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# Another one - OG Quant DS 116

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Another one - OG Quant DS 116 [#permalink]

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26 Oct 2006, 05:24
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Question: If x is a positive integer, is sqrt{x} an integer?

(1) sqrt{4x} is an integer
(2) sqrt{3x} is not an integer

I tried using a purely algebraic method to prove/disprove this. And I so far have been able to. The OG answer is rather intuitive, but not mechanical.

Here's my explaination:

Since sqrt{4x} is an integer, I can assume that sqrt{4x}=C , where C is an integer.
- sqprt{4x}=C
- sqrt{4}.sqrt{x}=C
- 2.sqrt{x}=C
- sqrt{x} = (1/2)(C)

Since C is an integer, and not necessarily a multiple of 2, it is not definitive that sqrt{x} is an integer, as we are unable to mathematically prove that C/2 is an integer.

In words, its just as intuitive, we have no proof that the integer that results from the square root of {4x} is a multiple of 2.

Let me give an alternate path to failure:

sqrt{4x} = integer. (no info provided on nature of integer, so i assume for all values)

so let sqrt{4X]=9
- sqrt{4}.sqrt{x}=9
- 2.sqrt{x}=9
- sqrt{x}=9/2 --------->which is not an integer !!!

So how is it that OG can state that "Since sqrt{4x} is an integer, it follows that 4x must be the square of an integer. Clearly 4 is the square of 2. For 4x to be square of an integer as well, x must also be the square."

what say you guys? Is there a loophole in my understanding ?? Or is there a better way to explain this?

Thanks guys,
Dennis

Last edited by dennisquah on 26 Oct 2006, 09:54, edited 1 time in total.
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Re: Another one - OG Quant DS 116 [#permalink]

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26 Oct 2006, 07:17
sqrt (4x) is an integer, that means 2sqrt(x) is an integer.

for 2 sqrt(x) to be an integer, sqrt(x) will need to be either an integer or of the form a.5 (eg 1.5, 2.5 etc)

sqrt( x) can not be of the form a.5, because then x won't be an integer. So sqrt(x) has to be an integer.
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26 Oct 2006, 09:56
AH cripes !!

Ok i think i gotta take a break ... am losing it now ...

thanks mst, its all good now.

Cheers
26 Oct 2006, 09:56
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