Question

SOME ONE HELP ME I REALLY NEED A SOLUTION FOR THIS PROBLEM WITH A CLEAR EXPLANATION

Answer 1

I'm going to answer this question using logic.  Let x be Jenny's favorite number.  We are going to square root this number, $\sqrt{x}$ and then multiply it by $\sqrt{2}$.

This product needs to be an integer.  How is that obtainable? To be an integer, we need to get rid of the nasty $\sqrt{2}$ part.  The only way I can think of to get rid of it, is to multiply it by $\sqrt{2}$, because $\sqrt{2} *\sqrt{2} = 2$.  Thus we have some conditions we need to fulfill when choosing Jenny's favorite number.

When we take the square root of Jenny's favorite number, x, it must contain a perfect square and a 2 in its prime factorization.  For example, 8 works because 8 = 2 x 2 x 2, or 2² x 2.  

You notice 8 is made up of a perfect square multiplied by 2.  So when we take the square root of 8, we get:

$\sqrt{8} = \sqrt{2^{2}*2 } = \sqrt{ 2^{2} }*\sqrt{2} = \sqrt{4} *\sqrt{2} = 2*\sqrt{2}$

So 8 is the same thing as $2\sqrt{2}$

So when we multiply this by $\sqrt{2}$, we will get an integer! So as long as Jenny's favorite number consists of a perfect square and two in its prime factorization, we will have an integer!

So possible choices are: 2,8,18.

Why does 18 work? Because $\sqrt{18} = \sqrt{9}* \sqrt{2} = 3\sqrt{2}$

When we multiply this by $\sqrt{2}$, we get 6, which is an integer.

b) Suppose instead of multiplying by $\sqrt{2}$, we divided by $\sqrt{2}$. Is the resulting quotient still an integer?

YES, because we can get rid of the $\sqrt{2}$ part by dividing by $\sqrt{2}$ as well.  This leaves only the "perfect square" part left in our square root, and obviously a perfect square is an integer when we square root it.

I hope that made sense! (⌐■_■)