1. ## New to proofs...

If $a \leq b$ and $b \leq a+c$, then $|a-b| \leq c$.

Ok this is what I need to prove. I am fairly new to this so i'm not too sure what i have done is right, but here is what i have.

$a \leq b \leq a+c$ (subtract a)

$0 \leq b-a \leq c$ (multiply by -1)

$0 \geq a-b \geq -c$ so (??? not sure after this point ???) $c \geq 0$. therefor $c \geq 0 \geq a-b \geq -c$ so $c \geq a-b \geq -c$ which equals $|a-b| \leq c$.

Please if i have don something wrong correct me or if there is a better way of going about this that would be helpful too. Thanks for you help.

2. You have gotten to the point $\displaystyle 0 \leq b-a \leq c$.

Since $\displaystyle b-a$ is a positive value, it's equal to its absolute value.

So $\displaystyle |b - a| \leq c$.

But $\displaystyle |b-a| = |a -b|$.

Therefore $\displaystyle |a - b| \leq c$.

Q.E.D.