Composition relation.

**scorpion007** · March 8th 2009, 04:19 AM

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?

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

**Plato** · March 8th 2009, 04:30 AM

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:
$R_2 \circ R_1 = \left\{ {\left( {s,u} \right):\left( {\exists t} \right)\left[ {s|t\;\& \,t \le u} \right]} \right\}$

**scorpion007** · March 8th 2009, 04:34 AM

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

Thank you.

**scorpion007** · March 8th 2009, 04:37 AM

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

**Plato** · March 8th 2009, 04:39 AM

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.

**scorpion007** · March 8th 2009, 05:10 AM

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?

**Plato** · March 8th 2009, 05:27 AM

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 $R_2 \circ R_1$ .

**scorpion007** · March 8th 2009, 03:39 PM

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.

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