Divisibility

Section 4.1 Divisibility

We begin by introducing terminology.

Definition 4.1.

Suppose that for integers \(a\) and \(b\text{,}\) there is an integer \(q\) such that \(a = b \cdot q\text{.}\) Then \(b\) divides \(a\text{.}\)

By Definition 4.1 if \(b\) divides \(a\text{,}\) then \(a = b \cdot q\) for some integer \(q\text{.}\) Then we have \(a = b \cdot q + 0\) so that in particular \(a \fmod b = 0\text{.}\) If \(b\) does not divide \(a\text{,}\) then \(a \fmod b\ne 0\text{.}\) It follows immediately that if \(b\) divides \(a\) then \(b\le a\text{.}\)

Problem 4.2. Determine divisibility.

For the given values of \(a\) and \(b\text{,}\) determine whether or not \(b\) divides \(a\text{.}\) If \(b\) divides \(a\text{,}\) determine the integer \(q\) such that \(a = b \cdot q\text{.}\)

\(a=30\) and \(b=10\)
\(a=2\) and \(b=46\)
\(a=29\) and \(b=4\)

Solution.

In each case we consider the remainder \(a\fmod b\) of the division \(a\) and \(b\text{.}\)

We compute \(30 \fmod 10 = 0\text{.}\) So \(10\) divides \(30\text{.}\) Furthermore, we have that \(30\fdiv 10=3\) so \(30=10\cdot 3\text{.}\)
We compute \(2 \fmod 46 = 2\ne 0\text{.}\) So \(46\) does not divide \(2\text{.}\) (Be careful not to mix up \(a\) and \(b\) during the division or the conclusion. The order matters. It turns out that \(2\) does divide \(46\) since \(46 = 2\cdot 23\text{.}\))
We compute \(29 \fmod 4 = 1\ne 0\text{.}\) So \(4\) does not divide \(29\text{.}\)

Use the method from the solution of Problem 4.2 to determine divisibility in Checkpoint 4.3.

Checkpoint 4.3. Determine divisibility.

Compute the remainder and complete the statement about divisibility

Because \(34 \bmod 15\)= we have that \(15\)

select
divides
does not divide

\(34\text{.}\)

Because \(51 \bmod 7\)= we have that \(7\)

select
divides
does not divide

\(51\text{.}\)

Because \(9 \bmod 3\)= we have that \(3\)

select
divides
does not divide

\(9\text{.}\)

Because \(20 \bmod 4\)= we have that \(4\)

select
divides
does not divide

\(20\text{.}\)

Because \(60 \bmod 17\)= we have that \(17\)

select
divides
does not divide

\(60\text{.}\)

Because \(18 \bmod 16\)= we have that \(16\)

select
divides
does not divide

\(18\text{.}\)

There are several other formulations for \(b\) divides \(a\text{.}\)

Definition 4.4.

Let \(a\) and \(b\) be integers. If \(b\) divides \(a\) we also say:

\(a\) is divisible by \(b\)
\(a\) is a multiple of \(b\)
\(b\) is a divisor of \(a\)
\(b\) is a factor of \(a\)

In Checkpoint 4.5 we give several statements about divisibility formulated in various ways. Decide which statements are true and which statements are false.

Checkpoint 4.5. Decide which statements are correct.

Enter T or F depending on whether the statement is a true proposition or not. (You must enter T or F -- True and False will not work.)

20 is a factor of 220
220 divides 20
20 divides 220
220 is a divisor of 20
20 is a factor of 220
20 is a divisor of 220

If a number divides two other numbers, it divides their sum.

Theorem 4.6.

Let \(b\) be a natural number and let \(a\) and \(c\) be integers. If \(b\) divides \(a\) and \(b\) divides \(c\text{,}\) then \(b\) divides \(a+c\text{.}\)

Proof.

As \(b\) divides \(a\text{,}\) there is an integer \(q\) such that \(a=b\cdot q\text{.}\) As \(b\) divides \(c\text{,}\) there is an integer \(s\) such that \(c=b\cdot s\text{.}\) With substitution and the distributive property we obtain

\begin{equation*} a+c=(b\cdot q)+(b\cdot s)=b\cdot(q+s)\text{.} \end{equation*}

Thus \(a+c\) is a multiple of \(b\) which means that \(b\) divides \(a+c\text{.}\)

Example 4.7.

Let \(b:=10\) and \(a:=100\) and \(c:=1000\text{.}\) Then \(b\) divides \(a\) and \(b\) divides \(c\text{.}\) Also \(b\) divides \(a+c=1100\text{.}\)

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