Proving "there exists" statements

This post was originally published on WordPress
https://explainingmaths.wordpress.com/2021/01/09/proving-there-exists-statements/

Here is another question I was asked on my first-year pure maths Piazza discussion forum.

When proving there exists statements, is it enough to give just one example or do you have to prove it using the definitions, etc.?

Here is my reply:

Note: I use \(\mathbb{Q}^c\) to denote the set of irrational numbers, so
            \(\mathbb{Q}^c = \mathbb{R}\setminus \mathbb{Q}\,.\)

Hi,
The answer is really "yes, as long as you know what it means to give one example".
Let me give two examples, one relatively easy, one more complicated.

First a fairly straightforward example. In practice coursework one, we discovered that the following statement is true:
(S1)      There exists \(x \in \mathbb{Q}^c\) such that \(x^2 \in \mathbb{Q}\).
This statement can be proven true using one specific example, e.g. \(x=\sqrt 2\): it is standard that \(\sqrt 2\) is irrational, and \((\sqrt 2)^2 = 2 \in \mathbb{Q}\).
Notice that we could use any other letter in place of \(x\) and the meaning of (S1) does not change. But we wouldn't usually use \(m\), \(n\) or \(k\) (for example) as those letters usually represent integers.

Now for a slightly more complicated example. Consider the following statement:
(S2)      \(\forall x \in \mathbb{Q}, \exists n \in \mathbb{N}: nx \in \mathbb{Z}\)
In words, in an optionally expanded form, this says
 (S2b)      For every rational number \(x\), it is true that there exists at least one positive integer \(n\) such that \(nx\) is an integer.
(There are many other ways to express this, all of which mean the same thing.)

This is really claiming that an infinite number of "There exists" statements are all true. These statements depend on the rational number \(x\) involved. If all of these statements are true, then they can each be proven true using one example of an integer \(n\). But the \(n\) you need will depend on \(x\). And it would not be enough to just check a few values of \(x\), because the "outer" statement is a "for all" statement. So we would need a general proof, valid for all rational numbers \(x\), which shows, for each rational number \(x\), how we can choose a suitable value of \(n\).

Now (S2) and (S2b) are true, and each of these "there exists" statements can be proven true using one example \(n\), but where the \(n\) you choose depends on \(x\).

Here is how a proof of (S2) goes.

Let \(x \in \mathbb{Q}\). Then there are \(m \in \mathbb{Z}\) and \(n \in \mathbb{N}\) such that \(x=m/n\). But then \(n \in\mathbb{N}\) and \(nx=m \in \mathbb{Z}\). Thus it is true that \(\exists n \in \mathbb{N}: nx \in \mathbb{Z}\).
Since \(x\) was an arbitrary element of \(\mathbb{Q}\), this shows that
          \(\forall x \in \mathbb{Q}, \exists n \in \mathbb{N}: nx \in \mathbb{Z}\),
i.e. (S2) is true, as required.

Notice that our "one example" here depends on \(x\), and we could use any positive integer that works as a denominator for \(x\). If you wanted to, you could use lowest terms, and that would then give you the smallest positive integer \(n\) that works. But you don't have to do this here.

Best wishes,
Dr Feinstein