Between what integers does #sqrt132# lie?

2 Answers
Jan 13, 2017

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

Jan 13, 2017

Answer:

#11# and #12#

Explanation:

Note that:

#132 = 11xx12#

and:

#11xx11 < 11xx12 < 12xx12#

Hence:

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

That is:

#11 < sqrt(132) < 12#

#color(white)()#
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+...)))))))#