Theorem 7.10. For every a, b E N exactly one of the following must hold: (1) a = b (2) a> b (3) b> a.

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Prove the following piece of Thm 7.10: For
every a, b E N at most of one of the three
alternatives of Thm 7.10 holds. Hint: Use Th
Transcribed Image Text:Prove the following piece of Thm 7.10: For every a, b E N at most of one of the three alternatives of Thm 7.10 holds. Hint: Use Th
Theorem 7.10. For every a, b € N exactly one of the following must
hold:
=
b
(1) a
(2) a > b
(3) b> a.
Proof. (i) It follows from Theorem 7.7 that for every a, b E N at most
one of the above conditions hold. Details left as HW.
(ii) For every a, b E N at least one of the above conditions hold: For
any a, let Ca be the set of natural numbers b such that at least one
of the conditions (1)-(3) holds. Then one can prove that 1 € Ca, and
that for any n if n € Ca then n + 1 € Ca. (Details may be discussed in
class or HW). Therefore, by the Principle of Mathematical Induction
Ca N. That means that for every a and b, at least one of the above
conditions (1)-(3) hold.
Transcribed Image Text:Theorem 7.10. For every a, b € N exactly one of the following must hold: = b (1) a (2) a > b (3) b> a. Proof. (i) It follows from Theorem 7.7 that for every a, b E N at most one of the above conditions hold. Details left as HW. (ii) For every a, b E N at least one of the above conditions hold: For any a, let Ca be the set of natural numbers b such that at least one of the conditions (1)-(3) holds. Then one can prove that 1 € Ca, and that for any n if n € Ca then n + 1 € Ca. (Details may be discussed in class or HW). Therefore, by the Principle of Mathematical Induction Ca N. That means that for every a and b, at least one of the above conditions (1)-(3) hold.
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