Jumpstart - Questions and Answers Week 1

Question 1 (Matthew Hirning, incoming Fall 2018)

I'm having a little bit of trouble wrapping my head around the idea of construction of the integers and rationals using the equivalence classes on cross products of other number systems. Specifically, the idea of the rationals being a subset of $\mathbb{Z} \times \mathbb{N}$. In the lecture notes, in proving that the rationals are countable, we use the fact that $\mathbb{Q} = (\mathbb{Z} \times \mathbb{N})/\sim$ and also $(\mathbb{Z} \times \mathbb{N}) /\sim \,\subset \mathbb{Z} \times \mathbb{N}$. In my mind, however, the elements of $\mathbb{Z} \times \mathbb{N}/\sim$ are equivalence classes which are sets themselves. So I get that the equivalence classes partition $\mathbb{Z} \times \mathbb{N}$ but each of the classes aren't elements of $\mathbb{Z} \times \mathbb{N}$ to begin with, right? I'd appreciate it if someone could help me frame this construction correctly.

Actually I had the same question at the first glance, then I realized that this is just a kind of notation. For instance, $\mathbb{Z} \times \mathbb{N} / \sim$ is used to denote rational numbers, and in rational numbers we treat $\frac{1}{2}$ and $\frac{2}{4}$ as the same number, so they are partitioned into "one number" (or "one class", if you wish), and all the equivalence classes are sets of infinitely many numbers.

Question 2 (Daniel Morrison, incoming Fall 2018)

This is a bit of a broad question, but I was intrigued by the construction of the reals in Lecture 0 and I wanted to know more about it. Specifically, I can see how defining reals in terms of Cauchy sequences of rationals might lead to $\mathbb{R}$ being complete, but I feel like there is a bit of a gap between these results since we need Cauchy sequences of real numbers (not just rational numbers) to converge. However in trying to close that gap I ran into some issues like how would we define an order on $\mathbb{R}$ using the Cauchy sequences, defining a distance function on $\mathbb{R}$, and even then a sequence of real numbers would be a sequence of sequences of rational numbers which is very confusing. Can anyone point me in the right direction for constructing these necesary parts and give me an idea how to think about it?

In your question you are really identifying the main issues that need to be sorted out to "complete" the construction only sketched in the lecture.

(i) Define an order on $ \mathbb{R}$. It is enough to define what is meant by $x>0$ since $x<y$ can then be defined by the validity of $y-x>0$. We define $x>0$ by the validity of $$ x\neq 0 \text{ and } \exists\: N\in \mathbb{N} \text{ s.t. } \forall\: (x_n)_{n\in \mathbb{N}}\in x\:\exists\: M\in \mathbb{N} \text{ and } x_n>\frac{1}{N}\:\forall\: n\geq M. $$

(ii) Define a distance on $ \mathbb{R}$. To do this, we only need to find an absolute value in $ \mathbb{R}$ that extends that of $ \mathbb{Q}$. We do this by setting $$ |x|=\big | [(x_n)_{n\in \mathbb{N}}]\big |:=\big[(|x_n|)_{n\in \mathbb{N}}\big]. $$ You should verify that this definition makes sense (i.e. that $(|x_n|)_{n\in \mathbb{N}}$ is a Cauchy sequence), is independent of the choice of representative, and delivers a function which enjoys the properties of an absolute value (non-negativity, positive homogeneity, and triangle inequality).

(iii) Show density of $ \mathbb{Q}$ and completeness of $\mathbb{R}$. Show first that, given $x\in \mathbb{R}$ and $N\in \mathbb{N}$, there is $q\in \mathbb{Q}$ such that $$ |x-q|<\frac{1}{N}. $$ Remember that $ \mathbb{Q}\ni q\equiv \big[ (q,q,\dots)\big]$. Then use this to generate a Cauchy sequence of rationals, starting with a Cauchy sequence of reals, that stays "close'' to the latter. Finally show that, in fact, the sequence of reals converges to this Cauchy sequence of rationals.

Question 3 (Greg Zitelli, Course's Teaching Assistant 2018)

For every $n \in \mathbb{N}$, is it true that $\sqrt{n+1}+\sqrt{n-1}$ is irrational?

Yes. Suppose there was some $n\in\mathbb{N}$ so that $\sqrt{n+1}+\sqrt{n-1}$ is rational. Then $$ \bigl(\sqrt{n+1}+\sqrt{n-1}\bigr)\bigl(\sqrt{n+1}-\sqrt{n-1}\bigr) = (n+1)-(n-1) = 2, $$ so $\sqrt{n+1}-\sqrt{n-1}$ must also be rational. It then follows that both $\sqrt{n+1}$ and $\sqrt{n-1}$ are rational. But if the square root of a natural number is rational, it must be an integer since if $\sqrt{m}=p/q$ with $p,q$ relatively prime then $p^2/q^2\in\mathbb{N}$. Hence $q^2|p^2$ and so $q=1$ and $\sqrt{m}$ is an integer. Thus $\sqrt{n+1}=a$ and $\sqrt{n-1}=b$ for some $a,b\in\mathbb{N}$. But there are no solutions to $(a+b)(a-b)=2$ in $\mathbb{N}$ so by contradiction $\sqrt{n+1}+\sqrt{n-1}$ is irrational for every $n\in\mathbb{N}$.