What wrong with this proof?
$(-1)=(-1)^{\frac{2}{2}}=(-1)^{2\times \frac{1}{2}}=\sqrt{1}=1$ then $1=-1$
Answer
$x^{\frac{1}{2}}$ is a multiple-valued "function", since in general $x$ has two square roots. One could also write:
$$\sqrt1=-1$$
I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...
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