if a = b then in step 2 you are multipling by zero… but I can see how this looks true that a = b.
Look at this example whereby somebody tries to prove 2=1 in a similar fashion:
How would you prove that 2 = 1?
If a = b (so I say) [a = b]
And we multiply both sides by a
Then we’ll see that a2 [a2 = ab]
When with ab compared
Are the same. Remove b2. OK? [a2− b2 = ab − b2]
Both sides we will factorize. See?
Now each side contains a − b. [(a+b)(a − b) = b(a − b)]
We’ll divide through by a
Minus b and olé
a + b = b. Oh whoopee! [a + b = b]
But since I said a = b
b + b = b you’ll agree? [b + b = b]
So if b = 1
Then this sum I have done [1 + 1 = 1]
Proves that 2 = 1. Q.E.D.
Written by PeterW
(Just in case you’re wondering – the above proof is incorrect because in step 5, we divided by (a – b) which is 0 since a = b)
a2, b2 means a squared, b squared just to clarify
[http://www.onlinemathlearning.com/algebra-math-riddles.html]
So it is kinda like this but backwards Image may be NSFW.
Clik here to view.
