-3
$\begingroup$

I need to find the flaw in the following proof:

$a,b\in\mathbb{R}$\ $\left\{ 0 \right\} $ such that $a=b$

1) Multiplying both sides by $a$ yields the equality: $a^2=ab$

2) Subtracting $b^2$ from both sides yields the equality $a^2-b^2=ab-b^2$

3) Then, $a^2-b^2=ab-b^2\Rightarrow (a-b)(a+b)=b(a-b)$

4) Then, dividing both sides of $(a-b)(a+b)=b(a-b)$ by $(a-b)$ yields: $a+b=b$

5) Substituting $a=b$ yields $2b=b$

5) Therefore, $2=1$


My thoughts and ideas on this:

It seems to me that step $5$ has an error. It should be that $a+b=b\Rightarrow a=0$. However, we initially said that $a,b$ cannot be $0$.

$\endgroup$
2
  • $\begingroup$ Is this really still going around? I first saw this in a 'Murderous Maths' book when I was 10 years old, and even then I felt insulted by it. $\endgroup$
    – user85798
    Oct 23, 2015 at 12:53
  • $\begingroup$ Why don't you try it with actual numbers? Start with $5=5$, say. Multiply both sides by $a$, or $5$, yields $25=25$; subtracting $b^2$ or $5^2$ from both sides yields $0=0$. Dividing by $a-b=5-5=0$… uh-oh! $\endgroup$ Oct 23, 2015 at 12:59

4 Answers 4

3
$\begingroup$

No, the problem is at step 4. Since you have $a=b$, you divide by $(a-b)=0$. You can't divide by zero.

$\endgroup$
0
$\begingroup$

Addition, subtraction and multiplication are defined for any two elements of $\mathbb{R}$. However when you divide you should be careful since division by 0 is not defined. So the only step you should re-examine is step 4.

$\endgroup$
0
$\begingroup$

$a=b \implies a-b=0 \implies \text{dividing by }(a−b)$ yields undefined result.

$\endgroup$
0
$\begingroup$

In step 4) both sides are divided by $a-b$ which is forbidden if $a-b=0$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .