Another Proof that 1=2, can you find the flaw?

**Stephen347** · November 27th 2012, 09:15 AM

This false proof brings to light an assumption that we tend to take for granted. Try to see if you can figure it out, and where the logical fallacy takes place in the steps below. To start, we agree that (1)(1) = (-1)(-1)

Then,

1. $\frac{(1)(1)}{(1)(-1)}\ =\ \frac{(-1)(-1)}{(1)(-1)}$ - Dividing both sides by (1)(-1).

2. $\frac{1}{-1}\ =\ \frac{-1}{1}$ - By simplifying.

3. $\sqrt{\frac{1}{-1}}\ =\ \sqrt{\frac{-1}{1}}$ - Taking the positive square root.

4. $\frac{\sqrt{1}}{\sqrt{-1}}\ =\ \frac{\sqrt{-1}}{\sqrt{1}}$ - By simplifying.

5. $\frac{1}{i}\ =\ \frac{i}{1}$ - Simplifying with the imaginary number $i$ .

6. $1^2\ =\ i^2$ - By cross-multiplying.

7. $1\ =\ -1$ - Simplifying.

8. $2\ =\ 0$ - Adding 1 to both sides.

9. $1\ =\ 0$ - Dividing both sides by 2.

10. $2\ =\ 1$ - Adding one to both sides.

**ebaines** · November 27th 2012, 10:23 AM

This has always been one of my favorite "proofs" for 1 = -1. The trick is at step 4: it is not necessarily true that $\sqrt{\frac a b} = \frac {\sqrt a}{\sqrt b}$ for negative b; you have to consider that $\sqrt {-1} = \pm i$ . So that 4th step should really be $\sqrt{\frac 1 {-1}} = \frac {\sqrt 1}{- \sqrt {-1}} = \frac 1 {-i}$ and in step 5 you get

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

**Stephen347** · November 27th 2012, 11:19 AM

Right, we cannot assume that $\sqrt{\frac{a}{b}}\ =\ \frac{\sqrt{a}}{\sqrt{b}}$ if a or b are negative. And while you are right that $i\ =\ \pm \sqrt{-1}$ , how do you know when to substitue $i$ and $-i$ ? I think a slight better way to reconcile this problem is to include absolute values (moduli for complex numbers). i.e.

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

$\left|\sqrt{\frac{1}{-1}}\right|\ =\ \left|\sqrt{\frac{-1}{1}}\right|$

$\frac{\left|\sqrt{1}\right|}{\left|\sqrt{-1}\right|}\ =\ \frac{\left|\sqrt{-1}\right|}{\left|\sqrt{1}\right|}$

$\frac{1}{\left|\pm i\right|}\ =\ \frac{\left|\pm i\right|}{1}$

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

$1\ =\ 1$

And hence no contradiction. In other words,

$\left|\sqrt{ab}\right|\ =\ \left|\sqrt{a}\right| \left|\sqrt{b}\right|$ for all a and b.

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