$A = \{1; 2; 3; 4; 5; 6\}$ and define $aRb$ if a has remainder ≤ 1 when divided by b. Give the domain and range of $R$.

Question

$A = \{1; 2; 3; 4; 5; 6\}$ 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.

Answer 1

$\{1,6\}$ consists of two elements $1$ and $6$.

$A=\{i \in \mathbb{N}: 1 \le 1 \le 6\}=\{1,2,3,4,5,6\}$ consists of $6$ elements. Hence they are different.

Answer 2

The domain is the set of numbers that can appear as $a$ in $aRb$. You said yourself that $4R3$ is valid, so clearly there're things missing in your domain. In fact you can consider $xRx$ for every $x\in\{1,2,3,4,5,6\}$ since $x$ divided by itself has a remainder of $0\le 1$, so the domain is the whole set.

On the other hand, it happens the same with the range: it's the set of numbers that can appear as $b$ in $aRb$. Given that you could consider $xRx$ with $x$ being any number, your range is the whole set too.