to be $n_0$ , so that $|a_n - a_{n_0}| \leq \varepsilon/2$ for all $n \geq N$ . We can rewrite this using Exercise 5.4.6 so that $a_{n_0}-\varepsilon/2 \leq a_n \leq a_{n_0}+\varepsilon/2$ for all $n \geq N$ . Now we have two cases, depending on the sign of $a_{n_0} - L$ .

If $a_{n_0} - L$ is non-negative, then $a_{n_0} - L = |a_{n_0} - L| > \varepsilon$ . We also have $a_{n_0} -\varepsilon/2 \leq a_n \leq a_{n_0}+\varepsilon/2$ for all $n \geq N$ , so in particular $a_{n_0}-\varepsilon/2 \leq a_n$ for all $n \geq N$ . Adding this inequality to the inequality $\varepsilon < a_{n_0} - L$ we obtain $a_{n_0} + \varepsilon/2 < a_{n_0} + a_n - L$ . This simplifies to $a_n > L+\varepsilon/2$ , and holds for all $n \geq N$ . By Exercise 5.4.8 (formally, we replace $(a_n)_{n=m}^\infty$ by an equivalent sequence $(b_n)_{n=m}^\infty$ by changing the finite start of the sequence so that the inequality $b_n > L+\varepsilon/2$ holds for all $n \geq m$ ), this implies $L \geq L + \varepsilon/2$ , a contradiction since $\varepsilon/2$ is positive.

If $a_{n_0} - L$ is negative, then $-(a_{n_0} - L) = |a_{n_0} - L| > \varepsilon$ , i.e. $\varepsilon < -a_{n_0} + L$ . We also have $a_{n_0}-\varepsilon/2 \leq a_n \leq a_{n_0}+\varepsilon/2$ for all $n \geq N$ , so in particular $a_n \leq a_{n_0}+\varepsilon/2$ for all $n \geq N$ . Adding these two inequalities, we have $\varepsilon + a_n < L + \varepsilon/2$ , i.e. $a_n < L - \varepsilon/2$ , which holds for all $n \geq N$ . Using Exercise 5.4.8 we obtain $L \leq L - \varepsilon/2$ , a contradiction.

Both possibilities lead to a contradiction, so $(a_n)_{n=m}^\infty$ must converge to $L$ .

Solutions

Exercise 6.1.5

May 30, 2020 Issa Rice Leave a comment

Exercise statement

Prove Proposition 6.1.12.

Proposition 6.1.12 (Convergent sequences are Cauchy). Suppose that $(a_n)_{n=m}^\infty$ is a convergent sequence of real numbers. Then $(a_n)_{n=m}^\infty$ is also a Cauchy sequence.

Hints

Use the triangle inequality, or Proposition 4.3.7.

How to think about the exercise

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

Model solution

Let $(a_n)_{n=m}^\infty$ be a convergent sequence of real numbers. We want to show that $(a_n)_{n=m}^\infty$ is Cauchy, so let $\varepsilon > 0$ be a real number. We must find an $N \geq m$ such that $|a_j - a_k| \leq \varepsilon$ for all $j,k \geq N$ . Since $(a_n)_{n=m}^\infty$ is convergent, it converges to some real number $L$ , and since $\varepsilon/2$ is positive, we have some $N \geq m$ such that $|a_n - L| \leq \varepsilon/2$ for all $n \geq N$ . Choose this $N$ . Then if $j,k \geq N$ , we have both $|a_j - L| \leq \varepsilon/2$ and $|a_k - L| \leq \varepsilon/2$ . By the triangle inequality, we have $d(a_j,a_k) \leq d(a_j,L) + d(L,a_k)$ . Thus we have $|a_j - a_k| \leq |a_j-L| + |a_k -L| \leq \varepsilon/2 + \varepsilon/2 = \varepsilon$ . In other words, for all $j,k \geq N$ we have $|a_j - a_k|\leq \varepsilon$ as required.

Solutions

Exercise 6.1.4

May 29, 2020 Issa Rice 6 Comments

Exercise statement

Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers, let $c$ be a real number, and let $k \geq 0$ be a non-negative integer. Show that $(a_n)_{n=m}^\infty$ converges to $c$ if and only if $(a_{n+k})_{n=m}^\infty$ converges to $c$ .

Hints

None.

How to think about the exercise

If this exercise seems confusing to you, I would encourage you to define $b_n := a_{n+k}$ and then show that $(a_n)_{n=m}^\infty$ converges to $c$ if and only if $(b_n)_{n=m}^\infty$ converges to $c$ .

Model solution

Suppose first that $(a_n)_{n=m}^\infty$ converges to $c$ . We want to show that $(a_{n+k})_{n=m}^\infty$ converges to $c$ , so let $\varepsilon > 0$ be a real number. Since $(a_n)_{n=m}^\infty$ converges to $c$ , there is some $N \geq m$ such that $|a_n - c| \leq \varepsilon$ for all $n \geq N$ . We want to find an $N' \geq m$ such that $|a_{n+k} - c|\leq\varepsilon$ for all $n \geq N'$ . Choose $N' := N$ . If $n \geq N' = N$ , then $n+k \geq n \geq N$ , so we have $|a_{n+k} - c| \leq \varepsilon$ . Thus $(a_{n+k})_{n=m}^\infty$ converges to $c$ .

