What is the fallacy of this proof?

Question

I recently was working with square roots and came across this-

$({\sqrt -1})$$=-1^\frac12$$=-1^\frac24$$=(-1^2)^\frac14$$=1^\frac14$$=1$

I understand that this is not true,but despite repeated attempts failed to prove it wrong.Can someone please point out the fallacy in this proof.Thanks in advance.

You could simplify your fallacy: $-1=(-1)^{1}=(-1)^{\frac{2}{2}}=((-1)^2)^{\frac{1}{2}}=1^{\frac{1}{2}}=1$. ;-) — Blackbird, Aug 13 '15 at 09:26

score 3 · Answer 1 · answered Aug 13 '15 at 09:21

3

The property that $(a^n)^m=(a^m)^n=a^{mn}$ is only defined for real numbers. But $\sqrt{-1}=i$ is a complex number.

answered Aug 13 '15 at 09:21

Aditya Agarwal

4,847
3
30
56

1

See my comment above. ;-) – Blackbird Aug 13 '15 at 09:27
Only for real positive numbers if one wants the property to be true for all $m,n\in\mathbb{R}$. – A.Γ. Aug 13 '15 at 11:58

score 1 · Answer 2 · answered Aug 13 '15 at 09:21

1

The fourth root of $1$ has four solutions: $1$, $-1$, $j$ and $-j$. The last two solutions are identical to the two solutions of the square root of $-1$.

answered Aug 13 '15 at 09:21

Bernhard

1,191

What is $j$? Are you talking about $\iota$? – Aditya Agarwal Aug 13 '15 at 09:23
Yes, exactly. Different notation, apparently. – Bernhard Aug 13 '15 at 09:27
@Bernhard-Last two solutions may be identical but the first two are also solutions...why are they not acceptable? – Soham Aug 28 '15 at 14:53
The reason why the first two solutions are not correct is given by the comment of Aditya Agarwal: The product rule for powers is only valid for positive real numbers. – Bernhard Aug 29 '15 at 18:23

What is the fallacy of this proof?

2 Answers2