Answer
- E is Reflexive
- E is Symmetric
- E is Transitive
Work Step by Step
--E is reflexive:
[We must show that for all integers m, m E m.]Suppose m is any integer. Since m−m =0 and 2|0, we have that 2|(m−m). Consequently, m E m by definition of E.
-- E is symmetric:
[We must show that for all integers m and n, if m E n then n E m.]Suppose m and n are any integers such that m En. By definition of E, this means that 2|(m−n), and so, by definition of divisibility, m−n =2 k for some integer k. Now n−m =− (m−n). Hence, by substitution,
n−m =− (2k) =2(−k). It follows that 2|(n−m) by definition of divisibility (since −k is an integer), and thus n E m by definition of E.
--E is transitive:
[We must show that for all integers m,n and p if m E n and n E p then m E p.] Suppose m,n, and p are any integers such that m E n and n E p. By definition of E this means that 2|(m−n) and 2|(n− p), and so, by definition of divisibility, m−n =2 k for some integer k and n− p =2 l for some integer l. Now m− p = (m−n)+(n− p). Hence, by substitution, m− p =2 k+ 2 l =2(k+l). It follows that 2|(m− p) by definition of divisibility(since k+l is an integer),and the s m E p by definition of E.
