How to prove existence of solutions without constructing one?

Solution 1:

Here is a typical example of a non-constructive existence proof:

We want to prove that there exists a rational number $x^y$ such that both $x$ and $y$ are irrational. A way to do this is first to let $x=y=\sqrt 2$ (because we know that $\sqrt 2$ is irrational). If $\sqrt 2 ^{\sqrt 2}$ is rational, we have have a rational number on the desired form. On the other hand, if $\sqrt 2 ^{\sqrt 2}$ is irrational, let $x=\sqrt 2 ^{\sqrt 2}$ and $y=\sqrt 2$. Then $x^y=(\sqrt 2 ^{\sqrt 2})^{\sqrt 2} = \sqrt 2^2 = 2$, which is obviously rational.

We have proven that there exists a rational number $x^y$ such that both $x$ and $y$ are irrational. Either with $x=y=\sqrt 2$ or $x=\sqrt 2 ^{\sqrt 2}, y= \sqrt 2$, but we don't know which.

Edit:
Trying to generalize this idea. Suppose you want to prove a proposition. Find a set of cases such that only one of the cases has to be true for the proposition to be true. A constructive proof would be to find a specific case that is true. For a non-constructive proof, it is enough show that it is not possible that all the cases are false.

Solution 2:

A very common technique is by counting and comparing cardinalities. For example, one can prove that transcendental numbers must exist by noting that there is a countable number of algebraic numbers (since there is a countable number of polynomials), but the reals are uncountable by Cantor's argument. This shows that most numbers are transcendental without producing a single one of them.