$a=b \implies b=0$ puzzle. What is wrong?

Question

I found the below online with no explanation what is wrong, but I am guessing it must be a condition or assumption, but can't see what the issue is.

a = b
ab = b²
ab - a² = b² - a²
a(b - a) = (b + a)(b - a)
a = b + a
0 = b

Can anyone see what the issue is?

Answer 1

You cannot cancel $b-a$ like that since you have started with $a=b$ and hence $b-a =0$.

Answer 2

Division is only valid when you are not dividing by $0$. You divide by $b-a$, which is equal to $0$ since $a=b$

Answer 3

The way to disprove basically any of this kind of puzzle is to step through it with actual numbers. Where does it first go wrong?

Let $b=a=1$. Then:

• $1=1$
• $1\times 1 = 1^2$
• $1 \times 1 - 1^2 = 1^2 - 1^2$
• $1 (1-1) = (1+1)(1-1)$
• $1 = 1+1$ (wait, what?)
• $0=1$ (abort! abort!)

Answer 4

In the 4'th line you canceled out $(b-a)$. This $(b-a)$ term can be canceled out only when $(b-a)\neq0$, but according to your first assumption: $(a=b)\Rightarrow(b-a)=0$, so according to your assumption you can't cancel out $(b-a)$ like this.