Conversely suppose that $(a_{n+k})_{n=m}^\infty$ converges to $c$ . We want to show $(a_n)_{n=m}^\infty$ converges to $c$ , so let $\varepsilon > 0$ be a real number. Since $(a_{n+k})_{n=m}^\infty$ converges to $c$ , we have an $N \geq m$ such that $|a_{n+k} - c|\leq \varepsilon$ for all $n \geq N$ . We want to find an $N' \geq m$ such that $|a_n - c|\leq \varepsilon$ for all $n \geq N'$ . Pick $N' := N+k$ . If $n \geq N' = N+k$ , then $n-k \geq N$ so we have $|a_{(n-k)+k} - c| \leq \varepsilon$ . But $a_{(n-k)+k} = a_n$ so this means $|a_n - c| \leq \varepsilon$ . Thus $(a_n)_{n=m}^\infty$ converges to $c$ as required.

Solutions

Exercise 6.1.3

May 28, 2020 Issa Rice 2 Comments

Exercise statement

Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers, let $c$ be a real number, and let $m' \geq m$ be an integer. Show that $(a_n)_{n=m}^\infty$ converges to $c$ if and only if $(a_n)_{n=m'}^\infty$ converges to $c$ .

Hints

None.

How to think about the exercise

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

Model solution

Suppose first that $(a_n)_{n=m}^\infty$ converges to $c$ . We want to show that $(a_n)_{n=m'}^\infty$ converges to $c$ , so let $\varepsilon > 0$ be a real number. Since $(a_n)_{n=m}^\infty$ converges to $c$ , we know that there is some $N \geq m$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$ -close to $c$ . Define $N' := N+m'$ . Then $N' \geq m'$ , and if $n\geq N'$ then $n \geq N$ so $|a_n - c|\leq \varepsilon$ . Thus $(a_n)_{n=N'}^\infty$ is $\varepsilon$ -close to $c$ , which means that $(a_n)_{n=m'}^\infty$ is eventually $\varepsilon$ -close to $c$ . Since $\varepsilon > 0$ was arbitrary, this shows that $(a_n)_{n=m'}^\infty$ converges to $c$ , as desired.

Conversely suppose that $(a_n)_{n=m'}^\infty$ converges to $c$ . We want to show that $(a_n)_{n=m}^\infty$ converges to $c$ , so let $\varepsilon > 0$ be a real number. We know that $(a_n)_{n=m'}^\infty$ is eventually $\varepsilon$ -close to $c$ , so there is an $N\geq m'$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$ -close to $c$ . We have $N\geq m' \geq m$ , so this same $N$ will work: $(a_n)_{n=m}^\infty$ converges to $c$ as desired.

Solutions

Exercise 6.1.2

May 27, 2020 Issa Rice Leave a comment

Exercise statement

Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers, and let $L$ be a real number. Show that $(a_n)_{n=m}^\infty$ converges to $L$ if and only if, given any real $\varepsilon > 0$ , one can find an $N \geq m$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$ .

Hints

None.

How to think about the exercise

In most real analysis texts, the result of this exercise is the definition of convergence. This means that this exercise is only an exercise because Tao’s definition of convergence “wraps up” all the concepts behind terminology like eventual $\varepsilon$ -closeness.

I think this can be a pretty tricky exercise, even though it’s very simple. It can seem like we’re just shuffling symbols in a meaningless way. But there is meaning.

Model solution

Suppose that $(a_n)_{n=m}^\infty$ converges to $L$ . Let $\varepsilon > 0$ be a real number. Since $(a_n)_{n=m}^\infty$ converges to $L$ , we see that $(a_n)_{n=m}^\infty$ is eventually $\varepsilon$ -close to $L$ . This means that there exists an $N \geq m$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$ -close to $L$ . But saying $(a_n)_{n=N}^\infty$ is $\varepsilon$ -close to $L$ is the same thing as saying that $|a_n - L| \leq \varepsilon$ for every $n \geq N$ . In other words, we have found an $N \geq m$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$ , as desired.

Conversely suppose that given any real $\varepsilon > 0$ , one can find an $N \geq m$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$ . We want to show that $(a_n)_{n=m}^\infty$ converges to $L$ . Let $\varepsilon > 0$ be a real number. We must show that $(a_n)_{n=m}^\infty$ is eventually $\varepsilon$ -close to $L$ , i.e. we must find some $N \geq m$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$ -close to $L$ . We know that we can find an $N \geq m$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$ , so pick this $N$ . We will show that $(a_n)_{n=N}^\infty$ is $\varepsilon$ -close to $L$ . Indeed, if $n \geq N$ , we have $|a_n - L| \leq \varepsilon$ , as desired.

Solutions

Exercise 6.1.1

May 26, 2020 Issa Rice 4 Comments

Exercise statement

Let $(a_n)_{n=0}^\infty$ be a sequence of real numbers, such that $a_{n+1} > a_n$ for each natural number $n$ . Prove that whenever $n$ and $m$ are natural numbers such that $m > n$ , then we have $a_m > a_n$ . (We refer to these sequences as increasing sequences.)

Hints

Use induction.

How to think about the exercise

It should be pretty clear what is going on. If $m > n$ , then we have $a_m > a_{m-1} > a_{m-2} > \cdots > a_{n+2} > a_{n+1} > a_n$ ; each inequality follows because the indexes differ by $1$ . So now $a_m > a_n$ by transitivity of inequality. The whole point of the proof is to justify what we mean by “ $\cdots$ ”.

