Answer
- **\((m+n) \mod d = a+b\)?**
Not necessarily. It holds when \(a+b < d\), but if \(a+b \ge d\), then \((m+n) \mod d \neq a+b\) (since the remainder must be less than \(d\)).
- **\((m+n) \mod d = (a+b) \mod d\)?**
Yes. The proof shows that for any integers \(m, n, a, b,\) and \(d>0\) with \(m \mod d = a\) and \(n \mod d = b\), we have
\[
(m+n) \mod d = (a+b) \mod d.
\]
Work Step by Step
Let \( m, n, a, b, d \) be integers with \( d > 0 \) and suppose
\[
m \mod d = a \quad \text{and} \quad n \mod d = b.
\]
This means that there exist integers \( q \) and \( r \) such that
\[
m = qd + a \quad \text{and} \quad n = rd + b,
\]
with \(0 \le a, b < d\).
### Part 1: Is \((m+n) \mod d = a+b\)?
Compute \(m+n\):
\[
m+n = qd + a + rd + b = (q+r)d + (a+b).
\]
When we take \((m+n) \mod d\), we are interested in the remainder when \(m+n\) is divided by \(d\). The term \((q+r)d\) is divisible by \(d\), so
\[
(m+n) \mod d = (a+b) \mod d.
\]
Notice that if \(a+b < d\), then \((a+b) \mod d = a+b\). However, if \(a+b \ge d\), then \((a+b) \mod d\) is not equal to \(a+b\) but rather to \(a+b-d\) (or more generally, the unique number in the set \(\{0,1,\ldots,d-1\}\) that is congruent to \(a+b\) modulo \(d\)).
Thus, in general, **\((m+n) \mod d\) is not necessarily equal to \(a+b\)**; it equals \(a+b\) only in the case when \(a+b < d\).
### Part 2: Is \((m+n) \mod d = (a+b) \mod d\)?
From the computation above, we obtained
\[
(m+n) = (q+r)d + (a+b).
\]
Taking modulo \(d\), we get
\[
(m+n) \mod d = (a+b) \mod d.
\]
This is true regardless of the size of \(a+b\).
