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Homework Statement
I am working my way through the Theorem on Division by a Prime. "Let p be a prime number.Then for all integers x and y, if p divides xy, then p divides x or p divides y.
The proof is being done by complete induction.
Proof. Let x be a whole number p a prime numberk and y be an integer. Then let P(x) be the sentence "If p divides xy, then p divides x or p divides y.
Base Case: P(0) says if p divides (0*y) then p divides 0 or p divides y. This is true since anything divides 0.
Inductive Step: Assume P(0),...,P(x) are all true. We wish to show that P(x+1) is also true. Now x+1<p or x+1>=p.
Case 1: Assume x+1<p. So by the division lemma, p=q(x+1)+r for some integer q and r in {0,1,...,x}. Now multiply by y such that py=q(x+1)y+ry. Then ry=py-q(x+1)y. Thus p divides ry. Now r satisfies the induction hypothesis so P(r) is true. Thus p divides r or p divides y. If p divides r then since 0<=r<x+1<p we know that r=0. So p=q(x+1). Since p is prime then q=1 or x+1=1. But q does not = 1 since p>x+1. Thus x+1=1 and so (x+1)y=y and so since p divides (x+1)y, p must divide y. So P(x+1) is true.
Case 2: Assume x+1>=p. By the division lemma, x+1=qp+r for some integer q and r in {0,1,...,p-1}. Since p<=x+1, then p-1<=x so r in {0,1,...,x}. Now multiply by y such that (x+1)y=qpy+ry so ry=(x+1)y-qpy. Thus p divides ry. By the inductive hypothesis p must divide r or p must divide y. If p divides r then since x+1=qp+r, we must have p divides x+1. So P(x+1) is true.
Conclusion: Therefore, by complete mathematical induction, P(x) is true for all whole numbers x so it is true for all integers x.
Homework Equations
The part that I don't understand is py=q(x+1)y+ry so ry=py-q(x+1)y, so p divides ry.
The Attempt at a Solution
I don't understand how we can say that p divides ry. I know that if p divides a number say x, then there exists an integer k such that x=pk so we can say y is that integer but here we also have the extra term q(x+1)y. Doesn't that interfere with the division of p?
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