Model solution

Let $n$ be a natural number. We will show by induction that $a_{n+k} > a_n$ for all positive integers $k$ . For the base case, $k=1$ (see here for why induction starting at a base case of $1$ is valid), we must show that $a_{n+1} > a_n$ . This follows from the assumption that $a_{n+1} > a_n$ for each natural number $n$ .

Now suppose inductively that $a_{n+k} > a_n$ . We want to show that $a_{n+k+1} > a_n$ . Since $n+k$ is a natural number, we have $a_{(n+k)+1} > a_{n+k}$ . By inductive hypothesis, we have $a_{n+k} > a_n$ . Thus by transitivity of inequality we have $a_{n+k+1} > a_n$ , which closes the induction.

Let $m$ be a natural number such that $m > n$ . By Proposition 2.2.12(f), we can write $m = n+k$ for some positive integer $k$ . By what we showed above, we have $a_m = a_{n+k} > a_n$ as required.

Solutions

Exercise 5.6.3

May 25, 2020 Issa Rice Leave a comment

Exercise statement

If $x$ is a real number, show that $|x| = (x^2)^{1/2}$ .

Hints

None.

How to think about the exercise

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

Model solution

We have three cases, by the trichotomy of real numbers. If $x=0$ then $|x| = 0$ by definition of absolute value, and $(x^2)^{1/2} = 0^{1/2} = 0$ . (If $y = 0^{1/2}$ we can square both sides to obtain $y^2 = 0$ ; if $y$ is positive or negative $y^2$ is positive, so $y=0$ . Alternatively we can go back to the definition of $n$ th roots like we did in the proof of Lemma 5.6.6(c).) Thus $|x| = (x^2)^{1/2}$ .

If $x > 0$ is positive, then $|x| = x$ . Also, $(x^2)^{1/2} = x$ by Lemma 5.6.6(b). Both are equal to $x$ , so we are done.

If $x < 0$ is negative, then by Proposition 5.4.4 $-x$ is positive. Thus $((-x)^2)^{1/2} = -x$ by Lemma 5.6.6(b). We also know $x^2 = (-x)^2$ . Thus $(x^2)^{1/2} = ((-x)^2)^{1/2} = -x$ . By definition of absolute value, $|x| = -x$ . Both are equal to $-x$ , so we are done.

Solutions

Exercise 5.6.2

May 25, 2020 Issa Rice 11 Comments

Exercise statement

Prove Lemma 5.6.9.

Lemma 5.6.9. Let $x,y > 0$ be positive reals, and let $q,r$ be rationals.

(a) $x^q$ is a positive real.
(b) $x^{q+r} = x^q x^r$ and $(x^q)^r = x^{qr}$ .
(c) $x^{-q} = 1/x^q$ .
(d) If $q > 0$ , then $x > y$ if and only if $x^q > y^q$ .
(e) If $x > 1$ , then $x^q > x^r$ if and only if $q > r$ . If $x < 1$ , then $x^q > x^r$ if and only if $q < r$ .
(f) $(xy)^q = x^q y^q$ .

Hints

You should rely mainly on Lemma 5.6.6 and on algebra.

How to think about the exercise

There is basically one main trick that gets used over and over in this exercise, namely the idea to “integerize the exponent” (this isn’t a standard term as far as I know; I’ve named it in analogy with how in algebra one “rationalizes the denominator”). We don’t know how to directly manipulate rational exponents, so the idea is to introduce extra integers to cancel out the denominators in the exponents, and then later get rid of these extra integers using the cancellation law $x^n = y^n \implies x=y$ .

Model solution

(a) Write $q = a/b$ for an integer $a$ and positive integer $b$ . Thus $x^q = (x^{1/b})^a$ . Since $x$ is positive, by Lemma 5.6.6(c) we see that $x^{1/b}$ is positive. By Proposition 4.3.12(b) (which we can use thanks to Proposition 5.6.3) we see that $(x^{1/b})^a$ is positive.

(b) First we show $x^{q+r} = x^q x^r$ . Write $q=a/b$ and $r=c/d$ for integers $a,c$ and positive integers $b,d$ . We have $x^{q+r} = (x^{1/(bd)})^{ad+bc}$ by definition of rational exponentiation, so $(x^{q+r})^{bd} = ((x^{1/(bd)})^{ad+bc})^{bd}$ . Using Proposition 4.3.12(a) and Lemma 5.6.6(a) we have

$\begin{aligned}((x^{1/(bd)})^{ad+bc})^{bd} &= (x^{1/(bd)})^{(ad+bc)(bd)} \\ &= (x^{1/(bd)})^{(bd)(ad+bc)} \\ &= ((x^{1/(bd)})^{bd})^{ad+bc} \\ &= x^{ad+bc}\end{aligned}$

Hence, $(x^{q+r})^{bd} = x^{ad+bc}$ .

Similarly, we can compute $(x^q x^r)^{bd}$ . We have

$\begin{aligned}(x^q x^r)^{bd} &= ((x^{1/b})^a (x^{1/d})^c)^{bd} \\ &= (x^{1/b})^{abd} (x^{1/d})^{cbd} \\ &= ((x^{1/b})^b)^{ad} ((x^{1/d})^d)^{cb} \\ &= x^{ad} x^{cb} \\ &= x^{ad + bc}\end{aligned}$

This shows that $(x^{q+r})^{bd} = (x^q x^r)^{bd}$ . By Proposition 4.3.12(c) we thus have $x^{q+r} = x^q x^r$ as required.

