Between what integers does $sqrt132$ lie?

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Answer 1 · 2017-01-13T21:12:15

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

Answer 2 · 2017-01-13T22:16:19

$11$ and $12$

Note that:

$132 = 11xx12$

and:

$11xx11 < 11xx12 < 12xx12$

Hence:

$sqrt(11xx11) < sqrt(11xx12) < sqrt(12xx12)$

That is:

$11 < sqrt(132) < 12$

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Footnote

Since $132 = 11xx12$ is of the form $n(n+1)$ (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:

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

Between what integers does sqrt132sqrt132 lie?