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OG integer with square root problem

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OG integer with square root problem

by Redhorsep » Sat Nov 26, 2011 9:08 am
Hi,

I have a problem understanding one statement sufficient, please explain, thanks!

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If x is a positive integer, is square root of (x) an interger?

1. square root of (4x) is integer
2. square root of (3x) is not an integer






spoiler: so statement 1 is sufficient, but I don't know why so: Because if you simplify it, isn't it equal to 2 time square root of (x), which is an integer, but can't I argue that square root of (x) can equal to 1.5, but still make the product of 2 times 1.5 an integer?
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by shankar.ashwin » Sat Nov 26, 2011 10:33 am
Interesting question you pose..

But note that only numbers of the form XXX.5 when multiplied with 2 will give you an integer (i.e have '0.5' after the decimal point multiplied by 2 would end in .0)

And in such case, the number (XXX.5) when squared will always result in a number of the type (XXXX.25)

If you take 1.5 - 1.5^2 = 2.25.

But we are given X is an integer in the question stem. SO any number of the type XXXX.25 is not an integer. Hence if square root of (4x) is integer, we can be certain to deduce 'x' will be a perfect square and hence Sqrt(x) will be an integer.
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by tpr-becky » Sat Nov 26, 2011 2:18 pm
The problem with your reasoning about the square root of x able to be 1.5 is that the stem defines x as an integer 1.5(1.5) = 2.25 (not an integer)- which integer would yield a square root that ends in a .5? The decimal has to be .5 for 2(sqrtx) to be an integer. When you square any number that ends in a 0.5 you will always get a number with 0.25 as an end, which is not an integer.
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by knight247 » Sat Nov 26, 2011 3:02 pm
@Redhorse

https://en.wikipedia.org/wiki/Irrational ... uare_roots

Open the above link and read the first couple of lines. The square root of a non-perfect square is ALWAYS irrational. So having a sq root of 1.5 is impossible.
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