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# 2.4 Arithmetic of divisibility

• • Contributed by Pamini Thangarajah
• Associate Professor (Mathematics & Computing) at Mount Royal University

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Theorem: Divisibility theorem I

Let $$a,b,c \in \mathbb{Z}$$ such that $$a+b=c$$. If $$d \in \mathbb{Z_+}$$ divides any two of $$a,b$$ and $$c$$, then $$d$$ divides the third one.

Proof:

Proof:(by cases)

Case 1: Suppose $$d \mid a$$ and $$d \mid b$$. We shall show that $$d \mid c$$.

Since $$d \mid a$$ and $$d \mid b$$, $$a = dm$$ and $$b = dk$$ , for some $$m, k \in \mathbb{Z}$$.

Consider, $$c = a + b = d(m + k).$$

Since $$(m + k )\in \mathbb{Z}$$, $$d \mid c$$.

Case 2: Suppose $$d \mid a$$ and $$d \mid c$$. We shall show that $$d \mid b$$.

Since $$d \mid a$$ and $$d \mid c$$, $$a = dm$$ and $$c = dk$$, for some $$m, k \in \mathbb{Z}$$.

Consider, $$b = c - a= d(k - m).$$

Since $$(k - m)\in \mathbb{Z}$$,$$d \mid b$$.

Case 3: Suppose $$d \mid b$$ and $$d \mid c$$. We shall show that $$d \mid a$$.

Since $$d \mid b$$ and $$d \mid c$$, $$b = dm$$ and $$c = dk$$, with $$m, k \in \mathbb{Z}$$.

Consider, $$a = c - b= d(k - m).$$

Since $$(k - m)\in \mathbb{Z}$$, $$d \mid a$$.

Having examined all possible cases, given a + b = c, then if $$d \in \mathbb{Z_+}$$ divides any two of $$a,b$$ and $$c$$, then $$d$$ divides the third one. $$\Box$$

Exercise $$\PageIndex{1}$$

Let $$a,b,c \in \mathbb{Z_+}$$ such that $$a+b=c$$. If $$d \in \mathbb{Z_+}$$ divides any two of $$a,b$$ and $$c$$, then $$d$$ divides the third one.

use cases to prove.

Theorem: Divisibility theorem II

Let $$a,b,c \in \mathbb{Z}$$ such that $$a \mid b$$. Then $$a \mid bc$$.

Proof:

Let $$a,b,c \in \mathbb{Z}$$ such that $$a \mid b$$.

We shall show that $$a \mid bc$$.

Consider that since a | b, b = ak, $$k \in \mathbb{Z}$$.

Further consider bc = a(ck).

Since $$ck \in \mathbb{Z}$$, a | bc.$$\Box$$

Theorem:

Let $$a,b,c,d \in \mathbb{Z}$$. Then

1. if $$a \mid b$$ and $$a \mid c$$ then $$a \mid (b+c)$$.
2. if $$a \mid b$$ and $$a \mid c$$ then $$a \mid (bc)$$.
3. if $$a \mid b$$ and $$c \mid d$$ then $$(ac) \mid (bd)$$.
Proof:

Proof of 1:

Let $$a,b,c,d \in \mathbb{Z}$$.

We shall show that if $$a \mid b$$ and $$a \mid c$$ then $$a \mid (b + c)$$.

Since $$a | b, b = ak, \,k \in \mathbb{Z}$$ and since $$a | c, c = am, \, m \in \mathbb{Z}$$.

Consider, $$b + c = a(k + m)$$.

Since $$k + m \in \mathbb{Z}$$, $$a | (b + c)$$.$$\Box$$

Proof of 2:

Let $$a,b,c,d \in \mathbb{Z}$$.

We shall show that if $$a \mid b$$ and $$a \mid c$$ then $$a \mid (bc)$$.

Since $$a | b, b = ak, \,k \in \mathbb{Z}$$ and since $$a | c, c = am, \, m \in \mathbb{Z}$$.

Consider $$bc = a(akm).$$

Since $$akm \in \mathbb{Z}$$, $$a | (bc)$$.$$\Box$$

Proof of 3:

Let $$a,b,c,d \in \mathbb{Z}$$.

We shall show that if $$a \mid b$$ and $$c \mid d$$ then $$(ac) \mid (bd)$$.

Since $$a | b, b = ak, \,k \in \mathbb{Z}$$ and since $$a | c, c = am, \, m \in \mathbb{Z}$$.

Consider $$bd = (ak)(cm) = ac(km).$$

Since $$km \in \mathbb{Z}$$,$$(ac) | (bd)$$.$$\Box$$

Example $$\PageIndex{1}$$:

Let $$a,b,c,d \in \mathbb{Z}$$ such that $$a \mid b$$ and $$c \mid d$$. Is it always true that $$(a+c) \mid (b+d)$$ ?

In other words, if a teacher was able to share 6 apples among 6 boys and 8 oranges among 4 girls equally, is it possible for the teacher to share the 14 fruits with the 10 children equally?