It's not proved yet.

State whether $(\pi)^e$ is a rational number or an irrational number with proof.

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#NumberTheory #RationalNumbers #IrrationalNumbers #Proof #Puremathematics

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Even proving $e+\pi , e\pi$ is irrational/rational is unsolved,despite the fact that, at-least one of them has to be irrational![You can show this easily] .

Shivang Jindal - 6 years, 2 months ago

More strongly, at least one of $e+\pi, e\pi$ is transcendental. Proof here: $x^2-(e+\pi)x+e\pi$ has roots $e,\pi$ . Assume for contradiction both $e+\pi, e\pi$ are algebraic. Then $e,\pi$ are algebraic (since they're roots of a polynomial with algebraic coefficients), contradiction, since $e,\pi$ are transcendental.

mathh mathh - 5 years, 10 months ago

Proving that a number is an irrational is extremely difficult. As Shivang Jindal has mentioned, there a lot of "at least one of them is irrational" cases, but to prove which one is in the stars with respect to difficulty. Even the irrationality of $\pi$ is very difficult to prove.

Raghav Vaidyanathan - 6 years, 2 months ago

It's transcendal number.

Dev Rajyaguru - 4 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$