Question

Between what integers does `sqrt132` lie?

Answer 1

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

Answer 2

`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+...)))))))`