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I am currently learning about properties of radicals in a high school math class, and I am curious as to why this property holds true?

$\sqrt[n]{x} \cdot \sqrt[n]{y} = \sqrt[n]{x \cdot y}$

Could anyone intuitively explain this to me?

Nick
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  • This is the general property of exponentiantion: $$(a \cdot b)^n=a^n \cdot b^n$$ Now apply it you your expression: you will just get: $$x \cdot y=x \cdot y$$ because by definition $$(\sqrt[n]{a})^n=a$$ – Yuriy S Mar 16 '16 at 20:24
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    Important note: if $n$ is even then this is only true for $x,y>0$. – Wojowu Mar 16 '16 at 20:26
  • Yuriy S I understand that you can do this: 
        $\sqrt[n]{x} = x^{\frac{1}{n}}$

    Is this what you are telling me to apply?

     – Nick Mar 16 '16 at 20:26
  • Yes, and see @Wojowu comment. Of course I meant, $x,y>0$ for even $n$ – Yuriy S Mar 16 '16 at 20:30
  • Yuriy S Ah so basically I can just make that expression:
    $(\sqrt[n]{x} \cdot \sqrt[n]{y})^{n} = (\sqrt[n]{x \cdot y})^{n}$ to yield $x \cdot y = x \cdot y$
    I guess that's pretty intuitive, although I wish there was something more visual.     – Nick Mar 16 '16 at 20:35
  • Wojowu
    Is this because if $n$ was even then the answer for both the radicals in the left expression could be either positive or negative, therefor I would have 4 different possible answers?     – Nick Mar 16 '16 at 20:37

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If $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are defined and real for real $a,b$, this is true.

Note that $(xy)^n = x^ny^n$. So if $x=\sqrt[n]{a}$ and $y=\sqrt[n]{b}$ then $$(xy)^n=\left(\sqrt[n]{a}\sqrt[n]{b}\right)^n = (\sqrt[n]{a})^n(\sqrt[n]{b})^n=ab$$

So $z=\sqrt[n]{a}\sqrt[n]{b}$ satisfies $z^n= ab$. So $z=\sqrt[n]{ab}$ by definition. (You'll need some additional arguments that $z$ is the right sign when $n$ is even, but that is easy.)

Cameron Williams
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Thomas Andrews
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