Next we show $(x^q)^r = x^{qr}$ . We have

$\begin{aligned}((x^q)^r)^{bd} &= ((((x^{1/b})^a)^{1/d})^c)^{bd} \\ &= ((((x^{1/b})^a)^{1/d})^d)^{cb} \\ &= ((x^{1/b})^a)^{cb} \\ &= ((x^{1/b})^b)^{ac} \\ &= x^{ac}\end{aligned}$

using Proposition 4.3.12(a) and Lemma 5.6.6(a). Similarly we have

$(x^{qr})^{bd} = ((x^{1/(bd)})^{ac})^{bd} = ((x^{1/(bd)})^{bd})^{ac} = x^{ac}$

Thus $((x^q)^r)^{bd} = (x^{qr})^{bd}$ , so by Proposition 4.3.12(c) we have $(x^q)^r = x^{qr}$ as desired.

(c) We first show that $x^{-n} = 1/x^n$ for an integer $n$ , regardless of the sign of $n$ . If $n$ is positive, then this follows by Definition 5.6.2. If $n=0$ , then $-n=0$ so both sides are equal to $1$ . If $n$ is negative, then $n=-m$ for some positive integer $m$ . We have $x^{-n} = x^{m}$ and $1/x^n = 1/x^{-m} = 1/(1/x^m) = x^m = x^{-n}$ .

Now we show that $x^{-q} = 1/x^q$ . Write $q = a/b$ for an integer $a$ and positive integer $b$ . Then we have $x^{-q} = x^{-a/b} = (x^{1/b})^{-a} = 1/(x^{1/b})^a = 1/x^q$ as required.

(d) Suppose $q > 0$ . Thus we can write $q = a/b$ for positive integers $a,b$ . Suppose first that $x > y$ . Then by Lemma 5.6.6 parts (d) and (c) we have $x^{1/b} > y^{1/b} > 0$ . By Proposition 4.3.10(c) we have $(x^{1/b})^a > (y^{1/b})^a$ . By definition of rational exponentiation, this means $x^q > y^q$ as required.

Conversely suppose $x^q > y^q$ . This means that $(x^{1/b})^a > (y^{1/b})^a$ . From Lemma 5.6.6(c) and Proposition 4.3.12(b) we see that both $(x^{1/b})^a, (y^{1/b})^a$ are positive, i.e. we have $(x^{1/b})^a > (y^{1/b})^a > 0$ . Thus by Lemma 5.6.6(d) we have $((x^{1/b})^a)^{1/a} > ((y^{1/b})^a)^{1/a}$ , which means $x^{1/b} > y^{1/b}$ . Applying Lemma 5.6.6(d) again, we have $x > y$ as required.

(e) We can write $q = a/b$ and $r = c/d$ for integers $a,c$ and positive integers $b,d$ .

Suppose first that $x > 1$ . We must show $x^q > x^r$ if and only if $q > r$ . First suppose $q > r$ , i.e. $a/b > c/d$ . Since $b,d$ are positive, we can multiply by $bd$ on both sides to obtain $ad > bc$ . This implies $ad - bc > 0$ . Since $x > 1$ , by Lemma 5.6.6(d) we have $x^{1/(bd)} > 1^{1/(bd)}$ . Since $1^{bd} = 1$ , by Lemma 5.6.6(b) we have $1 = 1^{1/(bd)}$ . Thus $x^{1/(bd)} > 1$ . We can show by induction that $(x^{1/(bd)})^n > 1$ for all positive integers $n$ (the base case is already given to us, and if $(x^{1/(bd)})^n > 1$ then $(x^{1/(bd)})^{n+1} = (x^{1/(bd)})^n (x^{1/(bd)}) > (x^{1/(bd)})^n 1 > 1$ ). Since $ad - bc$ is a positive integer, we have $(x^{1/(bd)})^{ad} (x^{1/(bd)})^{-bc} = (x^{1/(bd)})^{ad-bc} > 1$ , where the equality follows from Proposition 4.3.12(a). By Proposition 4.3.12(b), we know that $(x^{1/(bd)})^{bc}$ is positive, so we can multiply through by it to obtain $(x^{1/(bd)})^{ad} > (x^{1/(bd)})^{bc}$ . But $(x^{1/(bd)})^{ad} = (((x^{1/b})^{1/d})^d)^a = (x^{1/b})^a = x^q$ by Lemma 5.6.6(g), Proposition 4.3.12(a), and Lemma 5.6.6(a). Similarly, we have $(x^{1/(bd)})^{bc} = x^r$ . Hence $x^q > x^r$ .

Conversely suppose $x^q > x^r$ . By part (a), we know $x^q, x^r$ are positive, so we can apply Proposition 4.3.10(c) to conclude $(x^q)^{bd} > (x^r)^{bd}$ . We have $(x^q)^{bd} = ((x^{1/b})^a)^{bd} = ((x^{1/b})^b)^{ad} = x^{ad}$ and similarly $(x^r)^{bd} = (x^{c/d})^{bd} = x^{cb}$ . Thus $x^{ad} > x^{bc}$ . Since $x^{bc}$ is positive, we can divide by it on both sides to obtain $x^{ad-bc} > 1$ . We know that $x > 1$ so $x^n > 1$ for a positive integer $n$ . If $m = -n$ is a negative integer, then $1 = x^n/x^n > 1/x^n = x^{-n} = x^m$ . Thus if $ad-bc$ negative, we would have $x^{ad-bc} < 1$ , a contradiction. If $ad-bc=0$ we would have $x^{ad-bc} = 1$ , a contradiction. Thus we are left to conclude $ad-bc > 0$ , i.e. $ad > bc$ . Since $b,d$ are positive, we can divide through by $bd$ to obtain $q=a/b > c/d=r$ as required.

