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anybody here a math major?


1800bigk

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Let a and b be nonzero integers. Prove that there is a

natural number m such that

(i) ajm and bjm, and

(ii) if c is an integer such that ajc and bjc, then mjc.

Proof. Let S = fn 2 N : ajn and bjng. Since ab 2 S, S 6= ;. Thus,

by the Least-Natural-Number-Principle, there is a smallest element of

S. Let m be this smallest element. Now suppose ajc and bjc. Then

c 2 S and so c m. So by the division algorithm for integers, there

exist integers p; r such that 0 r < m such that c = mq+r. Since ajm

and bjm, then ajr and bjr. If c 6= 0, then r is an element of S smaller

than m. Thus, r = 0 and mjc.

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yeah its a math course its called Real Analysis, its a real pain in the ass.

putting the opposite of what I want to prove first is called proof by contrapositve ,and with this question since what I have to prove is compound, it makes it messy to do it that way.

Edited by 1800bigk
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