0
$\begingroup$

This bothers me for a while:

Proof $a = b$ $$a=b$$ $$b=a$$ $$a+b=b+a$$ $$a+b=a+b$$ $$0 = 0$$

This looks like a proof for $a=b$, but it shouldn't work like this. But i can't put my finger on why it's wrong.

$\endgroup$
  • 1
    $\begingroup$ if you start from $0=0$ you can't assume that a=b and cancel a and b in step 2. And if you're trying to show a=b and you start with a=b and accept things as is then you've only shown a=b when a=b=0 not when a,b can be anything. If you started with 0=0 everything is good up until step 4 when for no reason $a+b=b+a$ turns into $b=a$. Also there are no justifications for any of these steps. $\endgroup$ – Eleven-Eleven Apr 22 '13 at 13:26
  • 5
    $\begingroup$ This (among other things) proves that $a=b$ assuming $a=b$. There is a simpler way to do this: just note that $a=b$. $\endgroup$ – Dejan Govc Apr 22 '13 at 14:00
  • 1
    $\begingroup$ More simply: $\ a = b\:\Rightarrow\: 0 = 0\:$ by multiplying both sides by $\,0.\:$ Because that operation is not invertible, you cannot reverse the implication, i.e. the converse is not true. However, it is true that $\:a = b\iff ac = bc\:$ if $\:c \ne 0\:$ in a field (or if $\:c\:$ is cancellable in a ring). $\endgroup$ – Math Gems Apr 22 '13 at 14:06
23
$\begingroup$

This merely proves that if $a=b$ then $0=0$. Try proving it the other way around and you'll see that you can't reverse the steps.

$\endgroup$
  • $\begingroup$ Yes, all of the responses get at aspects of the difficulty with this "proof". It is not "reversible" exactly because it does follow that if a number is equal to itself (a = b and b = a), then 0 must equal 0. It does not in turn follow that a = b = 0 for all values of $a$. $\endgroup$ – colormegone Apr 22 '13 at 13:35
  • 1
    $\begingroup$ @RecklessReckoner I don't think the deduction is that $a=b=0$, only that $a=b$. There is an intuitive technique for checking identities to manipulate them until you get a "true" identity. But, as you say, that only works when the steps are reversible. $\endgroup$ – Thomas Andrews Apr 22 '13 at 13:38
  • $\begingroup$ if we don't reverse it and assume that a=b you can't go from $a+b=a+b$ to $0=0$ unless we've already said that $a=b=0$ $\endgroup$ – Eleven-Eleven Apr 22 '13 at 13:41
  • 2
    $\begingroup$ No, we get that by adding $-(a+b)$ to both sides. @ChristopherErnst $\endgroup$ – Thomas Andrews Apr 22 '13 at 13:42
  • $\begingroup$ @ThomasAndrews I'm saying that it is attempted to claim that it must follow somehow that $a = b = 0$ (everything is equal), which really comes from absolutely nowhere among this list of perfectly true equations. Another way of attacking this is that this claims at the start that this is supposed to be a proof that $a = b$, but then immediately begins by assuming the conclusion... $\endgroup$ – colormegone Apr 22 '13 at 13:55
8
$\begingroup$

If you reverse the steps, there is no viable rule or theory that lets you go from $$a+b=b+a$$ to $$b=a$$

$\endgroup$
  • 1
    $\begingroup$ Well put. Just because $3+2=2+3$, does not mean $2=3$ !! $\endgroup$ – goblin Aug 1 '13 at 2:28
3
$\begingroup$

There's nothing wrong with this. You started by assuming that $a=b$, so your proof only holds if, indeed, $a$ is equal to $b$.

$\endgroup$
2
$\begingroup$

Make the proof complete by adding either $\;\Rightarrow\;$, $\;\Leftarrow\;$, or $\;\Leftrightarrow\;$ inbetween each two expressions. Then you will see that you have proved $\;a = b \;\Rightarrow\; 0 = 0\;$.

$\endgroup$
0
$\begingroup$

It starts off with what it attempts to prove.

Prove $a=b$, is not supposed to start off with $\text{Let }a=b$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.