Now suppose $0 < x < 1$ . (This argument is very similar to the previous case so we will quite brief, skipping citations of previous results; I may come back to add these in later.) We must show that $x^q > x^r$ if and only if $q < r$ . First suppose that $q < r$ . Thus $a/b < c/d$ , so $ad < bc$ , which means $0 < bc-ad$ . Since $0 < x < 1$ we have $x^{1/(bd)} < 1^{1/(bd)} = 1$ . An induction argument similar to the one above shows that $(x^{1/(bd)})^{bc-ad} < 1$ so $(x^{1/(bd)})^{bc} < (x^{1/(bd)})^{ad}$ . After rewriting this is $x^r < x^q$ as required.

Conversely suppose that $x^q > x^r$ . By part (a), both sides are positive, so we have $(x^q)^{bd} > (x^r)^{bd}$ , i.e. $x^{ad} > x^{bc}$ . This rearranges to $x^{ad-bc} > 1$ . Since $0 < x < 1$ , we can show by induction that for positive $n$ that $x^n < 1$ ; show that for $n=1$ that $x^n = 1$ ; and show that for negative $n$ , i.e. $n = -m$ for positive $m$ that $x^n = x^{-m} = 1/x^m$ , and since $x^m < 1$ we have $1=x^m x^{-m} < 1x^{-m} = x^n$ . Since $x^{ad-bc} > 1$ this means $ad-bc < 0$ , i.e. $ad < bc$ . Thus $a/b < c/d$ , which means $q < r$ as required.

(f) We must show $(xy)^q = x^q y^q$ . Write $q = a/b$ , where $a$ is an integer and $b$ is a positive integer. We have $((xy)^q)^b = (((xy)^{1/b})^a)^b = (((xy)^{1/b})^b)^a = (xy)^a$ . Also, $(x^q y^q)^b = (x^q)^b (y^q)^b = ((x^{1/b})^a)^b ((y^{1/b})^a)^b = x^a y^a = (xy)^a$ . Thus $((xy)^q)^b = (x^q y^q)^b$ . By Proposition 4.3.12(c) we conclude $(xy)^q = x^q y^q$ as desired.

Solutions

Exercise 5.6.1

May 24, 2020 Issa Rice 22 Comments

Exercise statement

Prove Lemma 5.6.6.

Lemma 5.6.6. Let $x,y \geq 0$ be non-negative reals, and let $n, m \geq 1$ be positive integers.

(a) If $y = x^{1/n}$ , then $y^n = x$ .
(b) Conversely, if $y^n = x$ , then $y = x^{1/n}$ .
(c) $x^{1/n}$ is a non-negative real number, and is positive if and only if $x$ is positive.
(d) We have $x > y$ if and only if $x^{1/n} > y^{1/n}$ .
(e) If $x > 1$ , then $x^{1/k}$ is a decreasing (i.e. $x^{1/k} > x^{1/l}$ whenever $l > k$ ) function of $k$ . If $0 < x < 1$ , then $x^{1/k}$ is an increasing function of $k$ . If $x=1$ , then $x^{1/k}$ for all $k$ . Here $k$ ranges over the positive integers.
(f) We have $(xy)^{1/n} = x^{1/n} y^{1/n}$ .
(g) We have $(x^{1/n})^{1/m} = x^{1/nm}$ .

Hints

Review the proof of Proposition 5.5.12.
You will find proof by contradiction a useful tool, especially when combined with the trichotomy of order in Proposition 5.4.7 and Proposition 5.4.12.
The earlier parts of the lemma can be used to prove later parts of the lemma.
With part (e), first show that if $x > 1$ then $x^{1/n} > 1$ , and if $x < 1$ then $x^{1/n} < 1$ .

How to think about the exercise

Some miscellaneous comments:

One of the things you have to do in this exercise is decide when you need to go back to the definition of $x^{1/n}$ and when you can just get by with algebraic manipulations. That’s part of the challenge here, and there isn’t a short rule for making this decision, so you just have to stumble around.
This exercise is annoying and it took me several days to write this up. I’m not satisfied with how this post turned out, and I hope to come back to it later to clean it up more.
Part (a) of this exercise is basically showing that the $n$ th root exists. The book showed in Lemma 5.6.5 that the notation $x^{1/n}$ makes sense (i.e. assigns a unique real number given $x$ ) but the notation, however suggestive it may be, does not show that $x^{1/n}$ in fact has the meaning we want it to have. So part (a) is the work we need to justify calling $x^{1/n}$ the $n$ th root.
Later on, once we have the tools of continuous functions (in particular, the intermediate value theorem, Theorem 9.7.1) we can give a much simpler proof of part (a); see Remark 9.7.3.
There’s something tricky/sneaky going on where we fix epsilon, but then we keep changing it around as we make it smaller to meet our requirements. What would the argument look like if we were to formalize it in first-order logic? See here for some further thoughts.

Model solution

(a) We will do a proof similar to the one in Proposition 5.5.12.

