How do you find the perfect square between 60 and 70?

2 Answers
Apr 16, 2015

I think this is pretty clear:

For all #c# satisfying these:
#sqrt(60) < c < sqrt(70) and -sqrt(70) > c > -sqrt(60)#
#c^2# will be between #60# and #70#.

Since there is no natural number between #sqrt(60)# and #sqrt(70)# there won't be any #c in NN#

Jul 10, 2015

Answer:

#ceil(sqrt(60))^2 = floor(sqrt(70))^2 = 8^2 = 64#

Explanation:

#7 < sqrt(60) < 8 < sqrt(70) < 9#

So #ceil(sqrt(60)) = 8 = floor(sqrt(70))#

and #60 < 8^2 = 64 < 70#

#64# is the only perfect square integer between #60# and #70#, but there are an infinite number of square rational numbers between #60# and #70#.

In fact for any integer #n >= 22#, we find

#64 < (64(n+1)^2) / n^2 < 70#