For $a, b\in R$, define $a$ ~ $b$ to mean that $|a-b|<1$. I am trying to prove that this relation is symmetric, and it seems quite clear to me but I am having difficulty constructing my argument.
If $a$ ~ $b$ and $b<a$ then $|a-b|=a-b<1$ and $|b-a|=a-b<1$ meaning $b$ ~ $a$
Similarly, if $a<b$ then $|a-b|=b-a<1$ and $|b-a|=b-a<1$ also meaning $b$ ~ $a$
Does this suffice as it seems quite trivial, but maybe the proof is just trivial?
Thanks!
You needn't split the proof in cases. It suffices to say that if $a\sim b$ then $$|b-a|=|a-b|<1$$ and therefore $b\sim a$.