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Perfect square?

by Holocene
Tags: perfect, square
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Holocene
#1
Jan7-08, 12:07 AM
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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]
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rocomath
#2
Jan7-08, 12:30 AM
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http://en.wikipedia.org/wiki/Perfect_square

Thus a perfect square always has a square root that has no decimal expansion.
HallsofIvy
#3
Jan7-08, 07:00 AM
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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.

dodo
#4
Jan7-08, 11:30 AM
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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.
HallsofIvy
#5
Jan7-08, 11:33 AM
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Quote Quote by Dodo View Post
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.
My calculator is under the impression that 772= 5929, not 5229.
dodo
#6
Jan7-08, 11:34 AM
P: 688
Arrgh! Sorry, I just magnified your typo. :D
HallsofIvy
#7
Jan7-08, 01:40 PM
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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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