Why does $\sqrt{x} / y =\sqrt{x/y/y}$? Sorry for the awkward title, hard to to sum a mathematical problem with words alone.
Having said that, I recently learned that the root of any value, $x$, and then that over value $y$, is identical to the root of $x/y/y$, as in $\sqrt{x/y/y}$.
For e.g, $\sqrt{35} / 7 = \sqrt{5/7}$, and I cannot logically deduce why. So far, I know that:
$$
35 = 7 \times 5 \\
\frac{\sqrt{35}}{7} = \frac{\sqrt{7} \times \sqrt{5}}{7}
$$
And somehow this being equal to:
√(5/7)

So it seems that we divided the numerator, √35 by √5 to give √7. However, 7/√7 is not √7! Yet, apparently it is?
So, a fraction is of course equal to itself if scaled up or down by the same amount, in that 10/2 is the same as 5/1, or 5. So scaling down √35 by √5 makes sense, as it only leaves √7, but it seems that the same cannot be said for the denominator, 7. 7/√7 = √7!? Yet, the two expressions really are the same, √35/7 and √(5/7).
Thanks for any help in advance.
P.S This works for any values, so it's no coincidence.
 A: The typesetting is difficult to understand, perhaps this is what you're looking far :
$$
\dfrac{\sqrt{\vphantom{b} x}}{y}=\dfrac{\sqrt{\vphantom{b} x}}{\sqrt{y^2}}=\sqrt{\dfrac{\vphantom{b} x}{y^2}}.
$$
So for your example with $x=35$ and $y=7$ we get :
$$
\dfrac{\sqrt{35}}{7}=\sqrt{\dfrac{\color{royalblue}{35}}{7^2}}=\sqrt{\dfrac{\color{royalblue}{7\cdot5}}{7\cdot7}}=\sqrt{\dfrac57}.
$$
A: $$\frac{\sqrt{35}}{7} = \frac{\sqrt5 \cdot \sqrt 7}{7} = \sqrt{5}\cdot \frac{\sqrt 7}{(\sqrt {7})^2} =\frac{\sqrt 5}{\sqrt 7} = \sqrt{\frac 57}$$
A: I expect you are aware that $$\frac {a\cdot b}b = a $$ because we can cancel the $b$ in the numerator and denominator.
The piece you seem to be missing is that $$ \frac a {\sqrt a} =\frac {\sqrt a\sqrt a}{\sqrt a} = \sqrt a $$ where in the last step we canceled $\sqrt a $ from the numerator and denominator. 
Or put another way, you know that if $xy=z$ then $x=\frac zy $. Take $x=\sqrt a, y=\sqrt a, z=a$. Then $$\sqrt a\sqrt a= a ,$$ so dividing by $\sqrt a $ on both sides we get $\sqrt a =\frac a {\sqrt a}$.
