# Math Help - Composition relation.

1. ## Composition relation.

What is the composition of the relation | (divides) and $\le$?

My general interpretation:

Let $R_1$ be the | relation, and $R_2$ be $\le$. $R_1 \circ R_2$ should be the set of all points $(s, u)$ where $s$ divides some $t \le u$.

If this is correct, (I'm not sure) how would I express this as set-builder notation?

$R_1 \circ R_2 = \lbrace (s, u)\; |\; s | t,\; t \le u \rbrace$ ?

2. Originally Posted by scorpion007
What is the composition of the relation | (divides) and $\le$?My general interpretation:
Let $R_1$ be the | relation, and $R_2$ be $\le$. $R_1 \circ R_2$ should be the set of all points $(s, u)$ where $s$ divides some $t \le u$.
If this is correct, (I'm not sure) how would I express this as set-builder notation?
$R_1 \circ R_2 = \lbrace (s, u)\; |\; s | t,\; t \le u \rbrace$ ?
That is very good. Except look at the order of composition. I would add the bit about 'some t' as:
$R_2 \circ R_1 = \left\{ {\left( {s,u} \right):\left( {\exists t} \right)\left[ {s|t\;\& \,t \le u} \right]} \right\}$

3. Ah, I see. Is there a reason why $\exists t$ needs to be parenthesised?

Thank you.

4. I'm sorry, why is the order of composition reversed here?

5. Originally Posted by scorpion007
Ah, I see. Is there a reason why $\exists t$ needs to be parenthesised?
Style
Be sure to see my edit about order of composition
It sould be $R_2 \circ R_1$, remember right to left.

6. Hmm.. My book states the following regarding a composition:

Given relations $R_1 \subset S \times T$ and $R_2 \subset T \times U$, the composition $R_1 \circ R_2$ consists of all pairs $(s, u) \in S \times U$ for which there is a $t \in T$ with $(s, t) \in R_1$ and $(t, u) \in R_2$.
From that definition, I do not see why the composition of | and $\le$ should be $R_2 \circ R_1$ given $R_1 = |$ and $R_2 = \le$ as previously.

Perhaps I missed something?

7. Think of function composition. Suppose that $f(x) = x^2 \;\& \,g(x) = x + 1$.
In the composition $f \circ g(x) = f\left( {g(x)} \right)$ ‘do the $g$ function’ first then we ‘do the $f$ function’:
we first add one then square the result $(x+1)^2$.

On the other hand $g \circ f(x)$ reverses the order: we first square it and add one to the result $(x^2+1)$.

That also true of relations. The order is from right to left.
So doing the divisor first and then the less than or equal to, we get $R_2 \circ R_1$.

8. Oh, so $R_1 \circ R_2$ is "All points (s, u) where $s \le t$ for some t | u? Which is not what we're after. I guess that makes sense, thanks.