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

Question

I'm trying to show that $\sqrt{ab} = \sqrt{a} \sqrt{b}$ where $a, b > 0$

For contradiction assume $$\sqrt{ab} \not= \sqrt{a} \sqrt{b}$$

$$ \sqrt{ab} - \sqrt{a} \sqrt{b} \not=0 $$

$$ \left(\sqrt{ab} + \sqrt{a} \sqrt{b}\right) \left(\sqrt{ab} - \sqrt{a} \sqrt{b}\right) \not=0 $$

$$ ab - ab \not=0 $$

$$ ab \not= ab $$

If this logic is incomplete/ wrong, could you post a link of the correct proof?

I assume you're referring to the square root of a non-negative number. This only works when $a$ and $b$ are non-zero (positive). However, you can handle the zero case separately. — Ben Grossmann, Nov 17 '14 at 06:22

Suzu Hirose · Answer 1 · 2014-11-17T06:34:42.190

1

The proof you have assumes that $\sqrt{ab}+\sqrt{a}\sqrt{b}\neq 0$ without explicitly stating/proving that. But if you want to work from that contradiction, squaring both sides of the (in)equation $$ \sqrt{ab} \neq \sqrt{a} \sqrt{b} $$ and using commutativity of multiplication would do it, so the reason for that whole business isn't really clear.

edited Nov 17 '14 at 06:34

answered Nov 17 '14 at 06:26

Suzu Hirose

11,660

Noting lab's comment above, this assumes that $\sqrt{a^2} = a$ whenever $\sqrt a$ is defined. – Ben Grossmann Nov 17 '14 at 06:31
@Omnomnomnom this is true, I suppose my answer should have been a comment on the method used by OP, which seems to make a mountain out of a molehill. – Suzu Hirose Nov 17 '14 at 06:33

score 1 · Accepted Answer · answered Nov 17 '14 at 06:41

You can prove it only if you know the following fact:

If $a\geq 0$ then there exists exactly one real number $\xi \geq 0$ (called the square root of $a$) such that $\xi^2=a$

If $\xi\geq 0, \eta\geq 0$ such that $\xi^2=a, \eta^2=b$ then $$ (\xi \eta)^2= \xi^2\eta^2=ab $$ and since you know that $\xi \eta \geq 0$ this is (!) the square root of $ab$

score 1 · Answer 3 · answered Nov 19 '14 at 15:38

This is partly a comment on the answer given by Blah, but also an answer in its own right. The OP's proof is basically correct; it lacks only the explicit statement that $\sqrt{ab}+\sqrt a\sqrt b\not=0$ if $a,b\gt0$. Moreover the OP's proof essentially contains a proof of the fact Blah says you need to know.

Let me re-express what's being asked for. Suppose $p$ is a positive number such that $p^2=ab$. Likewise suppose $r$ and $s$ are positive numbers such that $r^2=a$ and $s^2=b$. We wish to prove $p=rs$. The OP's proof starts by assuming $p\not=rs$, which means $p-rs\not=0$. Multiplying this by the positive (hence non-zero) quantity $p+rs$, we get

$$(p+rs)(p-rs)=p^2-(rs)^2\not=0$$

But

$$p^2-(rs)^2=p^2-r^2s^2=ab-ab=0$$

which is a contradiction. Hence we must have $p=rs$. (In particular, by setting $b=1$ and taking $s=1$ to be "a" square root of $b$, we find that $p=r$ if $p^2$ and $r^2$ both equal $a$.)

One final remark (mostly to the OP). Note that instead of a string of formulas, I wrote things out in more or less complete sentences, making explicit the logical connections. I would strongly encourage the OP to do so as well, especially when first learning to write proofs (which I assume is the case here).

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

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