Digraphs and Relations Warm Up

The Divisibility Relation Let “|” be the binary relation on N×N such that a|b (“a divides b”) iff there is an n ∈ N such that a∙n=b. Examples: –2|4 but not 2|3 and not 4|2 –1|a for any a since 1∙a=a –What about 0|a? –What about a|0? Show that “|” is a partial order but not a total order. What does that mean? Reflexive, transitive, antisymmetric But not true that for any a and b, either a|b or b|a

a|b iff for some n ∈ N, a∙n = b Reflexive? a|a for any a since a∙1=a. Transitive? If a|b and b|c, then there exist n, m ∈ N such that a∙n=b and b∙m=c. Then a∙(nm)=c so a|c. Antisymmetric? Suppose a|b and a≠b. We want to say “then a<b” but that is not right! Why? If b≠0 then a<b (why?) so it cannot be that b|a. If b=0 then NOT b|a since 0|a only if a=0.

So “|” is a partial order. It is not a total order because, for example, neither 2|3 nor 3|2 is true.

Some Properties of “|”as a partial order There is an infinite set of numbers no two of which bear “|” to each other –Namely the prime numbers For every pair of numbers a and b, there is a number c such that a|c and b|c and c is minimal (that is, if a|d and b|d then c≤d). –Namely the least common multiple of a and b.

FINIS