# I need a proof that -1 DOES NOT equal 1

1. Nov 22, 2013

### anis91

hey everybody, once i saw a thread here (didn't want to revive it) about an equation that proves that 1=-1, it was proved wrong ofc, but at the end, someone posted this:

" -1=(-1)^1
=(-1)^2*1/2
=[(-1)^2]^1/2
=(1)^1/2
=√1
=1 "
yet no one replied to it, can someone show me which is the "trippy" step here? the one that misuses an algebra rule? (e.g. a rule that can only be applied to positive numbers etc..."

thank you.

2. Nov 22, 2013

### Office_Shredder

Staff Emeritus
We get this question a lot, there's actually a thread in the FAQ devoted to answering questions like it

The main point is that whne you write
$$\left( -1 \right)^{2/2} = \left( (-1)^2 \right)^{1/2}$$
you have performed an operation which is not actually valid. Taking exponents of negative numbers is tricky and you have to be more careful than when you are working with exponents of positive numbers. In general
$$x^{ab} =\left( x^{a} \right)^{b}$$
is something that can only be applied when x is a positive number.

3. Nov 22, 2013

### anis91

okay thanks alot! i appreciate it! ^^

4. Nov 22, 2013

### quietrain

hello office_shredder,

may i know why the indices rule is invalid for negative numbers? i tried for example, (-2^6) and split them up to [-2^(2*3)] = 4^3 and i still yielded 64.

where does taking exponents of negative numbers breakdown?

thanks!

5. Nov 22, 2013

### Staff: Mentor

No it doesn't. The parentheses you have in (-2^6) don't do anything and might as well not be there. (-2^6) is exactly the same as -2^6 which is the same as -(2^6) or -64.

If you want to raise -2 to the 6th power, you have to write it as (-2)^6.
It breaks down when the exponent is fractional and represents an even root (i.e., square root, fourth root, and so on). There is no problem when the exponent is an integer unless you happen to be taking 0 to a negative power.

6. Nov 22, 2013

### quietrain

yes that was sloppy of me :D

anyway, is the even root fractional exponent the only case whereby this rule breaks down ?

7. Nov 22, 2013

### Staff: Mentor

Yes, since odd roots (cube root, fifth root, and so on) can have negative arguments. For example, $\sqrt[3]{-27} = -3$ and $\sqrt[5]{-32} = -2$.

If you have an expression such as (-27)2/3, you can write it either as [(-27)2]1/3 or as [(-27)1/3]2, both of which are equal to 9.

The first expression simplifies to (729)1/3 = 9, and the second expression simplifies to (-3)2, which is also 9.

8. Nov 22, 2013

It'd be cooler if someone managed to "prove" that i=√1

9. Nov 23, 2013

### quietrain

that was very insightful, thank you

10. Nov 23, 2013

### Zeda

I won't try that, but I thought of this yesterday (breaking the same rule as above):

$\sqrt{x}$
Now to factor out a -1:
$=i\sqrt{-x}$
And to factor out another -1:
$=i*i\sqrt{x}$
$=-\sqrt{x}$

$\Rightarrow \sqrt{x}=-\sqrt{x}$

:P The issue is that even functions are not 1-1, meaning they can map multiple inputs to the same output. Naturally, the inverse function would have to map backwards, but it would have to be split off to multiple values. That is why, for example, $\sqrt{9}=\{3,-3\}$

11. Nov 24, 2013

### jbriggs444

It is a standard notational convention that $\sqrt{x}$ where x is a non-negative real number always refers to the positive root.

12. Nov 24, 2013

### hilbert2

If you want a rigorous proof for this kind of statements, you need to use the axioms of real numbers: http://math.berkeley.edu/~talaska/h1b/axioms-real-numbers.pdf .

First you add $1$ to both sides of the equation $1=-1$ and get $1+1=0$. Next you use the order axioms to show that $1+1>1>0$, which is a contradiction and proves that $1$ can't equal $-1$ (for real numbers $a$ and $b$, the inequalities $a>b$ and $a=b$ can't both be true).

13. Nov 29, 2013

### 1MileCrash

√9 is 3, √9 is not -3, and √x is a mapping from one real to precisely oneother real; a function.

14. Nov 29, 2013

### Zeda

I know, I should have been more clear by being more confusing :P I was using √ to represent the function that, given the output of f(x)=x2, would return x. When I was working on a little project dealing with sine and cosine, I would often have a function squared on one side, where the otherside, after taking the square root, was indeed negative and the positive square root would cause the fully reduced form to fail.