Square root math and squaring I'm having some trouble making sense of this.
$\sqrt{\dfrac 12 \operatorname{in.}^2} =
\dfrac{1}{\sqrt 2} \operatorname{in.} =
\dfrac{1}{\sqrt 2} \operatorname{in.} \times \dfrac{\sqrt 2}{\sqrt 2} =
\dfrac{\sqrt 2}{2} \operatorname{in.}$
My question is in regards to the first two steps of the equation.  How does one remember that taking the square root of a multi term requires you to distribute the square root to both terms? Is there a more intuitive explanation to this?
Like I know that:
$\sqrt{4*4}$ is 2*2 but is there a more intuitive way to remember this?
 A: The way I think about it is that square root is no different to the power of one half and powers are multiplication in which the order doesn't matter. E.g.
$$\sqrt{4\times4}=\left(4\times4\right)^{\frac{1}{2}}=4^\frac{1}{2}\times4^\frac{1}{2}=\sqrt{4}\times\sqrt{4}=2\times2$$
Although all that happens in head automatically.
A: You simply need to know that the square root function distributes over multiplication, so $$\sqrt{abc\cdots} = \sqrt{a}\sqrt{b}\sqrt{c} \cdots$$
A: Well, we know that:
$$a=\sqrt a\sqrt a\tag1$$
almost by definition. Also:
$$a=\sqrt{aa}\tag2$$
for positive $a$. (From now on, I'll assume $a$ and $b$ are positive.)
So:
\begin{align}
\sqrt{ab}&=\sqrt{(\sqrt a\sqrt a)(\sqrt b\sqrt b)}\\
&=\sqrt{(\sqrt a)(\sqrt a)(\sqrt b)(\sqrt b)}\\
&=\sqrt{(\sqrt a)(\sqrt b)(\sqrt a)(\sqrt b)}\\
&=\sqrt{(\sqrt a\sqrt b)(\sqrt a\sqrt b)}\\
&=\sqrt a\sqrt b
\end{align}
I used equation $(1)$ for the first step, associativity and commutativity of multiplication for the next three steps, and then equation $(2)$ at the end.
A: Let $a, b \geq 0$.  Each non-negative real number has a unique non-negative square root.  Let $x = \sqrt{a}$; let $y = \sqrt{b}$. Observe that 
\begin{align*}
ab & = x^2y^2\\
   & = x \cdot x \cdot y \cdot y\\
   & = x \cdot y \cdot x \cdot y\\
   & = xy \cdot xy\\ 
   & = (xy)^2\\
   & = (\sqrt{a}\sqrt{b})^2
\end{align*}
Taking square roots yields $\sqrt{ab} = \sqrt{a}\sqrt{b}$.
A: Completely formally, if $x$ and $y$ are nonnegative real numbers and $\alpha$ is a real number, then
$$(xy)^\alpha = x^\alpha \; y^\alpha$$
So
\begin{align}
  \sqrt{\dfrac 12 \operatorname{in.}^2}
  &= \left( \dfrac 12 \operatorname{in.}^2 \right)^{1/2} \\
  &= \left( \dfrac 12 \right)^{1/2} (\operatorname{in.}^2)^{1/2} \\
  &= \sqrt{ \dfrac 12} \; \sqrt{\operatorname{in.}^2} \\
  &= \dfrac{1}{\sqrt 2} \; \operatorname{in.}
\end{align}
