Theorem ? : All numbers are equal to zero.?

Question

Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0
What do you think?

Furthermore if a + b = b, and a = b, then b + b = b, and 2b = b, which mean that 2 = 1.

Answer 1

Ah this has to all be zero.
You are definitely correct because your equations are correct.
All equals zero.

Answer 2

no.

2

Answer 3

You have a divide by zero problem there that throws everything out of whack.

a - b if a=b is zero. zero divided by zero is undefined.

a = b
okay
a^2 = ab
Doing good so far
a^2 - b^2 = ab - b^2
Ok, that makes sense
(a + b)(a - b) = b(a - b)
Hold up there, charlie. (a+b)(a-b) if a=b is zero. So we have zero equals zero.
a + b = b
Here you divided out the (a-b) part. Not a valid operation. Cannot divide by zero.

Answer 4

You divided both sides by (a-b). However, because a=b, you are dividing by zero. This renders the rest of the 'proof' irrelevant.

Answer 5

in step 4 you are dividing by (a-b) which = 0 as a=b
this will result in any further answers being incorrect and is a big no no in maths.

Answer 6

the factor (a -- b) can't divide the two sides as it is equal to zero and division by zero is undefined.

Answer 7

im confuzzled... no number is equal 2 zero but zero

Answer 8

(a + b)(a - b) = b(a - b)

you cannot go to the next step if a =b ( you cannot simplify by (a -b) , i.e , you are dividing by zero.
a + b = b

Answer 9

Oh wow I haven't seen that one before...