# Between what integers does #sqrt132# lie?

##### 2 Answers

Jan 13, 2017

Between 11 and 12

Please remember that

132 is between those two squares.

Jan 13, 2017

#### Explanation:

Note that:

#132 = 11xx12#

and:

#11xx11 < 11xx12 < 12xx12#

Hence:

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

That is:

#11 < sqrt(132) < 12#

**Footnote**

Since

#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+...)))))))#