#sqrt167# lies between which two successive integers?

1 Answer
Mar 1, 2017

Answer:

12 and 13

Explanation:

Let #n# be the lower of the successive integers.

Thus: #n< sqrt167<(n+1)# #[n in ZZ]#

#sqrt167 ~= 12.9228#

#:. n = #the interger part of #12.9338 = 12#

Hence: #12< sqrt167<13#

To check this result:

Consider #12^2 = 144# and #13^2 = 169#

and #144<167<169#