If $\sqrt{28x}$ is an integer is $\sqrt{7x}$ always an integer?

Question

If $\sqrt{28x}$ is an integer is $\sqrt{7x}$ an integer? I have a book that says no, but I cannot think of an example of the contrary... Not looking for a full proof here just wanting to see a counterexample or logic showing that the book is wrong...

Answer 1

I cannot think of an example of the contrary $\dots$ just wanting to see a counterexample

Would $~x=\dfrac1{28}~$ satisfy your curiosity ?

Answer 2

Note that $\sqrt{28x}=\sqrt{4\times 7x}=2\sqrt{7x}$. Can you see what might happen so that $\sqrt{28x}$ is an integer while $\sqrt{7x}$ is not?

Answer 3

Consider the first part $\sqrt{28 \, x} = \pm n$, $n$ being an integer. Then one finds $$x = \frac{n^{2}}{28}.$$ Now consider the second part $$\sqrt{7 \, x} = \sqrt{ \frac{7 \, n^{2}}{28} } = \sqrt{\frac{n^{2}}{4}} = \pm \frac{n}{2}.$$ This shows that if $n$ is not an even value then the value of $\sqrt{7 \, x}$ is not an integer.

Answer 4

Suppose x is an integer.

Then 28 * x is always even, since 28 * x = 2 * 14x.

The square root of an even number, is never an odd number. (Check that odd * odd = odd)

So if you take examples where x is an integer, sqrt(28x) is never odd, and therefore there exists no counterexample.

So x must be a fraction to find a counterexample, (like 1/28 or 9/28).