# The square root

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!

pwsnafu
Hello everyone,

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

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.

but I can't figure out why there's an apparent contradiction...

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.

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• 1 person
arildno
Homework Helper
Gold Member
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Do you mean: $a: \sqrt{(-1)^{2}} b: (\sqrt{-1})^{2}$

Curious3141
Homework Helper
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!

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.

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arildno
Homework Helper
Gold Member
Dearly Missed
remember that:
$$\sqrt{x^{2}}=|x|$$
not x.

• 1 person
Mark44
Mentor
What is the square root of a square of a negative number equal to? For example: $\sqrt{-1}^{2}$
Based on what VC described rather than the notation used, this is the problem: ##\sqrt{(-1)^2}##. This expression simplifies to 1.

• 1 person
Mark44
Mentor
Do you mean: $a: \sqrt{(-1)^{2}} b: (\sqrt{-1})^{2}$
Actually, the notation used was clear and unambiguous, but did not agree with his description.

• 1 person
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: