What's wrong with this 1 = -1 proof?

Question

Stared at this proof for 10 minutes, perhaps even more. Still a quite stumped, but I'm pretty sure the answer is staring me right in the face.

Okay, so we know that $i^2 = -1$.

Dividing both sides by $i$:

$$i = - \frac{1}{i}$$

Squaring both sides:

$$i^2 = -\frac{1^2}{i^2}$$

Obviously $i^2 = -1$, as previously shown, so therefore:

$$-1 = -\frac{1}{-1}$$

Both negatives become a positive, so we're left with:

$$-1 = \frac{1}{1}$$

Which simplifies to:

$$-1 = 1$$

I'm not quite sure what's wrong here. Unless I'm there's an important step I skipped, I don't really see any problem here.

"Squaring both sides"... you should know that if you square $-a$ you get $a^2$. Recognize then that squaring $(-\frac{1}{i})$ gives $(-\frac{1}{i})^2$ i.e. $\frac{1^2}{i^2}$. — JMoravitz, Sep 07 '18 at 05:06
As you play with $-1=1$ "(fake) proofs", you should be careful about several things. One of which is how you use various identities involving roots or exponents which in reality are only guaranteed to work for positive real numbers. See this post for one such example. — JMoravitz, Sep 07 '18 at 05:08
"Minus times minus equals plus: The reason for this we need not discuss" — Angina Seng, Sep 07 '18 at 05:10
Possible duplicate of Why $\sqrt{-1 \times {-1}} \neq \sqrt{-1}^2$? — A. Pongrácz, Sep 07 '18 at 05:56

score 7 · Accepted Answer · answered Sep 07 '18 at 05:05

7

The square of $-x$ is $x^2$, not $-(x^2)$. Your error is in the "squaring both sides" step.

answered Sep 07 '18 at 05:05

score 1 · Answer 2 · answered Sep 07 '18 at 05:15

1

You're doing mistake in the "squaring both side" step. You're squaring like this -(1/i)²

It should be done like this (-1/i)²

answered Sep 07 '18 at 05:15

Abhishek VERMA

score 0 · Answer 3 · answered Sep 07 '18 at 05:09

0

$\left(\frac{-1}{i}\right)^2=\frac{(-1)^2}{i^2}=\frac{1}{-1}=-1$

answered Sep 07 '18 at 05:09

gandalf61

3 Answers3