Proving "-1 = 1" Wrong: Ideas?

[Rectifier]

Hey there!
These is this falsidical paradox that I can't seem to prove wrong.

## -1 = (-1)^1 = (-1)^\frac{1}{1}= (-1)^\frac{2}{2} = (-1)^{\frac{2}{1} \cdot \frac{1}{2}} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^{\frac{1}{2}} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^{\frac{1}{2}} = (1)^{\frac{1}{2}} = 1 ##

Any ideas?

 

[Gil]

You have (1)^1/2, and it can be +1 or -1. The mistake you do is that you sould consider ((-1)²)^1/2=|-1| = 1

 

[micromass]

You have (1)^1/2, and it can be +1 or -1.

No it can't. It's always 1.

 

[symbolipoint]

Someone will say it eventually. [tex] i^2=-1[/tex]
[tex](i^2)^(1/2)=+-(-1)^(1/2)[/tex], the text is there but it does not look right.
[tex]√(i^2)=+-√(-1)=i[/tex]

 

[Gil]

No it can't. It's always 1.

The square root of 1 is ±√1 = ±1, isn't it?

 

[micromass]

The square root of 1 is ±√1 = ±1, isn't it?

1 has two square roots: -1 and 1. But ##1^{1/2} = 1##.

 

[axmls]

1 has two square roots: -1 and 1. But ##1^{1/2} = 1##.

A number raised to the power of 1/2 is the exact same thing as a square root. So [itex](-1)^{1/2} = \pm 1[/itex]. Information is lost upon squaring. That's where I'd say the issue comes from.

Namely, the OP starts with a number [itex]x[/itex], and takes [tex]x = x^{(2)(1/2)} = (x^2)^{1/2}[/tex]

but information is lost when we square, because [tex] x = a \implies x^2=a^2[/tex]

but [tex]x^2 = a^2 \nRightarrow x = a[/tex]
Though someone else may see farther than me regarding this problem.

 

[Mark44]

The square root of 1 is ±√1 = ±1, isn't it?

micromass is correct. ##\sqrt{1} = +1##.

 

[Stephen Tashi]

## (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^{\frac{1}{2}} ##

It isn't in general true that [itex] x^{ab} = (x^a)^b [/itex].

 

[Mark44]

A number raised to the power of 1/2 is the exact same thing as a square root. So [itex](-1)^{1/2} = \pm 1[/itex].

No. The sqaure root of -1 is the imaginary unit i.

Information is lost upon squaring.

That's irrelvant to this question. We're taking the square root, not squaring something.

That's where I'd say the issue comes from.

Namely, the OP starts with a number [itex]x[/itex], and takes [tex]x = x^{(2)(1/2)} = (x^2)^{1/2}[/tex]

but information is lost when we square, because [tex] x = a \implies x^2=a^2[/tex]

but [tex]x^2 = a^2 \nRightarrow x = a[/tex]
Though someone else may see farther than me regarding this problem.

 

[Mark44]

There is a misconception about square roots that shows up here quite often. An expression such as ##\sqrt{4} = 2##, not ##\pm 2##. While it's true that 4 has two square roots, one positive and one negative, the symbol ##\sqrt{4}## represents the positive square root.

More generally, for any positive real number a, the expression ##\sqrt{a}## represents the positive number b such that b² = a.