The language of divisibility #

The following 7 statements are equivalent:

1. 18 is divisible by 6
2. 18 is a multiple of 6
3. 6 is a factor or divisor of 18
4. 6 goes into 18, or 6 divides 18
5. \( \frac{18}{6} \) is a whole number
6. 18 is equal to 6 times a whole number, which is the same as saying there is \( 18 = 6k \) for some \( k \in \mathbb{Z} \)
7. In the long division of 18 by 6, the remainder is 0.

Note: \( 6\ |\ 18 \) is read as "6 divides 18".

Some more statements that are equivalent:

1. 19 is not divisible by 6
2. 19 is not a multiple of 6
3. 6 is not a factor or divisor of 19
4. 6 does not go into 19, or 6 does not divide 18
5. \( \frac{19}{6} \) is not a whole number
6. 19 is not equal to 6 times a whole number, which is the same as saying \( 19 \not = 6k \) for any \( k \in \mathbb{Z} \)
7. In the long division of 19 by 6, the remainder is not 0

So, we can say that \( 6 \not | \ 19 \) .

Divisbility theorem #

Theorem. If \( 7 | a \) , and \( 7 | b \) , then \( 7 | a \pm b \) .

In the above theorem, it is understood that \( a, b \in \mathbb{Z} \) .

Proof. We are given that \( 7 | a \) and \( 7 | b \) , so \[\begin{aligned} a &= 7 k_1 \\ b &= 7 k_2 \end{aligned}\]

where \( k_1, k_2 \in \mathbb{Z} \) .

So,

\[\begin{aligned} a + b &= 7 k_1 + 7 k_2 \\ &= 7 ( k_1 + k_2 ) \end{aligned}\]

Note that \( k_1 + k_2 \) is also \( \in \mathbb{Z} \) .

We can also show the same with a minus, therefore we are done. By statement 6 at the top we are done.

\( \square \)

A more general divisbility theorem #

This is a generalization of the above theorem.

Theorem. If \( c | a \) , and \( c | b \) , then \( c | ma \pm nb \) , where \( a, b, c, m, n \in \mathbb{Z} \) .

Using the fundamental theorem of arithmetic #