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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

Answer is A.

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