What wrong with this proof?
$(-1)=(-1)^{\frac{2}{2}}=(-1)^{2\times \frac{1}{2}}=\sqrt{1}=1$ then $1=-1$
$\begingroup$
$\endgroup$
1
-
$\begingroup$ The issue is how you define $x \mapsto (x)^{\frac{1}{2}}$. $\endgroup$ – copper.hat Oct 31 '13 at 18:11
Add a comment
|
$\begingroup$
$\endgroup$
$x^{\frac{1}{2}}$ is a multiple-valued "function", since in general $x$ has two square roots. One could also write:
$$\sqrt1=-1$$
$\begingroup$
$\endgroup$
0
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
answered Oct 31 '13 at 18:14
$\begingroup$
$\endgroup$
$(-1)^{2 \times\frac{1}{2}}$ 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.
answered Nov 1 '13 at 1:34
