Between what integers does $\sqrt{132}$ $sqrt132$ lie?

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Between what integers does $\sqrt{132}$ $sqrt132$ lie?

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Answer 1 · 2017-01-13T21:12:15

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

Answer 2 · 2017-01-13T22:16:19

$11$ $11$ and $12$ $12$

Explanation:

Note that:

$132 = 11 \times 12$ $132 = 11xx12$

and:

$11 \times 11 < 11 \times 12 < 12 \times 12$ $11xx11 < 11xx12 < 12xx12$

Hence:

$\sqrt{11 \times 11} < \sqrt{11 \times 12} < \sqrt{12 \times 12}$ $sqrt(11xx11) < sqrt(11xx12) < sqrt(12xx12)$

That is:

$11 < \sqrt{132} < 12$ $11 < sqrt(132) < 12$

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Footnote

Since $132 = 11 \times 12$ $132 = 11xx12$ is of the form $n (n + 1)$ $n(n+1)$ (with $n = 11$ $n=11$ ), 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:

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

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