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

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Answer 1 · 2017-03-05T00:25:27

#7 < sqrt(51) < 8#

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)#