Is $a=b$ when $|a-b|=0$? If so, how can I prove it?

Question

I was looking at another proof in my book which seemed to use this fact for granted. I am pretty sure it's right, but how could I prove that?

from @Martin R 's comment you even get the direction you were asking for. — Peter Melech, Aug 05 '19 at 10:45
@coreyman 317 You are proving the reverse claim. Can't assume what you need to prove, can you. — uniquesolution, Aug 05 '19 at 10:45
I think you should have tried to do an attempt and shown it with the question. There's a good chance that you would have produced a correct proof by applying the definition of absolute value. — skyking, Aug 05 '19 at 10:51

score 4 · Accepted Answer · answered Aug 05 '19 at 10:45

4

The definition of absolute value is

$$ |y|=\begin{cases} y &\text{if }y\geq0\\ -y &\text{if }y\leq0\end{cases} $$

Therefore, if $|y|=c$, it follows that either $y=c$ or $-y=c$. Can you finish the proof from here?

answered Aug 05 '19 at 10:45

BallBoy

I would state that $c\ge0$ because clearly $|y|=-1$ does not have the solution $y=-1$ or $y=1$. – Peter Foreman Aug 05 '19 at 13:11
@PeterForeman Well, technically it's true as I stated it -- the conditional "if $|y|=c$, then..." is always true when the antecedent $|y|=c$ is false (as it must be if $c<0$). But in practice I agree $c\geq0$ is a good condition to check. – BallBoy Aug 05 '19 at 14:16

score 3 · Answer 2 · answered Aug 05 '19 at 10:45

3

If you wish, you can write two equations: if $|a-b|=0$, then either $$ a - b = 0, $$ or $$ b - a = 0, $$ which means $$ a = b, $$ or $$ b = a $$

answered Aug 05 '19 at 10:45

guest

score 1 · Answer 3 · answered Aug 05 '19 at 10:50

Well, your question seems to have a very fundamental answer.

$1. $ If you are aware of Normed Linear Spaces then the answer follows from the definition of norm $|\cdot|$ of $\Bbb R$.

$2.$ If you do not like the answer $1.$ then recall that $|x|=\begin{cases} x,\text{ if $x\ge0$}\\ -x,\text{ if $x< 0$}\end {cases}$

Here given that $|a-b|=0\implies a-b=0\text{ or }b-a=0$ in both case $a=b$.

3 Answers3