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If y is a positive integer is root(y) an integer?
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26 Jan 2011, 05:37

4

6

rxs0005 wrote:

If y is a positive integer is root (y) an integer

S1 root( 4* y) is not an integer

S2 root( 5* y) is an integer

I do not agree with OA

If \(y\) is a positive integer is \(\sqrt{y}\) an integer?

Note that as \(y\) is a positive integer then \(\sqrt{y}\) is either a positive integer or an irrational number. Also note that the question basically asks whether \(y\) is a perfect square.

(1) \(\sqrt{4*y}\) is not an integer --> \(\sqrt{4*y}=2*\sqrt{y}\neq{integer}\) --> \(\sqrt{y}\neq{integer}\). Sufficient.

(2) \(\sqrt{5*y}\) is an integer --> \(y\) cannot be a prefect square because if it is, for example if \(y=x^2\) for some positive integer \(x\) then \(\sqrt{5*y}=\sqrt{5*x^2}=x\sqrt{5}\neq{integer}\). Sufficient.

If \(y\) is a positive integer is \(\sqrt{y}\) an integer?

1) Means that we have \(\sqrt{4*y^{odd}}\) --> \(\sqrt{y^{odd}} \neq {integer}\) for \(y>1\) If \(y = 1\)then statement is not true.

2) \(\sqrt{5y}\) means that y has a \(5^{odd}\) combination in its prime box, which means we have \(\sqrt{5^{odd}*y}\), where y is some integer and \(\sqrt{5^{odd}*y} \neq {integer}\)

Re: If y is a positive integer is root(y) an integer?
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20 Feb 2018, 22:16

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