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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?

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    $\begingroup$ The second line $\left(x-\sqrt{\pi^2}\right)^2=\sqrt{e^2}$ is rational is nonsensical. How do you conclude that? If you could conclude that, you'd be done. $\endgroup$ Sep 27, 2013 at 16:00
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    $\begingroup$ Your "proof" is invalid. I'm pretty sure what you are trying to prove is still an open problem. Please do a simple google search before posting here. $\endgroup$ Sep 27, 2013 at 16:02
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    $\begingroup$ Is there a typo in the second step? If $x = \sqrt{\pi^2} + \sqrt{e^2}$ then $(x-\sqrt{\pi^2})^2 = e^2$. And how does it follow that this is rational? $\endgroup$ Sep 27, 2013 at 16:03
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    $\begingroup$ Also, in the second equation, you subtract $\pi$ from both sides and square, but you forget to square the right-hand side (you corrected this in the third line). Also, note that $x$ is rational, so we must have $x-\pi$ is irrational. We cannot say that $(x-\pi)^2$ is rational. In the last line, you should have a $+\pi^2$ instead of $-\pi^2$. Unfortunately, you have several mistakes in your attempt. $\endgroup$
    – Clayton
    Sep 27, 2013 at 16:06
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    $\begingroup$ What's with the serial downvoting and voting to close on a perfectly sincere question which shows research effort? $\endgroup$ Sep 27, 2013 at 19:17

1 Answer 1

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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?

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