The most beautiful solution ever!

First of all I want to say that this is not the namesake competition . We are not " you against me " but " infinity against infinity". This post is for all those who have either created or have perceived " the most beautiful solution ever " .I created this for the dreamers who may want to wander through the realms of mathematics in one alley having many shops . Please provide the link to that particular solution(s).

#TheBeautyOfMath

Easy Math Editor

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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}`	$\boxed{123}$

Comments

Claim: An irrational raised to the power of an irrational can be a rational.

Proof: Either $\sqrt{2}^\sqrt{2}$ is irrational or rational.

If it is rational, we are done.

Otherwise, ${\sqrt{2}^\sqrt{2}}^\sqrt{2} =\sqrt{2}^2 = 2$ is rational.

Agnishom Chattopadhyay - 5 years, 6 months ago

Nice example, but I think that the last equation should be written as

$\large (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{(\sqrt{2}\times \sqrt{2})} = \sqrt{2}^{2} = 2,$

since $\large \sqrt{2}^{\sqrt{2}^{\sqrt{2}}} = 1.7608.....$ is irrational.

Brian Charlesworth - 5 years, 6 months ago

Nice : )

Raven Herd - 5 years, 6 months ago

Don Zagier's one sentence proof of Fermat's theorem on sums of two squares is quite beautiful.

Prasun Biswas - 5 years, 5 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}`	$\boxed{123}$