Between which two consecutive integers is $\sqrt{51}$ $sqrt(51)$ ?

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Answer 1 · 2017-03-05T00:25:27

$7 < \sqrt{51} < 8$ $7 < sqrt(51) < 8$

Note that:

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

Hence:

$7 < \sqrt{51} < 8$ $7 < sqrt(51) < 8$

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Bonus

Since $51 = 7^{2} + 2$ $51 = 7^2+2$ is of the form $n^{2} + 2$ $n^2+2$ , 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:

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

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

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