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Mathematics LibreTexts

2.4 Arithmetic of divisibility

  • Page ID
    7428
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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.

    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 

    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?