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This warm-up exercise explores the divisibility relation defined on natural numbers (N×N) using the notation “|”. We examine how a divides b if there exists an n in natural numbers such that a·n = b. Examples include demonstrating that 2 divides 4 but not 3, highlighting the reflexive, transitive, and antisymmetric properties of this relation. Additionally, we differentiate between partial and total orders, noting why the divisibility relation is a partial order rather than a total order.
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The Divisibility Relation • Let “|” be the binary relation on N×N such that a|b (“a divides b”) iffthere 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|biff 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.
