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By Kenneth Kuttler

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Then if the equation has any rational solutions, these are of the form factor of a0 . ± factor of an Proof: Let pq be a rational solution. Dividing p and q by (p, q) if necessary, the fraction may be reduced to lowest terms such that (p, q) = 1. Substituting into the equation, an pn + an−1 pn−1 q + · · · + a1 pq n−1 + a0 q n = 0. 11. EXERCISES 35 Hence an pn = − an−1 pn−1 q + · · · + a0 q n and q divides the right side of the equation and therefore, q must divide the left side also. 3 q|an because it does not divide pn due to the fact that pn and q have no prime factors in common.

Therefore, r ≡ b − pa ≥ 0. If r ≥ a then b − pa ≥ a and so b ≥ (p + 1) a contradicting p + 1 ∈ S. Therefore, r < a as desired. To verify uniqueness of p and r, suppose pi and ri , i = 1, 2, both work and r2 > r1 . Then a little algebra shows r2 − r1 p1 − p2 = ∈ (0, 1) . 8. The case that r1 > r2 cannot occur either by similar reasoning. Thus r1 = r2 and it follows that p1 = p2 . This theorem is called the Euclidean algorithm when a and b are integers. 9 Exercises 1. The Archimedian property implies the rational numbers are dense in R.

S) or often sup (S) . b. (S) or inf (S) similarly. b. (S) means g is a lower bound for S and it is the largest of all lower bounds. If S is a nonempty subset of R which is not bounded above, this information is expressed by saying sup (S) = +∞ and if S is not bounded below, inf (S) = −∞. Every existence theorem in calculus depends on some form of the completeness axiom. 2 (completeness) Every nonempty set of real numbers which is bounded above has a least upper bound and every nonempty set of real numbers which is bounded below has a greatest lower bound.

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