# 2.4 Arithmetic of divisibility

- Page ID
- 7428

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

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:**-
(by cases)**Proof:****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.

**Answer**-
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

- if \(a \mid b \) and \(a \mid c\) then \(a \mid (b+c)\).
- if \(a \mid b \) and \(a \mid c\) then \(a \mid (bc)\).
- 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?