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If d is a positive integer is d^1/2 an intger

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If d is a positive integer is d^1/2 an intger [#permalink] New post 07 Nov 2010, 09:27
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If d is a positive integer is \sqrt{d} an integer?

(1) d is the square of an integer
(2) \sqrt{d} is the square of an integer


[Reveal] Spoiler:
I have rephrased the question to basically state, is d a perfect square?

1) If d is the square of an integer this is saying that, an integer, when multiplied by itself = d --> Sufficient

2) I had a little trouble rephrasing this to be more understandable, can anybody clarify?
[Reveal] Spoiler: OA
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Re: Quant Review 2nd Edition: DS 115 [#permalink] New post 07 Nov 2010, 10:08
2. The sqr(d) is a square of an integer, thus we do know that a square of a ninteger must be an integer- SQR(D) is an integer- sufficient.
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Re: Quant Review 2nd Edition: DS 115 [#permalink] New post 07 Nov 2010, 10:19
I dont know why i cant see this, Would you be able to help visualize with numbers..?
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Re: Quant Review 2nd Edition: DS 115 [#permalink] New post 07 Nov 2010, 11:39
Hi,

Correct me, If I am wrong,

Quote:
If d is a positive integer is \sqrt{d}an integer?


The Integer set contains, all the whole number and the negatives of all the natural number ie -1,-2,0,1,2

As per your statement, If Suppose 3 is a positive integer, \sqrt{3} will also be positive integer, which is not true as sqrt3 is approx 1.73, which by definition doesn't fall in integer set.

If & Only If d is a perfect square, the \sqrt{d} will a integer as in 121 will give +11, -11.

Do let me know, If I am missing something or mis-stated something.

Regards,
Tushar
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Re: Quant Review 2nd Edition: DS 115 [#permalink] New post 07 Nov 2010, 11:44
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jscott319 wrote:
If d is a positive integer is \sqrt{d}an integer?

1) d is the square of an integer
2) \sqrt{d} is the square of an integer


I have rephrased the question to basically state, is d a perfect square?

1) If d is the square of an integer this is saying that, an integer, when multiplied by itself = d --> Sufficient

2) I had a little trouble rephrasing this to be more understandable, can anybody clarify?


Given: d=integer. Question: is \sqrt{d}=integer? or as you correctly rephrased is d a perfect square: is d=integer^2?

(1) d is the square of an integer --> d=integer^2 --> directly gives an answer. Sufficient.
(2) \sqrt{d} is the square of an integer --> \sqrt{d}=integer^2=integer (as square of an integer is an integer). Sufficient.

Answer: D.

Similar questions:
i-cant-understand-how-the-oa-is-101475.html?hilit=irrational
algebra-ds-101464.html

Hope it helps.
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Re: Quant Review 2nd Edition: DS 115 [#permalink] New post 07 Nov 2010, 12:49
Thank you that does help. Not sure why I couldnt process it ha. Tricky wording. Thanks for the link as well +1
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Re: Quant Review 2nd Edition: DS 115 [#permalink] New post 07 Nov 2010, 15:07
Just put two values for positive integer variable D. For ex- D=15 or D=16.
Re: Quant Review 2nd Edition: DS 115   [#permalink] 07 Nov 2010, 15:07
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