# Thread: Is (-1) equal to (-1)^2/2 ?

1. ## Is (-1) equal to (-1)^2/2 ?

I want to ask whether (-1)^1 can be written as (-1)^2/2
and..

can (-1)^2/2 be written as [ under-root (-1)^2 ] since (-1)^2/2 = (-1)^2*1/2 and 1/2 power is under-root..

Please provide your answers keeping in view the limitations of the laws of Exponentations

2. You must understand that $\sqrt{x^2}=|x|$ for all real $x$.

3. i know that, im trying to figure out the limitations of Exponential laws.

4. You have to be careful with branch cut discontinuities. For example, you can't do this:

$1=\sqrt{1\cdot 1}=\sqrt{-1\cdot-1}=\sqrt{-1}\cdot\sqrt{-1}=i^{2}=-1.$

The problem happens right here:

$\sqrt{-1\cdot-1}=\sqrt{-1}\cdot\sqrt{-1}.$

That's not allowed. As I understand it, the square root function has a branch cut discontinuity on the negative real axis. In doing what I have just outlined, you're basically trying to come at the same point on the negative real axis from two different directions. And that doesn't work, because you're on two different sheets.

So, to answer your question, you can't do this:

$\left(\left(-1\right)^{2}\right)^{\tfrac{1}{2}}=\left(\left(-1\right)^{\tfrac{1}{2}}\right)^{2}.$

Is that what you were after?

5. Originally Posted by khamaar
I want to ask whether (-1)^1 can be written as (-1)^2/2
(-1)^2/2 = 1/2
(-1)^(2/2) = -1

AND: (-1)^1 = -1^1 (brackets not required)

6. Originally Posted by Wilmer
(-1)^(2/2) = -1
this is wrong, see Plato's post.

7. I think wilmer's making an observation about the lack of parentheses, and order of operations. If the caret operator takes precedence, then you evaluate the square first, producing +1, and then you divide by 2.

8. i made a mistake on the quote, it's fixed now.

9. Jolly good.

10. > (-1)^(2/2) = -1
Originally Posted by Krizalid
this is wrong, see Plato's post.
Huh?

11. It matters what order you do things, and how you interpret the expression. You could have

a. $(-1)^{2/2}\to ((-1)^{2})^{1/2}=1^{1/2}=1,$ or
b. $(-1)^{2/2}\to ((-1)^{1/2})^{2}=(i)^{2}=-1,$ or
c. $(-1)^{2/2}\to (-1)^{1}=-1.$

12. But not if the 2/2 is bracketed, surely; (2/2) is 1 and 1 only.

13. So you're distinguishing between $(-1)^{2/2}$ and $(-1)^{(2/2)}?$

14. As I said in my 1st post:
(-1)^2/2 = 1/2
(-1)^(2/2) = -1

1st one means (-1)^2 divided by 2
2nd one means (-1)^(2 divided by 2)

15. I would agree that (-1)^2/2=((-1)^{2})/2=1/2, since the exponential operator takes precedence over division. Both my calculator and Mathematica evaluate the expression the way you have described. The expression (-1)^2/2 isn't very good writing, I would argue. If this is what the OPer meant, it would have been better to write ((-1)^{2})/2.

I would also agree that (-1)^(2/2)=-1, since the parentheses elevate the precedence of the division over the exponentiation. However, you still run into subtleties, because very few people would even write (-1)^(2/2) unless they meant either ((-1)^2)^(1/2) or ((-1)^(1/2))^2. In that case, you run into the difficulties I outlined in post # 4.

In the context of the OP, especially the title of the current thread, I think you can infer that both the numerator and denominator of the expression 2/2 are in the exponent, thus ruling out the result of 1/2. I would agree, however, that technically speaking, the OPer wrote (-1)^2/2, which equals 1/2. The OPer wasn't as precise as he should have been.

The bottom line here is that it definitely pays to be very precise when raising negative numbers to various powers.

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