# The square root

1. Oct 5, 2013

### ViolentCorpse

Hello everyone,

What is the square root of a square of a negative number equal to? For example: $\sqrt{-1}^{2}$

It seems there are two possible ways of doing this, the problem is that I am getting two different answers using these two approaches i.e; We can first take the square of -1 and then take square root of the square of -1. This gives 1 as the answer. On the other hand, if we replace the radical sign by 1/2 as the exponent, we should then just be left with -1 as the answer. The correct answer, I think, is +1, but I can't figure out why there's an apparent contradiction...

Thank you!

2. Oct 5, 2013

### pwsnafu

As written, most mathematicians would (probably*) parse that as $(\sqrt{-1})^2$. The square root evaluates to either i or -i depending on where your branch cut is, but they both square to -1.

*: It's ambiguous though.

With complex numbers your assumptions of exponentiation derived from the reals breaks down. In this case you are assuming $(a^{b})^c = (a^c)^b$, and this is simply not true. Further $(a^b)^c \neq a^{bc}$ in general.

Last edited: Oct 5, 2013
3. Oct 5, 2013

### arildno

Do you mean: $a: \sqrt{(-1)^{2}} b: (\sqrt{-1})^{2}$

4. Oct 5, 2013

### Curious3141

Your notation is ambiguous, as others pointed out, but your verbal explanation of the problem is clear. You want to evaluate $\sqrt{(-1)^2}$ or ${((-1)^2)}^\frac{1}{2}$.

In essence, they're the same thing. Both exponentiation operations are of equal precedence, so you work from inside out. You're left with $\sqrt{1}$ or ${1}^\frac{1}{2}$. Those are just different ways of writing the same thing. The usual implication of exponentiation to a fractional power is the extraction of the principal root, which is the unique positive value, if it exists. In this case, that's simply $1$.

Doing it the second way, you might be tempted to use the law of exponents $(a^b)^c = a^{bc}$ to cancel the $2$ and the $\frac{1}{2}$, but this would be wrong. That law only applies (without qualification) when $a$ is non-negative.

5. Oct 5, 2013

### arildno

remember that:
$$\sqrt{x^{2}}=|x|$$
not x.

6. Oct 5, 2013

### Staff: Mentor

Based on what VC described rather than the notation used, this is the problem: $\sqrt{(-1)^2}$. This expression simplifies to 1.

7. Oct 5, 2013

### Staff: Mentor

Actually, the notation used was clear and unambiguous, but did not agree with his description.

8. Oct 5, 2013

### ViolentCorpse

This is what I meant: $\sqrt{(-1)^2}$.

I'm sorry for the ambiguity guys. I tried putting the parentheses around -1 before, but the code broke down for some reason, so I removed it.

I think the greatest gap in my understanding was that I didn't know that the rule (a$^{b}$)$^{c}$ = a$^{bc}$ couldn't be applied for a<0 (still a bit confused about this point. Isn't (-2$^{3}$)$^{2}$ = (-2)$^{6}$?

I knew that $\sqrt{x}^{2}$=$\left|x\right|$, but it felt like this was inconsistent with the former rule (law of exponents), thanks to my ignorance.

Thank you so much, all of you! I seriously appreciate your help more than you probably know.

Last edited: Oct 5, 2013