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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?

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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.

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You simply need to know that the square root function distributes over multiplication, so $$\sqrt{abc\cdots} = \sqrt{a}\sqrt{b}\sqrt{c} \cdots$$

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    $\begingroup$ I would warn against this, as it can produce contradictions. $1=\sqrt{1}=\sqrt{(-1)\cdot (-1)}\neq \sqrt{(-1)}\sqrt{(-1)}=-1$. You need additional constraints that each term is non-negative. $\endgroup$ – JMoravitz Nov 19 '15 at 16:27
  • $\begingroup$ Well isn't that just because you decided to take the negative roots in step 3? Wouldn't Ian Miller's method have the same flaw if he used -4 in step 2? $\endgroup$ – Jwan622 Nov 19 '15 at 16:31
  • $\begingroup$ It was my assumption that since the question was tagged (algebra-precalculus), that complex numbers and negative square roots would not be used. $\endgroup$ – MathMajor Nov 20 '15 at 5:27
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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.

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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 $\sqrt{ab} = \sqrt{a}\sqrt{b}$.

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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}

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