Skip to main content
Mathematics LibreTexts

2.4: Arithmetic of divisibility

  • Page ID
    7428
  • This page is a draft and is under active development. 

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Thinking out loud

    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?

    Theorem: Divisibility theorem I (BASIC)

    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, \( c=a+b=dm+dk=d(m+k).\)

    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. \( \Box\)

    Theorem: Divisibility theorem II (MULTIPLE)

    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: Arithmetic of Divisibility

    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?

    Example \(\PageIndex{2}\):

    Let \(a\) and \(b\) be positive integers such that \(7 | (a+2b+5)\) and \(7 | (b−9)\). Prove that \(7 | (a + b).\)

    Solution

    Let \(a, b ∈ ℤ+ s.t. 7 ∣ (a+2b+5) \) and \( 7 ∣ (b-9).\)

    Consider \( a+2b+5 =7(m), m \in \mathbb{Z}.\)

    Further consider \(b-9 =7(k), k \in \mathbb{Z}.\)

    Next consider \(a+2b+5 -(b-9)=7m-7k.\)

    \(a+b+14 =7(m-k).\)

    \( a+b=7(m-k-2), m-k-2 \in \mathbb{Z}.\)

    Thus, \(7 | (a + b).□ \)

    Example \(\PageIndex{3}\):

    Let \(a\) and \(b\) be positive integers. Prove or disprove the following statements:

    1. If \( a|b\) then \(a^2|b^3\).

    2. If \( a^2|b^3\) then \(a|b\).

    Solution

    1. This statement is true.

    Proof:

    Let \(a\) and \(b\) be positive integers such that  \( a|b\). Then \(b=am, m \in \mathbb{Z_+}\).

    Consider \(b^3=(am)^3=a^3m^3=a^2(am^3).\)

    Since \(am^3 \in \mathbb{Z_+}, a^2|b^3.\) 

    2. This statement is false. Counterexample:

    Choose \(a=3^3, b=3^2\). Then  \(a^2=3^6=b^3.\)

    Hence \( a^2|b^3\) but  \(a\not \mid b\).

     

     


    This page titled 2.4: Arithmetic of divisibility is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Pamini Thangarajah.

    • Was this article helpful?