# What is wrong with my logic?

1. Feb 16, 2007

### jason17349

if $$e^{\pi\imath}=-1$$ then:

$$-e^{\pi\imath}=1$$ and,

$$e^{2\pi\imath}=1$$

then:

$$-e^{\pi\imath}=e^{2\pi\imath}$$

$$\rightarrow e^{2\pi\imath}+e^{\pi\imath}=0$$

$$\rightarrow (e^{\pi\imath})^2+e^{\pi\imath}=0$$

$$\rightarrow (e^{\pi\imath}+1)e^{\pi\imath}=0$$

then:

$$e^{\pi\imath}=0$$

and

$$e^{\pi\imath}+1=0$$

Can somebody explain this contradiction to me?

Last edited: Feb 16, 2007
2. Feb 16, 2007

### D H

Staff Emeritus
From the very first line you wrote:
$$e^{\pi i} + 1 = 0$$
You divided by zero when you went from
$$(e^{\pi i} + 1)e^{\pi i} = 0$$
to
$$e^{\pi i} = 0$$

3. Feb 16, 2007

### jason17349

I thought if $$ab=0$$ then you could have two solutions a = 0 and b = 0?

4. Feb 16, 2007

### D H

Staff Emeritus
Think again. For what values of $x$ is $0\cdot x = 0$ true?

5. Feb 16, 2007

### jason17349

Whoops, sorry

6. Feb 16, 2007

### cristo

Staff Emeritus
You're thinking of something like if ab=0, then either a=0 or b=0. However, in this case, you know that $e^{\pi\imath}+1=0$, and so $(e^{\pi\imath}+1)e^{\pi\imath}=0$ tells us nothing about $e^{\pi i}$