What does this math symbol mean $|$?

Question

I came across this symbol in my discrete mathematics course.

For example: $6 | n(n+1)(n+2)$

Answer 1

The notation $a | b$ refers to $a$ divides $b$, i.e. there exists $w \in \mathbb{Z}$ such that $$ a \cdot q=b $$

Answer 2

$a|b$ means $a$ divides $b$, that is we can find $k \in \mathbb{Z}$ such that $b=ak$.

$$6|n(n+1)(n+2)$$ means we can find $k \in \mathbb{Z}$ such that $n(n+1)(n+2)=6k.$

A remark about the formula is that among $3$ consecutive numbers there is $a$ multiple of $3$ and between $2$ consecutive numbers, there is an even number.

Answer 3

Jonathan's answer and ajotaxe's comments are correct. $a\mid b$ means "$a$ divides $b$" (or "$a$ divides into $b$", or sometimes "$a$ divides evenly into $b$") which means there is an integer $c$ so that $b = c*a$ or in other words so $\frac ba$ is an integer.

So in you example the statement "$6|n(n+1)(n+2)$ for all $n \in \mathbb N$" is the statement that "For any natural number $n$, we will have that $6$ will divide into $n(n+1)(n+2)$". For example: if $n = 13$ then $13*14*15 = 2730$ and $6$ divides into $2730$ evenly because $2730 = 6*455$. Or in other words $\frac {13*14*15}{6} = 455\in \mathbb Z$

There is a small caveat. For $a|b$ it isn't necessary that $a$ and $b$ themselves have to be integers; just that $\frac ba$ is an integer. So for example. $\frac 18\mid \frac 34$ becasue $\frac 34 = 6*\frac 18$ and $6$ is an integer; of $\frac 23 \mid -8$ because $-8 = -12*\frac 23$ and $-12$ is an integer. But $\frac 13 \not \mid \frac 34$ because $\frac {\frac 34}{\frac 13} = \frac 94;$ or $\frac 34 = \frac 94*\frac 13$ and $ \frac 94$ is not an integer.