Answer
1. \((L \circ M)(12) = 4,\quad (M \circ L)(12) = 4.\)
2. \((L \circ M)(9) = 16,\quad (M \circ L)(9) = 1.\)
3. No, \(\;L \circ M\) is **not** the same as \(M \circ L\).
Work Step by Step
## 1. The Functions
\(L(a) = a^2.\)
\(M(a) = a \bmod 5.\)
For any integer \(a\), \(M(a)\) is the remainder when \(a\) is divided by $5$.
## 2. Compositions
### 2.1 \((L \circ M)(a)\)
\[
(L \circ M)(a) \;=\; L\bigl(M(a)\bigr)
\;=\; \bigl(M(a)\bigr)^2
\;=\; \bigl(a \bmod 5\bigr)^2.
\] Effectively, you first reduce \(a\) mod $5$, then square the result.
### 2.2 \((M \circ L)(a)\)
\[
(M \circ L)(a) \;=\; M\bigl(L(a)\bigr)
\;=\; (\,a^2) \bmod 5.
\]
That is, you first square \(a\), then take that result mod $5$.
## 3. Evaluate at $12$ and $9$
#### (a) \((L \circ M)(12)\)
1. \(M(12) = 12 \bmod 5 = 2.\)
2. Then \(L(2) = 2^2 = 4.\)
Hence \((L \circ M)(12) = 4.\)
#### (b) \((M \circ L)(12)\)
1. \(L(12) = 12^2 = 144.\)
2. Then \(M(144) = 144 \bmod 5 = 4.\)
Hence \((M \circ L)(12) = 4.\)
#### (c) \((L \circ M)(9)\)
1. \(M(9) = 9 \bmod 5 = 4.\)
2. Then \(L(4) = 4^2 = 16.\)
Hence \((L \circ M)(9) = 16.\)
#### (d) \((M \circ L)(9)\)
1. \(L(9) = 9^2 = 81.\)
2. Then \(M(81) = 81 \bmod 5 = 1.\)
Hence \((M \circ L)(9) = 1.\)
## 4. Comparison
- For \(a=12\), both compositions yield $4$.
- For \(a=9\), \((L \circ M)(9)=16\) while \((M \circ L)(9)=1.\)
Therefore, the two functions \((L \circ M)\) and \((M \circ L)\) do **not** always agree. In fact, just from the \(a=9\) example, we see:
\[
(L \circ M)(9) \;=\; 16
\quad\neq\quad
1 \;=\; (M \circ L)(9).
\] Hence, **\(L \circ M \neq M \circ L\)** as functions on \(\mathbb{Z}\).
