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Mathematics LibreTexts

2.4 Arithmetic of divisibility

 

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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:(by cases)

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

Consider that a = dm and b = dk, with \(m, k \in \mathbb{Z}\).

Further, consider that c = a + b

    = d(m + k).

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

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

Consider that a = dm and c = dk, with \(m, k \in \mathbb{Z}\).

Further, consider that b = c - a

    = d(k - m).

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

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

Consider that b = dm and c = dk, with \(m, k \in \mathbb{Z}\).

Further, consider that a = c - b

    = d(k - m).

Since \((k - m)\in \mathbb{Z}\), d | 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. \( \Box\)

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 of 1:

Let \(a,b,c,d  \in \mathbb{Z}\).  

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

Consider that since a | b, b = ak, \(k  \in \mathbb{Z}\) and since a | c, c = am, \(m  \in \mathbb{Z}\).

Further, consider that 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 \(a \mid (bc)\).

Consider that since a | b, b = ak, \(k  \in \mathbb{Z}\) and since a | c, c = am, \(m  \in \mathbb{Z}\).

Further, consider that 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 \( (ac) \mid (bd)\).

Consider that since a | b, b = ak, \(k  \in \mathbb{Z}\) and since c | d, d = cm, \(m  \in \mathbb{Z}\).

Further, consider that 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?