Suppose $y = x^{1/n}$ . We want to show $y^n = x$ . To do this, we will show that both $y^n < x$ and $y^n > x$ lead to contradictions.

First suppose $y^n < x$ . Let $0 < \varepsilon < 1$ be a small positive number. We claim that $(y+\varepsilon)^n \leq y^n + M\varepsilon$ for some positive constant $M$ which can depend on $y$ but does not depend on $\varepsilon$ . To show this, we will use induction on $n$ . For the base case $n=1$ , we have $(y+\varepsilon)^1 = y^1 + 1\varepsilon$ as required. Now suppose inductively that $(y+\varepsilon)^n \leq y^n + M\varepsilon$ for some positive $M$ . We want to show that $(y+\varepsilon)^{n+1} \leq y^{n+1} + M'\varepsilon$ for some positive $M'$ . We have

$\begin{aligned}(y+\varepsilon)^{n+1} &= (y+\varepsilon)(y+\varepsilon)^n \\ &\leq (y+\varepsilon)(y^n + M\varepsilon) \\ &= y^{n+1} + (y^n + yM + M\varepsilon)\varepsilon \\ &\leq y^{n+1} + (y^n + yM + M)\varepsilon\end{aligned}$

so taking $M' := y^n + yM + M$ closes the induction. (Later on, in Exercise 7.1.4, you will prove the binomial formula, which gives a much more precise expression for $(y+\varepsilon)^n$ . It’s possible to prove the binomial formula at this point without circularity, but it’s not obvious how to discover this formula, and it’s the kind of ingenuity that one isn’t expected to have here. So I chose to prove the wimpy bound above.)

Since $y^n < x$ , we can choose an $0 < \varepsilon < 1$ such that $y^n + M\varepsilon < x$ , thus $(y+\varepsilon)^n < x$ . This means that $y+\varepsilon \in \{z \in \mathbf R : z \geq 0 \text{ and } z^n \leq x\}$ . But this contradicts the fact that $y = x^{1/n}$ is an upper bound of this set.

Now suppose that $y^n > x$ . We can show by an induction argument similar to the one above that $(y-\varepsilon)^n \geq y^n - M\varepsilon$ for some positive $M$ that does not depend on $\varepsilon$ . Here are the details. For the base case, we have $(y-\varepsilon)^1 = y^1 - 1\varepsilon$ . Now suppose inductively that $(y-\varepsilon)^n \geq y^n - M\varepsilon$ for some positive $M$ that does not depend on $\varepsilon$ . We want to show that $(y-\varepsilon)^{n+1} \geq y^{n+1} - M'\varepsilon$ for some positive $M'$ that does not depend on $\varepsilon$ . Since $y^n > x \geq 0$ we see that $y > 0$ (because if $y=0$ then $y^n = 0$ , a contradiction). Thus $y-\varepsilon > 0$ as long as $\varepsilon > 0$ is small enough, which means that $(y-\varepsilon)(y-\varepsilon)^n \geq (y-\varepsilon)(y^n - M\varepsilon)$ by induction hypothesis. We have

$\begin{aligned}(y-\varepsilon)^{n+1} &= (y-\varepsilon)(y-\varepsilon)^n \\ &\geq (y-\varepsilon)(y^n - M\varepsilon) \\ &= y^{n+1} + (-y^n - yM + M\varepsilon)\varepsilon \\ &\geq y^{n+1} + (-y^n - yM)\varepsilon\end{aligned}$

where the last inequality follows since $M, \varepsilon$ are both positive. So we can take $M' := y^n + yM$ to close the induction.

By what we said above, we have $(y-\varepsilon)^n \geq y^n - M\varepsilon$ for some positive $M$ . Since $y^n > x$ , we have $y^n - M\varepsilon > x$ as long as $\varepsilon > 0$ is small enough. Hence $(y-\varepsilon)^n > x$ . This means $y-\varepsilon \geq z$ for all $z \in \{z \in \mathbf R : z \geq 0 \text{ and } z^n \leq x\}$ (because if $y-\varepsilon < z$ then $(y-\varepsilon)^n < z^n \leq x$ , a contradiction). Thus $y-\varepsilon$ is an upper bound for $\{z \in \mathbf R : z \geq 0 \text{ and } z^n \leq x\}$ , which contradicts the fact that $y=x^{1/n}$ is the least upper bound of the set.

Both $y^n < x$ and $y^n > x$ lead to contradictions, so we are left to conclude that $y^n = x$ .

(b) Suppose $y^n = x$ . Then we have $(y^n)^{1/n} = x^{1/n}$ (this is just the substitution axiom of equality; see Appendix A.7). Call this quantity $z$ , i.e. $z := (y^n)^{1/n} = x^{1/n}$ . By part (a), we have $z^n = y^n = x$ . By Proposition 5.6.3 (specifically, Proposition 4.3.12(c)), we have $z = y$ . Therefore, $y = z = x^{1/n}$ as required.

(c) First we show that $x^{1/n}$ is a non-negative real number. We have $0 \geq 0$ and $0^n = 0 \leq x$ , so $0$ is in the set $\{y \in \mathbf R : y \geq 0 \text{ and } y^n \leq x\}$ . Since $x^{1/n}$ is an upper bound for this set, we have $x^{1/n} \geq 0$ .

Now we show that $x^{1/n}$ is positive if and only if $x$ is positive.

Suppose $x$ is positive. Suppose for contradiction that $x^{1/n}$ is not positive. Since $x^{1/n}$ is non-negative, it therefore must be equal to zero. By part (a) we thus have $0^n = x$ , i.e. $x=0$ , a contradiction.

