# 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

I mean your result.

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

I only used multiplication and the square root of a positive number. If a and b are negative, the answer is a real, positive number, and your result is invalid. Assuming we don't take the principal root (always possible)...

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

Which is multivalued. So if the principal root gives you a real, positive number, and the general root gives you multiple values, you won't receive a singlevalued negative real number from your operations.

(Note: I did post the wrong Wikipedia section... They are talking about complex numbers, not negative numbers. Sorry for that mixup.)

21. May 8, 2010

### Martin Rattigan

This is the same example that Mark44 previously used. As I said regarding that example:

the result agrees with Xitami's formula, which matches my answer for $a,b$ both negative. So that is not a counter-example.

Both notations $z^{0.5}$ and $\sqrt{z}$ are defined as the principal root (precisely to avoid multiple values). It's always possible for you not to take the principal root, but in that case you're not talking about the problem.

So I think the one line answer I included a few posts back was all that was really required for this thread.

PS. Neither OP nor I said anything about the expressions on either side being real. I would say it was clear that OP was well aware that they would not be.

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

### Char. Limit

You're assuming that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if both a and b are negative. This isn't true. After all, here's another equation using your same logic (it's your logic, right?)

$$-1=i^2=i \times i=\sqrt{-1}\sqrt{-1}=\sqrt{-1 \times -1}=\sqrt{1}=1$$

And that's why $\sqrt{ab}$ does not equal $\sqrt{a}\sqrt{b}$ for negative a and b. Because if it did, -1 would equal 1.

Edit: Here's another place that shows the same logic, but with division. The relevant material is near the bottom.

http://www.math.toronto.edu/mathnet/falseProofs/guess11.html

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

### Martin Rattigan

Absolutely not. If you read what I wrote a little more carefully it says that $\sqrt{ab}=-\sqrt{a}\sqrt{b}$ if both $a$ and $b$ are negative. This gives you:

$-1=i^2=i\times i=\sqrt{-1}\sqrt{-1}=-\sqrt{-1\times -1}=-\sqrt{1}=-1$

No surprise there.

24. May 8, 2010

### Char. Limit

Ok, then tell me why every single calculator I've used on this iPod (six at last count) gave me real and positive numbers for every square root of the product of two negative numbers that I put in? According to you, those numbers should be negative. Also, shouldn't math be consistent? So why can't I multiply the numbers in the radicand to produce a positive product, when I can with ANY other number under that root sign? In other words, why does multiplying the numbers in the radicand and then taking the square root (all allowed formal operations) produce a positive number, but your method produces a negative number?

25. May 9, 2010

### Mentallic

How did you come up with this?

The way I see it, and why I agree with Char.Limit is that $$\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}$$ from $$\sqrt{a}\sqrt{b}=\sqrt{ab}$$

The problem arises from the order the process is undertaken. Repeating what wikipedia said that Char.Limit posted,
$$1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i^2=-1$$

or

$$1=\sqrt{1}=\sqrt{(-1)(-1)}$$ (now at this point we multiply under the brackets first, to go back to the previous equality, in which case we will get 1 again).