Exercise 2.3.4

March 15, 2020 Issa Rice 4 Comments

Exercise statement

Prove the identity $(a+b)^2 = a^2 + 2ab + b^2$ for all natural numbers $a,b$ .

Hints

Use Proposition 2.3.4 and Lemma 2.3.2.

How to think about the exercise

This exercise is basically just an application of all of the propositions and lemmas that appeared in this section (Section 2.3).

Exercises like this one test your ability to rigorously apply definitions and results. For each step, you want to cite the result which justifies it.

Model solution

Let $a,b$ be natural numbers. By definition of exponentiation (Definition 2.3.11), $(a+b)^2 = (a+b)^1 \times (a+b) = (a+b)(a+b)$ . By the distributive law (Proposition 2.3.4), we have $(a+b)(a+b) = (a+b)a + (a+b)b$ . By the distributive law again, $(a+b)a = aa + ba$ ; by the definition of exponentiation and commutativity of multiplication (Lemma 2.3.2) this is just $a^2 + ab$ . Similarly, by the distributive law, $(a+b)b = ab + bb = ab + b^2$ . Summarizing everything so far, we have $(a+b)^2 = a^2 + ab + ab + b^2$ . It now suffices to show that $ab + ab = 2ab$ . By the associativity of multiplication and definition of multiplication, $2ab = 2\times(ab) = (1\times (ab)) + ab = ab+ab$ .

Solutions

Exercise 6.3.3

March 15, 2020 Issa Rice 6 Comments

Exercise statement

Prove Proposition 6.3.8.

Proposition 6.3.8 (Monotone bounded sequences converge). Let $(a_n)_{n=1}^\infty$ be a sequence of real numbers which has some finite upper bound $M \in \mathbf R$ , and which is also increasing (i.e., $a_{n+1} \geq a_n$ for all $n \geq m$ ). Then $(a_n)_{n=m}^\infty$ is convergent, and in fact

$\displaystyle \lim_{n\to\infty} a_n = \sup(a_n)_{n=m}^\infty \leq M$

Hints

Use proposition 6.3.6.

How to think about the exercise

I should think of more things to say for this exercise. I remember having a really difficult time with it when I first encountered it, but now it seems so simple.

Model solution

Call the least upper bound of the sequence $\alpha := \sup(a_n)_{n=m}^\infty$ . Since the sequence is bounded above, $\alpha$ is a real number. Now, $\alpha$ is the least upper bound while $M$ is merely an upper bound, so we have $\alpha \leq M$ (proposition 6.3.6). To complete the proof, we must show $\lim_{n\to\infty} a_n = \alpha$ . Let $\varepsilon > 0$ , and observe that $\alpha - \varepsilon$ cannot be an upper bound, since it is less than the least upper bound; in other words, if $\alpha - \varepsilon$ were an upper bound, then we have found a smaller upper bound than $\alpha$ , which contradicts the fact that $\alpha$ is the least upper bound. Thus there exists $N \geq m$ such that $a_N > \alpha - \varepsilon$ (again, this is an application of proposition 6.3.6). Since the sequence is increasing, we have $a_n \geq a_N$ for all $n \geq N$ . Thus

$\displaystyle \alpha + \varepsilon > \alpha \geq a_n \geq a_N > \alpha - \varepsilon$

for all $n \geq N$ . This means $\varepsilon > a_n - \alpha > -\varepsilon$ so $|a_n - \alpha| < \varepsilon$ . Since $\varepsilon > 0$ was arbitrary, this shows that $\lim_{n\to\infty} a_n = \alpha$ as desired.

Solutions

Exercise 5.4.8

March 15, 2020 Issa Rice 18 Comments

Exercise statement

Let $(a_n)_{n=1}^\infty$ be a Cauchy sequence of rationals, and let $x$ be a real number. Show that if $a_n \leq x$ for all $n\geq 1$ , then $\mathrm{LIM}_{n\to\infty} a_n \leq x$ . Similarly, show that if $a_n \geq x$ for all $n \geq 1$ , then $\mathrm{LIM}_{n\to\infty} a_n \geq x$ .

Hints

Do a proof by contradiction.
Use Proposition 5.4.14.
Use Corollary 5.4.10 or Proposition 5.4.9.

How to think about the exercise

This exercise is a pretty straightforward application of the propositions/corollaries mentioned in the hint, so there is not much to discuss!

Model solution

Suppose $a_n \leq x$ for all $n \geq 1$ , and suppose for sake of contradiction that $\mathrm{LIM}_{n\to\infty} a_n > x$ . By Proposition 5.4.14, there exists a rational number $q$ such that $\mathrm{LIM}_{n\to\infty} a_n > q > x$ . Consider the Cauchy sequence $(q)_{n=1}^\infty$ . Since $a_n \leq x < q$ for all $n \geq 1$ , Corollary 5.4.10 tells us that $\mathrm{LIM}_{n\to\infty} a_n \leq \mathrm{LIM}_{n\to\infty} q =q$ . But now we have both $\mathrm{LIM}_{n\to\infty} a_n \leq q$ and $\mathrm{LIM}_{n\to\infty} a_n > q$ , a contradiction.