# MAT 112 Integers and Modern Applications for the Uninitiated

## Section4.1Divisibility

We begin by introducing terminology.

### Definition4.1.

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 $$b\le a\text{.}$$

### Problem4.2.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 $$a = b \cdot q\text{.}$$
1. $$a=30$$ and $$b=10$$
2. $$a=2$$ and $$b=46$$
3. $$a=29$$ and $$b=4$$
Solution.
In each case we consider the remainder $$a\fmod b$$ of the division $$a$$ and $$b\text{.}$$
1. We compute $$30 \fmod 10 = 0\text{.}$$ So $$10$$ divides $$30\text{.}$$ Furthermore, we have that $$30\fdiv 10=3$$ so $$30=10\cdot 3\text{.}$$
2. 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 $$46 = 2\cdot 23\text{.}$$)
3. 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.

### Checkpoint4.3.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 $$a\text{.}$$

### Definition4.4.

Let $$a$$ and $$b$$ be integers. If $$b$$ divides $$a$$ we also say:
1. $$a$$ is divisible by $$b$$
2. $$a$$ is a multiple of $$b$$
3. $$b$$ is a divisor of $$a$$
4. $$b$$ 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.

### Checkpoint4.5.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.)
1. 20 is a factor of 220
2. 220 divides 20
3. 20 divides 220
4. 220 is a divisor of 20
5. 20 is a factor of 220
6. 20 is a divisor of 220
If a number divides two other numbers, it divides their sum.

### 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 $$a+c\text{.}$$

### Example4.7.

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