# SImpe question about square root function

1. May 6, 2010

Is it a right mathematical operation (a*b)^0.5 = a^0.5 *b^0.5 when a and b either both are negative or one of them is negative.

2. May 6, 2010

No..

3. May 6, 2010

### Staff: Mentor

If a >= 0 and b >=0, $$\sqrt{ab} = \sqrt{a}\sqrt{b}$$

4. May 6, 2010

### Xitami

If a <= 0 and b <=0, $$\sqrt{ab} = -\sqrt{a}\sqrt{b}$$

5. May 6, 2010

### Staff: Mentor

No, that's not true at all.
$$\sqrt{(-2)(-8)} = \sqrt{16} = 4 = \sqrt{2}\sqrt{8}$$

6. May 7, 2010

### Mentallic

And finally for the last case, a<0 b$\geq$0

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

7. May 7, 2010

### Char. Limit

Actually, if only one of a and b are positive, we don't know if the answer is multiplied by i or by -i, I believe. That non-property is used to show the fallacy in a certain 1=2 argument.

8. May 7, 2010

### Martin Rattigan

Has this thread something to do with the recent release of Alice in Wonderland?

Mark44: $\sqrt{(-2)(-8)}=\sqrt{2}\sqrt{8}$ (true) $=(\sqrt{-2}\sqrt{-8})/i^2=-\sqrt{-2}\sqrt{-8}$ (as in Xitami's formula)

Or setting $b=1,\sqrt{a}\neq 0$:

Mentallic: $1=i$

Char Limit: $1=\pm i$

9. May 7, 2010

### Mentallic

Of course :yuck: I felt I can venture into this forbidden territory of complex numbers without any issues... I was wrong.

I'll have some fun here to try and explain where Martin's second expression could've arisen from.

a>0, b>0

$$\sqrt{ab}=\sqrt{(-a)(-b)}=\sqrt{-a}\sqrt{-b}$$

$$=\frac{i^2}{i^2}\sqrt{-a}\sqrt{-b}$$

$$=\frac{-1}{i^2}\left( i\sqrt{a}\right)\left( i\sqrt{b}\right)$$

$$=\frac{-1}{i^2}(i^2)\sqrt{a}\sqrt{b}$$

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

as required... I suppose :tongue2:

Disclaimer: The first line, while mathematically sound, isn't allowed to be used in this square root function.

I'm not sure what you're getting at here. We were assuming a<0 so $\sqrt{a}\neq 0$ is a given.

10. May 8, 2010

### Martin Rattigan

If by second expression you mean $\sqrt{2}\sqrt{8}$ it was copied from Mark44's post (as was the first expression $\sqrt{(-2)(-8)}$, but for some reason itex doesn't extend the square root sign - it should have read $\sqrt{(-2)(-8)}$).

The third expression comes from $\sqrt{-2}=i\sqrt{2},\sqrt{-8}=i\sqrt{8}$ and substitution in the second expression.

The fourth expression comes from replacing $i^2$ by $-1$ and looks very much like what Xitami said and Mark44's post is claiming to be wrong.

What I'm getting at is you're claiming $\sqrt{ab}=i\sqrt{a}\sqrt{b}$ when $a<0,b\geq 0$. Substituing $b=1$ (which is $\geq 0$) gives, for $a<0$

$\sqrt{a.1}=i\sqrt{a}\sqrt{1}$

or $\sqrt{a}=i\sqrt{a}$

and after division by $\sqrt{a}$ which is possible because, as you point out $a<0$ therefore $\sqrt{a}\neq 0$

$1=i$

11. May 8, 2010

### Staff: Mentor

The "some reason" is that you switched from [ tex] tags to [ itex] tags.

12. May 8, 2010

### Mentallic

but for a<0, $$\sqrt{a}=i\sqrt{-a}$$

13. May 8, 2010

### Staff: Mentor

This isn't right. Neither equality is valid under the given conditions.

Since the possiblility of b being zero doesn't add anything useful, let me assume that b > 0 for the rest of the discussion, and that a < 0 as before.

$$\sqrt{ab} \neq \sqrt{-a}\sqrt{b}$$
In the expression on the left, ab < 0 so its square root is imaginary. In the expression on the right, both radicands are positive, so each square root is real and positive, making the product real and positive.

$$\sqrt{-a}\sqrt{b} \neq i\sqrt{a}\sqrt{b}$$
For reasons already given, the left side above is real and positive. The right side above is also real, but negative. The reason for this is that
$$i\sqrt{a}\sqrt{b} = i^2 \sqrt{-a}\sqrt{b}= -\sqrt{-a}\sqrt{b}$$

It might be helpful to show this with numbers instead of variables, replacing a with -2 and b with 8.
$$i\sqrt{-2}\sqrt{8} = i^2 \sqrt{2}\sqrt{8}= -\sqrt{2}\sqrt{8}$$

I think we have succeeded in going well beyond what the OP asked, and might have further confused him/her with a lot of extraneous information. I hope we are done here.

14. May 8, 2010

### Martin Rattigan

I might very well agree with that. You can divide by $i\sqrt{-a}$ and prove $1=i$ that way if you like.

But if you want to leave the rabbit hole you'll need to change your original equation.

15. May 8, 2010

### Martin Rattigan

I thought that was more or less what I said.

I don't think there was any confusion up to and including Xitami's first (correct) post, but it went downhill from there. All that was required at that point was a correct formula for $\sqrt{ab}$ for real $a,b$ of opposite sign.

16. May 8, 2010

### Char. Limit

I think that what we have here is notation error...

For positive real a and b...

$$\sqrt{(-a)(b)}=\sqrt{-a}\sqrt{b}=i\sqrt{ab}$$

Of course, a similar argument cannot be done for -a and -b because the square root function can't be "split" with two negative radicands. No, it can't.

17. May 8, 2010

### Martin Rattigan

If you want to avoid confusion it's probably not a good idea to change the meaning of $a$ and $b$ from those in the original question (repeated under).

For real $a$ and $b$, $(ab)^{0.5}=a^{0.5}b^{0.5}$ unless both $a$ and $b$ are negative, in which case $(ab)^{0.5}=-a^{0.5}b^{0.5}$.

18. May 8, 2010

### Char. Limit

No, if a and b are both negative, than the result is invalid. And I already proved the result if either a or b is negative.

So if both a and b are negative, the result is invalid (actually multivalued).

If either a or b is negative, my result follows above.

19. May 8, 2010

### Martin Rattigan

From the original question:

What I was suggesting is that this is likely to lead to confusion.

By "the result" here, do you mean the result given in the original question or the result I gave in answer?

If the former then I would agree with you. That is the import of the minus sign in the last expression in my answer.

If the latter, can you give a specific pair of real values for a and b where you think the answer is incorrect?

With the usual definition of $z^w$ for complex $z,w$ viz. $z^w=e^{w Log(z)}$ the Log is taken to be the principal log, so the function is single valued, so I think your remarks about the result being multivalued are mistaken if by "the result" in this case you mean one of (ab)0.5, a0.5 or b0.5.

Last edited: May 8, 2010
20. May 8, 2010

### Char. Limit

$$\sqrt{(-2)(-8)}=\sqrt{16}=4$$
$$\sqrt{(-2)(-8)}=\sqrt{16}=4 , -4$$