Posted: 4/23/2007 12:27:41 AM EDT
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Suppose that P(n) is a propositional function. Determine for which nonnegative integers n the statement P(n) must be true if: a) P(0) is true; for all nonnegative integers n, if P(n) is true, then P(n+2) is true. b) P(0) is true; for all nonnegative integers n, if P(n) is true, then P(n+3) is true. c) P(0) and P(1) are true; for all nonnegative integers n, if P(n) and P(n+1) are true, then P(n+2) is true. d) P(0) is true; for all nonnegative integers n, if P(n) is true, then P(n+2) and P(n+3) is true. Find the flaw with teh following "proof" that a^n = 1 for all nonnegative integers n, whenever a is a nonzero real number. Base Step: a^0 = 1 is true by the definition of a^0. Inductive Step: a^(k+1) = (a^k * a^k) / a^(k-1) = (1 * 1) / 1 = 1 The well-ordering property can be used to show that there is a unique greatest common divisor of two positive integers. Let a and b be positive integers, and let S be the set of positive integers of the form: a *s + b * t where s and t are integers. a) Show that S is nonempty. b) Use the well-ordering property to show that S has a smallest element c. c) Show that if d is a common divisor of a and b, then d is a divisor of c. d) Show that c | a and c | b. Any hints of help would be much appreciated on this. |