Conversely, suppose $x^{1/n}$ is positive. Suppose for sake of contradiction that $x$ is not positive. Since $x \geq 0$ by hypothesis, this means $x = 0$ . By part (b), we have $x^{1/n} = 0^{1/n}$ . Consider the set $\{ y \in \mathbf R : y \geq 0 \text{ and } y^n \leq 0 \}$ . If $y > 0$ then $y^n > 0$ by Proposition 4.3.12(b) (which we can use by Proposition 5.6.3), so $y$ would not be in this set. Thus the only element of this set is $0$ , so $0^{1/n} = \sup\{0\} = 0$ , which means $x^{1/n} = 0^{1/n} = 0$ , a contradiction.

(d) Suppose $x > y$ . Suppose for sake of contradiction that $x^{1/n} \leq y^{1/n}$ . By part (c), we have $0 \leq x^{1/n} \leq y^{1/n}$ so we can apply Proposition 4.3.10(c) to obtain $0 \leq (x^{1/n})^n \leq (y^{1/n})^n$ . By part (a), $(x^{1/n})^n = x$ and $(y^{1/n})^n = y$ , so $x \leq y$ , a contradiction.

Conversely suppose $x^{1/n} > y^{1/n}$ . By part (c), $x^{1/n} > y^{1/n} \geq 0$ , so by Proposition 4.3.10(c) again, we have $x > y \geq 0$ as required.

(e) Let $l > k$ be two positive integers, and suppose that $x > 1$ . We want to show that $x^{1/k} > x^{1/l}$ . Suppose for sake of contradiction that $x^{1/k} \leq x^{1/l}$ . Then by part (a) and Proposition 4.3.10 (parts (a) and (c)) we have $x^l = ((x^{1/k})^k)^l = (x^{1/k})^{kl} \leq (x^{1/l})^{kl} = ((x^{1/l})^l)^k = x^k$ . But $l > k$ so $l = k + r$ for some positive integer $r$ . By Proposition 4.3.10(a) we have $x^l = x^{k+r} = x^k x^r$ . An easy induction shows that $x^r > 1$ so $x^k x^r > x^k$ , i.e. $x^l > x^k$ . This contradicts what we showed earlier, that $x^l \leq x^k$ . This contradiction shows that $x^{1/k} > x^{1/l}$ .

Now suppose $0 < x < 1$ . We want to show that $x^{1/l} > x^{1/k}$ . Suppose for sake of contradiction that $x^{1/l} \leq x^{1/k}$ . By Proposition 4.3.10 we have $x^k = (x^{1/l})^{kl} \leq (x^{1/k})^{kl} = x^l$ . As above, we have $l = k+r$ for a positive integer $r$ . This time, $0 < x < 1$ so $x^r < 1$ (this is again an easy induction proof). Thus we have $x^r x^k < x^k$ , i.e. $x^l < x^k$ . We showed earlier that $x^k \leq x^l$ , so this is a contradiction. This contradiction shows that $x^{1/l} > x^{1/k}$ as required.

Finally suppose $x=1$ . We have $1^k = 1$ so by part (b) we have $1 = 1^{1/k}$ as required.

(f) We want to show $(xy)^{1/n} = x^{1/n} y^{1/n}$ . By part (a), $((xy)^{1/n})^n = xy$ . By Proposition 4.3.10(a) and part (a) of this proof, $(x^{1/n} y^{1/n})^n = (x^{1/n})^n (y^{1/n})^n = xy$ . Thus $((xy)^{1/n})^n = (x^{1/n} y^{1/n})^n$ , so by Proposition 4.3.12(c) we have $(xy)^{1/n} = x^{1/n} y^{1/n}$ as required.

(g) We want to show $(x^{1/n})^{1/m} = x^{1/nm}$ . We have $((x^{1/n})^{1/m})^{nm} = (((x^{1/n})^{1/m})^m)^n = x$ by Proposition 4.3.10(a) and part (a) of this proof. We also have $(x^{1/nm})^{nm} = x$ by part (a). Thus $((x^{1/n})^{1/m})^{nm} = (x^{1/nm})^{nm}$ , so by Proposition 4.3.12(c) we have $(x^{1/n})^{1/m} = x^{1/nm}$ as required.

Solutions

Exercise 5.5.5

May 17, 2020 Issa Rice 2 Comments

Exercise statement

Establish an analogue of Proposition 5.4.14, in which “rational” is replaced by “irrational”.

Hints

Use Proposition 5.4.14, or copy its proof.

How to think about the exercise

This exercise might generate a feeling of “this is impossible” in you. After all, in the entire book so far, we have only met a single irrational number! How in the world are we to show that there are so many irrational numbers that we can find one between any two real numbers?

But actually, this very observation is a clue as to how to solve the exercise. If $\sqrt 2$ is the only irrational number we know about, then the solution will probably involve $\sqrt 2$ somehow. In particular, can we create other irrational numbers using $\sqrt 2$ ?

If $q\ne0$ is a rational number, then it seems like $q\sqrt 2$ is also irrational. Indeed, if $q\sqrt 2 = q'$ were rational, we could divide by $q$ to obtain $\sqrt 2 = q'/q$ ; since the rationals are closed under division, this would show that $\sqrt 2$ is rational, a contradiction. Therefore, we conclude that $q\sqrt 2$ is irrational.

