Jenny multiplies the square root of her favorite positive integer by \(\sqrt{2}\). Her product is an integer. a) Name three numbers that could be Jenny's favorite positive integer, and explain why each could possibly be Jenny's favorite integer. b) Suppose Jenny instead divided the square root of her favorite positive integer by \(\sqrt{2}\). Must the resulting quotient be an integer? (For this part, Jenny's favorite positive integer still satisfies the condition that if she multiplies its square root by \(\sqrt{2}\), then she will get an integer. We're asking about what will happen if she divided by \(\sqrt{2}\) instead.)

Guest Jul 19, 2017

#1**+2 **

----*edit*-----

After thinking about this some more...I think this is a much better and simpler way to put it...

Jenny's number = 2 * n^{2} , where n is an integer.

...I guess I'll leave the old answer there... even though I REALLY don't like it anymore haha

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The prime factorization of Jenny's favorite positive integer must contain an odd number of 2's and an even number of anything other than 2.

For example, it could be..

2 * 3 * 3 = 18 ,

2 * 2 * 2 * 5 * 5 = 200 ,

2 * 7 * 7 * 3 * 3 = 882 , or just

2 = 2 .

If it satisfies this condition, then the square root of that number can always be written like this..

\(\sqrt2*\sqrt{\text{any positive integer}^2} \\~\\ \sqrt2*\text{any positive integer}\)

And multiplying, or dividing, this by \(\sqrt2\) will result in an integer.

hectictar Jul 19, 2017

#1**+2 **

Best Answer

----*edit*-----

After thinking about this some more...I think this is a much better and simpler way to put it...

Jenny's number = 2 * n^{2} , where n is an integer.

...I guess I'll leave the old answer there... even though I REALLY don't like it anymore haha

---------------

The prime factorization of Jenny's favorite positive integer must contain an odd number of 2's and an even number of anything other than 2.

For example, it could be..

2 * 3 * 3 = 18 ,

2 * 2 * 2 * 5 * 5 = 200 ,

2 * 7 * 7 * 3 * 3 = 882 , or just

2 = 2 .

If it satisfies this condition, then the square root of that number can always be written like this..

\(\sqrt2*\sqrt{\text{any positive integer}^2} \\~\\ \sqrt2*\text{any positive integer}\)

And multiplying, or dividing, this by \(\sqrt2\) will result in an integer.

hectictar Jul 19, 2017