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