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What's wrong with this demonstration?:

$$A \iff 1 = 1^1$$ $$A \implies 1 = 1^\frac{2}{2}$$ $$A \implies 1 = (1^2)^\frac{1}{2}$$ $$A \implies 1 = ((-1)^2)^\frac{1}{2}$$ $$A \implies 1 = (-1)^\frac{2}{2}$$ $$A \implies 1 = (-1)^1 = -1$$

So finaly we get $$1 = -1$$

Fly by Night
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rullof
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  • Your violating order of operations. $((-1)^2)^{\frac{1}{2}}=(1)^{\frac{1}{2}}$. Parenthesis before exponents. – Wintermute Jan 01 '14 at 22:55

3 Answers3

5

The property $\forall p\in \mathbb Z\forall q\in \mathbb N\left(a^{p/q}=(a^p)^{1/q}\right)$ doesn't work for negative $a$.

Howevever $\forall p\in \mathbb Z\forall q\in \mathbb N\left(q\text{ is odd}\implies a^{p/q}=(a^p)^{1/q}=(a^{1/q})^p\right)$ is true for all $a\in \mathbb Z\setminus \{0\}$.

Git Gud
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  • +1 Nice answer GG. I've been reading up on differentiability and think I've got it sorted. Thanks for your help the other day. – Fly by Night Jan 01 '14 at 23:02
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    @FlybyNight Anytime and thanks. I had already up voted you, so no (+1) retribuition ^_^ – Git Gud Jan 01 '14 at 23:03
1

One massive problem is that you assume that $\left(x^2\right)^{1/2} \equiv x$.

Here's a counter example: If $x=-2$ then $x^2=(-2)^2=+4$. Finally $(+4)^{1/2} = \pm 2$.

Clearly $-2$ is not equal to both $+2$ and $-2$ at the same time. There must be a problem!

Note that there is a big difference between $x^{1/2}$ and $\sqrt{x}$. The symbol $x^{1/2}$ represents the solutions to the equation $y=x^2$ while the symbol $\sqrt{x}$ represents the unique, real, positive solution to $y=x^2$. In a similar way $\sqrt[4]{1}=1$ while $1^{1/4} = \{\,\pm 1,\pm\operatorname{i}\,\}.$

Fly by Night
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0

The core problem: the squaring function defined on the line is not 1-1.

ncmathsadist
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