What's wrong with that proof?

Question

What wrong with this proof?

$(-1)=(-1)^{\frac{2}{2}}=(-1)^{2\times \frac{1}{2}}=\sqrt{1}=1$ then $1=-1$

The issue is how you define $x \mapsto (x)^{\frac{1}{2}}$. – copper.hat Oct 31 '13 at 18:11 — copper.hat, Oct 31 '13 at 18:11

score 7 · Accepted Answer · answered Oct 31 '13 at 18:04

7

$x^{\frac{1}{2}}$ is a multiple-valued "function", since in general $x$ has two square roots. One could also write:

$$\sqrt1=-1$$

answered Oct 31 '13 at 18:04

Jack M

27,819
7
63
129

score 1 · Answer 2 · edited Jun 12 '20 at 10:38

1

enter image description here

it is fake suppose f(x)=x and g(x)=(x)^(2/2)

f(x) is not equal to g(x) domain of both is R but g(x)=|x| instead of f(x)=x

edited Jun 12 '20 at 10:38

Community

1

answered Oct 31 '13 at 18:14

Khosrotash

24,922

score 0 · Answer 3 · answered Nov 01 '13 at 01:34

$(-1)^{2 \times\frac{1}{2}}$ is not equal to $\sqrt{1}$. The correct way to simplify $(-1)^{2 \times\frac{1}{2}}$, using universally accepted orders of operation (because they work), is to start by simplifying the exponent ${2 \times\frac{1}{2}}$ to $1$. Then you get $(-1)^1$ , which is equal to $-1$, which is where you started.

What's wrong with that proof?

3 Answers3