*Originally posted by smw6869*

**Anyone know how to prove this to be true(even though it's not). A math teacher proved it to my math class. Everything made sense and no one could find the falacy.**

If there is a general formula involved in the proof, make sure division by zero isn't happening because that isn't allowed. So for example the classic case of the fallacy of 1=2 (which is very similar to yours)

*step 1*
**Let a = b**
*step 2 Multiply both sides by a:*
**aa = ab**
*step 3 which is the same as:*
**a^2 = ab** ( "a squared equals a times b" )

*step 4 Add the quantity ( a^2 - 2ab) to both sides:*
**a^2 + (a^2 - 2ab) = ab + (a^2 - 2ab)**
*step 5 simplifying both sides we get:*
**(a^2 + a^2) - 2ab = a^2 + (ab - 2ab)**
**2(a^2) - 2ab = a^2 - ab**
**2 (a^2 - ab) = a^2 - ab**
*step 6 divide both sides by (a^2 - ab):*
**2(a^2 - ab) / (a^2 - ab) = (a^2 - ab) / (a^2 -ab)**
*step 7 cancel out like terms in num.&denom:*
**2 = 1 !!!**
Two equals one? Impossible! What's the catch? The catch is, our very first assumption is Let a= b, and if that's true then the quantity (a^2 - ab) = (a^2 - a^2) = 0, and we are not allowed to divide by zero which is what we do in our "proof" in step 6. The proof is not allowed. No division by zero.

I bet this is the trick that is happening in your false proof of 2+2=5.

(sorry for all the edits, but I had to work out some formatting issues)