Are there other ways to construct irrational numbers? What about addition? If $q$ is again rational, then $q + \sqrt 2$ would seem to be irrational. Indeed, we can use the same reasoning as above: if we were to have $q + \sqrt 2 = q'$ for some rational $q'$ , then we could write $\sqrt 2 = q' - q$ as a rational number, a contradiction.

So suddenly the exercise doesn’t seem so impossible: we just need to find a multiple or shifted copy of $\sqrt 2$ between the two real numbers we are given.

Model solution 1

The easy solution is to apply Proposition 5.4.14.

Let $x < y$ be two real numbers. Then we can add $\sqrt 2$ to both sides to obtain $x + \sqrt 2 < y + \sqrt 2$ . By Proposition 5.4.14, there exists a rational number $q$ such that $x + \sqrt 2 < q < y + \sqrt 2$ . Now subtracting $\sqrt 2$ from each part we have $x < q - \sqrt 2 < y$ . It now remains to show that $q - \sqrt 2$ is irrational. Suppose for sake of contradiction that it is a rational number. Then we have $q - \sqrt 2 = q'$ for some rational $q'$ . Thus $\sqrt 2 = q-q'$ is a rational number (the rationals are closed under subtraction), a contradiction. So $q - \sqrt 2$ is irrational as required.

Model solution 2

The following is a more involved solution that does not rely on Proposition 5.4.14. (It’s valid to use Proposition 5.4.14, but you might still wonder if there’s a way to find an irrational number “by hand”.) The idea here is to break things into two steps: (1) find an irrational number between any two integers; (2) find two integers between any two real numbers which are scaled to be sufficiently far apart (i.e. given $x < y$ , we want to find two integers between $Mx$ and $My$ , where $M$ is a positive integer).

Let $x < y$ be two real numbers. Then $y - x > 0$ so by Exercise 5.4.4 there exists a positive integer $N$ such that $y - x > 1/N > 0$ . Multiplying through by $2N$ we have $2Ny - 2Nx > 2$ .

By Exercise 5.4.3, there exists some integer $n$ such that $n \leq 2Nx < n+1$ . We will show that $n \leq 2Nx < n+1 < n+2 < 2Ny$ . The first two inequalities we already have, and the third is clear from the way integers work, so we will only need to show the fourth inequality. Suppose for sake of contradiction that $2Ny \leq n+2$ . Then we have $n \leq 2Nx$ so $-2Nx \leq -n$ . Adding this inequality to $2Ny \leq n+2$ we obtain $2Ny - 2Nx \leq 2$ , which contradicts the fact that $2Ny - 2Nx > 2$ . Thus we have $n \leq 2Nx < n+1 < n+2 < 2Ny$ as required.

Next we show that $1 < \sqrt 2 < 2$ . Suppose for sake of contradiction that $\sqrt 2 \leq 1$ . Since $\sqrt 2$ is positive (Proposition 5.5.12 showed that a positive square root of two exists), by Proposition 5.4.7(e) we have $\sqrt 2 \sqrt 2 \leq \sqrt 2$ by multiplying through by $\sqrt 2$ . Therefore $2 = (\sqrt 2)^2 \leq \sqrt 2 \leq 1$ , a contradiction. Similarly, suppose for sake of contradiction that $2 \leq \sqrt 2$ . By multiplying through by $\sqrt 2$ we obtain $2\sqrt 2 \leq 2$ . By multiplying the original inequality through by $2$ we obtain $4 \leq 2\sqrt 2$ . Thus $4 \leq 2\sqrt 2 \leq 2$ , a contradiction. So we have $1 < \sqrt 2 < 2$ as required.

Adding $n$ to each part of $1 < \sqrt 2 < 2$ gives $n+1 < n+\sqrt 2 < n+2$ . Combining this with the inequality we had earlier (from the second paragraph) we obtain $2Nx < n+1 < n+\sqrt 2 < n+2 < 2Ny$ . Dividing through by $2N$ we have $x < \frac{n+\sqrt 2}{2N} < y$ . It will now suffice to show that $\frac{n+\sqrt 2}{2N}$ is irrational. Suppose for sake of contradiction that $\frac{n+\sqrt 2}{2N} = q$ is a rational number. Then we have $n+\sqrt 2 = 2Nq$ , so $\sqrt 2 = 2Nq - n$ . But $2Nq - n$ is a rational number, which means $\sqrt 2$ is also rational, a contradiction. Thus $\frac{n+\sqrt 2}{2N}$ is irrational as required.

Model solution 3

Let $x < y$ be two real numbers. We have two cases, depending on whether or not $x$ is a rational number.

Suppose first that $x$ is rational. Then $y-x$ is a positive real, and we also know that $\sqrt 2$ is positive. Thus by the Archimedean property we have some positive integer $M$ such that $M(y-x) > \sqrt 2$ , i.e. $y-x > \sqrt 2 / M > 0$ . Adding $x$ to each part we obtain $y > x + \sqrt 2 / M > x$ . Since $x$ is rational, $x + \sqrt 2 / M$ irrational because otherwise we could show that $\sqrt 2$ is a rational number.

Now suppose $x$ is irrational. Then by Exercise 5.4.4 we can find a positive integer $N$ such that $y-x > 1/N > 0$ . Adding $x$ to each part we obtain $y > x + 1/N > x$ . Since $x$ is irrational, we see that $x + 1/N$ is also irrational.