Is this right?

Question

So -
if a = b, then
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b) x (a - b) = b x (a - b)
(a + b) x (a - b) / (a - b) = b x (a - b) / (a - b)
a + b = b
since a = b, substitute b for a, then
b + b = b
2 x b = b
2 x b / b = b / b
2 = 1

Answer 1

It is not right.
One of your steps was to divide both sides of the equation by:
(a-b)
Now a-b = 0, and you can not divide by zero, otherwise you get weird results.
One such weird result is that 2=1.

Answer 2

If a=b, then you can't divide by (a-b) since a-b=0

That's an old trick "proof" you've got there for showing that 2=1. I saw it first when I was in high school, back in the 70s, but I'm sure it's been around much longer!

~ ♥ ~

Answer 3

Simply put - NO!
One of your sequence of steps attempts to rewrite a basic law of mathematics, which is why your 'proof' of 2=1 is fallacious.
The step in which you divide (a+b)x(a-b) by (a-b) and
(b) x (a-b) by (a-b) is the illegal operation because you are dividing by ZERO! If a=b, then a-b= zero. The result of dividing by zero is 'undefined'. You are not permitted to divide by zero. Because of your 'illegal' operation, you are able to 'prove' 2=1.

Answer 4

i think you divided by zero some where along there. you can actually prove that any number equals any other number is you substitute letters for numbers and solve. hope this helps?

Answer 5

No.
If a = b
then:
(a+b)(a-b) = (a+b) x 0 = b x 0 = 0

Answer 6

it's not right because (a-b) = 0
so in the equation (a+b) (a-b) = b (a - b)
then 0 = 0

Answer 7

I doubt it