
#1
Jan708, 12:07 AM

P: 231

Is a perfect square only "perfect" if the root is a whole number? Or does the term just dictate that decimals must eventually terminate?
For instance: [tex]\displaystyle{\sqrt{59.29} = 7.7}[/tex] 



#2
Jan708, 12:30 AM

P: 1,757

http://en.wikipedia.org/wiki/Perfect_square
Thus a perfect square always has a square root that has no decimal expansion. 



#3
Jan708, 07:00 AM

Math
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PF Gold
P: 38,882

A number always has a "decimal expansion". Integers just happen to have all 0s after the decimal point! A "perfect square" is the square of an integer.
52.29 is NOT a perfect square. 



#4
Jan708, 11:30 AM

P: 688

"Perfect" square?
Though 5229 is a perfect square. If you have a rational number as the result of a root, then the thing inside the root is the square of a rational (in this case, (77/10)^2 = 5229/100), so there are two perfect squares involved somewhere. Other than these musings, all the replies above are, of course, right.




#5
Jan708, 11:33 AM

Math
Emeritus
Sci Advisor
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PF Gold
P: 38,882





#6
Jan708, 11:34 AM

P: 688

Arrgh! Sorry, I just magnified your typo. :D




#7
Jan708, 01:40 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

Ooops! The original post does say "59.29", not "52.29" so you were right and I was wrong. I hate when that happens!



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