Rational square root implies perfect square

Hi, this was based on a thought I had: How could the square of a 'proper' rational (i.e. a rational that when in lowest terms, has a denominator other than 1) be an integer? So let $ \sqrt{n}=\frac{r}{s}$ be the square root of a positive integer $n$. W.l.o.g. \[ \frac{r}{s} \] is in lowest terms. By squaring we get \[r^2 =ns^2\]. Then there exist integers \[u\,v \] such that \[ ru +sv=1 \] Multiplying the preceding equation by $r$ we get
$r =r^2u +rsv =s(nsu +rv)$
Thus $s$ is a divisor of $r$ , and in particular of 1.
So we are left with LaTeX: $s =\pm 1$ Hence LaTeX: $r^2 =n$
i.e. $n$ is a perfect square.
Therefore if $n$ is not a perfect square, then LaTeX: $\sqrt{n}$ is irrational.

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}`	$\boxed{123}$