The wikipedia tells that it is not known that $\pi+e$ is irrational?
Immediately after reading this my mind came with this proof-
Let $x =\sqrt{\pi^2}+\sqrt{e^2}$ be rational, then
$ \quad (x-\sqrt{\pi^2})^2=\sqrt{e^2}$ is rational , now
$\quad x^2-2x\sqrt{\pi^2}-\pi^2=e^2$ is rational , now
$\quad \frac{e^2-x^2-\pi^2}{-2x} = \sqrt{\pi^2}=\pi$ is rational.
This is a contradiction as $\pi$ is irrational! Thus we prove by contradiction that $\pi+e $ is irrational.
Please tell me me if this is a valid proof? And also tell my mistake if I am wrong somewhere?
It's not clear how you can deduce the first step of your proof.
Let $a=\sqrt[4]{2}$ and $b=2-\sqrt[4]{2}$. Then from:
$$2=a+b$$
can we conclude that:
$$(2-b)^2=a^2=\sqrt{2}$$ is rational?
It's certainly the case that $(2-b)^2=a^2$. But why is it rational?