# A little problem

1. Sep 5, 2009

### PrincePhoenix

Why is 2 = 3 over here?

Suppose 2=3

Then subtract 5/2 from both sides.
2 - 5/2 = 3 - 5/2
4-5/2 = 6-5/2
-1/2 = 1/2

Take square root
(-1/2)2 = (1/2)2
1/4 = 1/4

What's wrong? why is it proved that 2 = 3?

Last edited: Sep 5, 2009
2. Sep 5, 2009

### Staff: Mentor

You didn't prove anything. Note that when you square both sides you effectively multiply the sides by different numbers.

3. Sep 5, 2009

### pbandjay

Maybe it is because I am not fully awake yet, but I do not follow your arithmetic.

2 - 5/4 = 3/4 not -1/2
3 - 5/4 = 7/4 not 1/2

4. Sep 5, 2009

### PrincePhoenix

So you mean that taking square was wrong? I thought I did everything on both sides of the equation. I mean which step is incorrect?

5. Sep 5, 2009

### PrincePhoenix

I corrected the mistake. Check again.

6. Sep 5, 2009

### jgens

When you square an equation, you run the risk of creating extraneous roots. For example, $-2^2 = 2^2 = 4$ but this does not prove that $-2 = 2$. Also, another mistake you made is assuming what you're trying to prove. To have a correct proof you would need to begin with the equaltiy $1/4 = 1/4$ and proceed to show that $2 = 3$ (which you can't do).

Edit: Just so it's clear, the biggest fallacy in your "proof" was the assumption that $2 = 3$. You can't assume what you're trying to prove.

7. Sep 5, 2009

### jgens

To absolutely clear, here's why you can't assume what you're proving . . .

Let $a,b \in \mathbb{R}$ and suppose that $a = b$. Then clearly we have that $0*a = 0*b = 0$. Hence, if $a$ and $b$ are any two real numbers, then they are equal to each other (using your same logic). A correct proof would need to start at the equality $0 = 0$ and work from there (which is impossible).

8. Sep 5, 2009

### PrincePhoenix

Thanks for the answers. I didn't want to PROVE anything here. This was just shown to me by a friend and I just wanted to know what was wrong.