# Between which two consecutive integers is sqrt(51) ?

Mar 5, 2017

$7 < \sqrt{51} < 8$

#### Explanation:

Note that:

${7}^{2} = 49 < 51 < 64 = {8}^{2}$

Hence:

$7 < \sqrt{51} < 8$

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Bonus

Since $51 = {7}^{2} + 2$ is of the form ${n}^{2} + 2$, its square root takes a simple form as a continued fraction:

sqrt(51) = [7;bar(7,14)] = 7+1/(7+1/(14+1/(7+1/(14+1/(7+...)))))

We can use this continued fraction expansion to give us rational approximations for $\sqrt{51}$

For example:

sqrt(51) ~~ [7;7] = 7+1/7 = 50/7 = 7.bar(142857)

sqrt(51) ~~ [7;7,14,7] = 7+1/(7+1/(14+1/7)) = 4999/700 = 7.14bar(142857)