Answer
- **\((m \cdot n) \mod d = a \cdot b\)?**
Not necessarily; this holds only if \( a \cdot b < d \). In general, \( a \cdot b \) might be larger than or equal to \( d \), so its remainder modulo \( d \) will be less than \( d \).
- **\((m \cdot n) \mod d = (a \cdot b) \mod d\)?**
Yes; the proof shows that the remainder of \( m \cdot n \) modulo \( d \) is exactly the remainder of \( a \cdot b \) when divided by \( d \).
Work Step by Step
Let \( m, n, a, b,\) and \( d \) be integers with \( d > 0 \), and suppose that
\[
m \mod d = a \quad \text{and} \quad n \mod d = b.
\]
This means we can write
\[
m = qd + a \quad \text{and} \quad n = rd + b,
\]
for some integers \( q \) and \( r \), with
\[
0 \le a < d \quad \text{and} \quad 0 \le b < d.
\]
### Multiplicative Property: Derivation
Consider the product \( m \cdot n \):
\[
m \cdot n = (qd + a)(rd + b) = qrd^2 + (qb + ra)d + ab.
\]
Notice that both \( qrd^2 \) and \( (qb + ra)d \) are divisible by \( d \). Thus, when we take the remainder upon division by \( d \), these terms vanish, and we obtain:
\[
(m \cdot n) \mod d = ab \mod d.
\]
### 1. Is \((m \cdot n) \mod d = a \cdot b\)?
The equality
\[
(m \cdot n) \mod d = a \cdot b
\]
would mean that the product \( a \cdot b \) itself is the remainder when dividing \( m \cdot n \) by \( d \). However, by the definition of the modulo operation, the remainder must be an integer between \( 0 \) and \( d-1 \).
- **If \( a \cdot b < d \):** Then indeed \( ab \mod d = ab \), and the equality holds.
- **If \( a \cdot b \ge d \):** Then \( ab \mod d \) is the unique integer in the set \( \{0,1,\dots, d-1\} \) that is congruent to \( ab \) modulo \( d \); in this case, \( ab \mod d \) is not equal to \( ab \).
Thus, in general, **\((m \cdot n) \mod d\) is not necessarily equal to \(a \cdot b\)**; it equals \(a \cdot b\) only when \(a \cdot b < d\).
**Example:**
Let \( d = 10 \), \( a = 7 \), and \( b = 6 \). Then \( a \cdot b = 42 \), but
\[
42 \mod 10 = 2.
\]
So \((m \cdot n) \mod 10 = 2 \neq 42\).
### 2. Is \((m \cdot n) \mod d = (a \cdot b) \mod d\)?
From our derivation,
\[
m \cdot n = \underbrace{[qrd^2 + (qb + ra)d]}_{\text{divisible by }d} + ab.
\]
Taking modulo \( d \), the divisible part drops out, leaving:
\[
(m \cdot n) \mod d = (ab) \mod d.
\]
This equality holds regardless of the size of \( a \cdot b \).
Thus, **\((m \cdot n) \mod d = (a \cdot b) \mod d\)** always.
