Answer
- **\(m = n\)?** No; for instance, \(7 \neq 2\) even though \(7 \mod 5 = 2 \mod 5\).
- **\(m - n\) is divisible by \(d\)?** Yes; by writing \(m\) and \(n\) in the form \(qd + r\) and \(pd + r\), we see that \(m - n = (q-p)d\), which is divisible by \(d\).
Work Step by Step
We are given that \(m\), \(n\), and \(d\) are integers with \(d > 0\) and that
\[
m \mod d = n \mod d.
\]
### Does it necessarily follow that \(m = n\)?
**Answer:** No, it does not necessarily follow that \(m = n\).
**Proof/Counterexample:**
When we say that \(m \mod d = n \mod d\), we mean that the remainders when \(m\) and \(n\) are divided by \(d\) are equal. However, \(m\) and \(n\) may differ by a multiple of \(d\). For example, let
\[
m = 7,\quad n = 2,\quad d = 5.
\]
Then,
\[
7 \mod 5 = 2\quad \text{and} \quad 2 \mod 5 = 2.
\]
Thus, \(7 \mod 5 = 2 \mod 5\), but clearly \(7 \ne 2\). Hence, \(m \mod d = n \mod d\) does not imply that \(m = n\).
### Does it necessarily follow that \(m - n\) is divisible by \(d\)?
**Answer:** Yes, \(m - n\) is divisible by \(d\).
**Proof:**
Since \(m \mod d = n \mod d\), we can express \(m\) and \(n\) in the form:
\[
m = qd + r \quad \text{and} \quad n = pd + r,
\]
where \(q\) and \(p\) are integers, and \(r\) is the common remainder with \(0 \le r < d\).
Subtract \(n\) from \(m\):
\[
m - n = (qd + r) - (pd + r) = (q - p)d.
\]
Since \(q - p\) is an integer, it follows that \(m - n\) is an integer multiple of \(d\), i.e., \(m - n\) is divisible by \(d\).
