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That's the question.

I try to explain this to myself I can't find any good resources.

Note: I have researched for about 15 mins and yet haven't found the answer. I do apologize for any inconveniences.

Bahar
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Square root of a number $a$ is defined as "number, which when multiplied by itself, gives $a$", also $\left(\sqrt{a}\right)^2 = a$.

We can clearly see that if $\sqrt{a}\times\sqrt{a} = a$ and $\sqrt{b}\times\sqrt{b} = b$, then by multiplying both sides we get $\sqrt{a}\times\sqrt{a}\times\sqrt{b}\times\sqrt{b} = ab$.

But we already know, that $\sqrt{ab}\times\sqrt{ab}=ab$, from the definition.

And now it is obvious - $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$. Similarly for other roots.

Puding
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    Just a small remark: The squareroot ist defined as "positive number [...]" –  Aug 17 '17 at 19:45
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    @PaulK You're correct in that a nonzero number generally has two square roots, but we define the notation $\sqrt{x}$ to refer to the positive square root of $x$ when $x$ is a positive real number. It would be incorrect to say that $\sqrt{9} = \pm 3$. – Michael L. Aug 17 '17 at 20:02
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The definition of $\sqrt{n}$ is "the number which gives $n$ when you square it. So notice that $(\sqrt{a}\sqrt{b})^2 = \sqrt{a}^2\sqrt{b}^2 = ab.$ Since you get $ab$ when you square $\sqrt{a}\sqrt{b}$, it must be that $\sqrt{a}\sqrt{b}$ is the square root of $ab$, which is to say $\sqrt{a}\sqrt{b} = \sqrt{ab}.$

Who's buried in Grant's tomb?

B. Goddard
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    Just a small remark: The squareroot ist defined as "the positive number which gives [...]" –  Aug 17 '17 at 19:46