Normally to prove something exists, we would give an example of such a thing, but that’s not entirely necessary.
rational and irrational numbers
We showed last week, that $\sqrt{2}$ is not a rational number.
We’ll need the following property of exponents
$$ (a^b)^c=a^{bc}\\ \text{``obvious" if b and c are the integers, but also true more generally} $$
<aside> 🔖 Ex.
There exist irrational numbers $x$ and $y$ such that $x^y$ is rational.
$$ \text{call this } q $$
</aside>
We can conclude that there is an example without actually finding one.
Observation,
$$ (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}=\sqrt{2}^{\sqrt{2}\times\sqrt{2}}=2 $$
Claim either $\sqrt{2}^{\sqrt{2}}$ is rational or it isn’t
$p$ is the proposition “$\sqrt{2}^{\sqrt{2}}$ is rational”
so $p\vee\neg q$ is true.
Case 1:
If p is true, then $x=\sqrt{2}, y=\sqrt{2}$ is our required example.
Case 2:
If p is false, then $x=\sqrt{2}^{\sqrt{2}}, y =\sqrt{2}$ is our example.
We have the observations
$p\rarr q\\ \neg q\rarr q\\ \therefore(p\rarr q)\wedge(\neg q \rarr q)$
which is logically equivalent to $(p\vee \neg p)\rarr q$
but also $(p\vee \neg p)$