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We prove p(x, y) : <=> x+y is even is an equivalence relation on N.

  1. p is clearly a binary relation on N.
  2. p is reflexive on N: Take arbitrary x in N. We have to show x+x is even, i.e, 2x is even.
  3. p is symmetric on N: Take x in N, y in N. We assume x+y is even. Then y+x is even.
  4. p is transitive on N: Take arbitrary x in N, y in N, and z in N. We assume
    (1) x+y is even /\  y+z is even.
    We have to show (2) x+z is even. From (1), we have some a in N and b in N such that
    (3) 2a = x+y /\  2b = y+z.
    Thus we know (2) because of
    x+z = (x+y) + (y+z) - 2y = 2a + 2b - 2y = 2(a+b-y).

Author: Wolfgang Schreiner
Last Modification: January 12, 2000

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