Show that if a, b, c, and d are integers, where a ≠ 0, such

that a ∣ c and b ∣ d, then ab ∣ cd.

c = ma

d = nb
cd = ma*nb = (mn)(ab)
so ab|cd

To prove that ab | cd, we need to show that cd is divisible by ab.

Given that a | c and b | d, we can write c = ka and d = lb for some integers k and l.

Now let's consider cd:

cd = (ka)(lb) (substituting the values of c and d)

cd = kl(ab) (rearranging the terms)

Since kl is an integer and ab is the product of integers, kl(ab) is also an integer.

Therefore, cd is divisible by ab, which implies that ab | cd.

To show that if a, b, c, and d are integers, where a ≠ 0, such that a ∣ c and b ∣ d, then ab ∣ cd, we need to prove that the product of a and b divides the product of c and d.

To prove this, we can use the definition of divisibility.

1. Since a ∣ c, there exists an integer k such that c = ak. This means that c is a multiple of a.

2. Similarly, because b ∣ d, there exists an integer m such that d = bm. This implies that d is a multiple of b.

3. Now, let's calculate the product cd:
cd = (ak)(bm)
= abkm

4. We can rewrite abkm as (ab)(km), which shows that ab is a factor of the product cd. In other words, ab ∣ cd.

Therefore, we have shown that if a, b, c, and d are integers, where a ≠ 0, such that a ∣ c and b ∣ d, then ab ∣ cd.