$A = \{1; 2; 3; 4; 5; 6\}$ and define $aR\mkern 1mub$ if $a$ has remainder ≤ 1 when divided by $b$.
E.g. $4R\mkern 1mu3$ since $ \ 4/3 \ $ has a remainder of $1$, but not $2R\mkern 1mu5$ since $2/5$ has a remainder of $2$.
Give the domain and range of $R$.
When I originally did this problem I reached the answer:
$\operatorname{Domain}: \{1,6\} \ $ $\operatorname{Range}: \{1,6\}$
However the memo gave the answer of:
$ \operatorname{Domain} (R) = A\ $ and $\ \operatorname{Range} (R) = A$
Can someone explain to me how the answers differ?
I was under the impression that the notation I used specified a range and therefore would make both answers identical.