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

1. Originally Posted by Ackbeet
The expression (-1)^2/2 isn't very good writing,....
That's EXACTLY my point too, Ack.
I was trying to let the Poster know this (as politely as possible!).

2. Gotcha.

3. I also think that if that Poster had right away been told to look up
"The Order of Operations: PEMDAS"
in order to explain to him that x^a/a means x^a divided by a,
and did he mean x^(a/a)?
then everything would have been smooth...

4. ## The poster has questions for ackbeet!

Thanks A lot for so much replies…But no one clearly understood my question!
I want to say that .. since,
METHOD (i) (-1)2/2 = {(-1)2}1/2 = (1)1/2 = 1 equation 1
METHOD (ii) (-1)2/2 = (-1)1 = -1 [as 2/2=1] equation 2
(ack said the same in post # 11)

From ‘equation 1’ and ‘equation 2’ it has been proved that
-1=1

MY QUESTIONS-
What is wrong with my above statements?
Which method is right?
Will I get a Nobel Prize for this?

NOTE:-
Im 18, so please explain terminologies which are abit complicated.. I have asked an ‘organized’ question this time…Hope it will go smooth……….

Regards
Islamia College Peshawar, Pakistan

5. What's wrong with your statements is essentially what I said in Post # 4. Imagine the unit circle in the complex plane. For the expression

$((-1)^{2})^{1/2}=(1)^{1/2}=1,$

you are approaching the point -1 from one direction (clockwise along the unit circle from below. That is, you start at -1. Squaring pushes you to +1 in the counter-clockwise direction along the unit circle. Then taking the square root pushes you back, clockwise, towards the point -1).

For the expression

$-1=(-1)^{1}=(-1)^{(2/2)}=((-1)^{1/2})^{2}=i^{2}=-1,$

you start at the point -1, and then the square root takes you to $i$ by going clockwise on the unit circle, and then squaring pushes you back to -1 by going counter-clockwise on the unit circle.

I've attached a little schematic showing what I mean.

So the issue is that for the left picture, corresponding to ((-1)^2)^(1/2), when you come back via the square root function, after having squared, you're on a different branch of the square root function than you are in the right picture when you initially go out.

6. Originally Posted by khamaar
Thanks A lot for so much replies…But no one clearly understood my question!
I want to say that .. since,
METHOD (i) (-1)2/2 = {(-1)2}1/2 = (1)1/2 = 1 equation 1
METHOD (ii) (-1)2/2 = (-1)1 = -1 [as 2/2=1] equation 2
@ khamaar, I ought not waste my time here after that uncalled for response you first gave me. (Good that it was removed) The fact is, I did understand your point about the ambiguity in notation and the answer I gave was too brief nonetheless is the correct one.

RULE: If both $m~\&~n$ are even integers and $\dfrac{m}{n}=\dfrac{j}{k}$ in lowest terms then $x^{\frac{m}{n}}= |x|^{\frac{j}{k}}$.

That rule was devised to avoid the very ambiguity that you pointed out.
Unfortunately the rule is often overlooked by unknowing instructors.

7. ## More questions...More confusion...

Iota x iota = -1 right?

(-1)1/2 x (-1)1/2 = (-1)2/2 Equation 1.

Now applying Rule Provided BY PLATO.

RULE: If both m & n are even integers and m/n=j/k in lowest terms then xm/n=|x|j/k

1/1 is the lowest form of 2/2 so,

(-1)2/2 = |(-1)|1 /1 = 1

And hence I have proved (if plato is correct) that (iota)2=1

MY QUESTIONS:-

1- 1-What have I done wrong here?
2- 2-Whats equal to (-1)2/2 … (-1) or (1)
3- 3-I clearly don’t understand that Graphical approach by Ackbeet. We haven’t done Circles in Complex plane yet in college.. So can you provide a link where I would learn them, so that I can get what you have told me.?

THANKS

8. The rule that I quoted applies only to real numbers.

For any real number $x^2\ge 0$.
The complex number $i^2=-1$ therefore $i$ is not a real number so the rule does apply.

Your mistake is in thinking that the rules of exponents apply across all number systems.
That sadly is not the case.
For complex exponentiation there is an entirely different definition.
For example if each of $w~\&~z$ is a complex number then $z^w=\exp(w\cdot\log(z))$, where even $\log(z)$ has different sort of definition.

Here is an interesting fact: $e^{i\pi}=-1$.

9. I was applying Your Rule to (-1)... I surely know that -1 comes on the Real Line.... That is.

(-1)^(2/2)

And in your rule you said
(x)^(m/n)

So i have taken 'x' as '-1' which is a real number.

10. Originally Posted by Wilmer
AND: (-1)^1 = -1^1 (brackets not required)
I would leave the brackets in since the precedence of unary minus and exponentiation is not consistently agreed upon (I would not like to say what Excel evaluates -1^2 as, or "C" for that matter though the latter is easier to look up but who ever looks at the manuals)

CB

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