# Between what integers does sqrt132 lie?

Jan 13, 2017

Between 11 and 12
Please remember that ${11}^{2} = 121 \mathmr{and} {12}^{2} = 144$
132 is between those two squares.

Jan 13, 2017

$11$ and $12$

#### Explanation:

Note that:

$132 = 11 \times 12$

and:

$11 \times 11 < 11 \times 12 < 12 \times 12$

Hence:

$\sqrt{11 \times 11} < \sqrt{11 \times 12} < \sqrt{12 \times 12}$

That is:

$11 < \sqrt{132} < 12$

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Footnote

Since $132 = 11 \times 12$ is of the form $n \left(n + 1\right)$ (with $n = 11$), its square root has a simple continued fraction form:

sqrt(132) = [11;bar(2,22)] = 11+1/(2+1/(22+1/(2+1/(22+1/(2+1/(22+1/(2+...)))))))

In general:

sqrt(n(n+1)) = [n;bar(2, 2n)] = n+1/(2+1/(2n+1/(2+1/(2n+1/(2+1/(2n+1/(2+...)))))))