Download e-book for kindle: A radical approach to algebra by Mary W Gray

By Mary W Gray

ISBN-10: 020102568X

ISBN-13: 9780201025682

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N(n+ l)]2 Prove by induction that (I 3. Let A, ,.. , A,, be subsets of a set. Prove by induction the following laws: (a) Rational, real, and complex numbers 35 (b) 4. If a,b,c E Z such that aib and aic, show that + nc, for m,n E Z. 5. Let a,b E Z. ) of a and b, denoted by [a, b], is defined to be a positive integer d such that (1) aidand bid, whenever aix and bix, then dix. factorizationsofa Suppose a = and b = and b as a product of primes, where 0. Prove that (a) [a,b] is unique. (b) [a,b] = where g, = max(e,,f).

N X S —. S. there exists a unique mappingf N —' S such that f(l)=a and f(n+ foralln€N. Integers, real numbers, and complex numbers 34 Proof We first show that if a mapping satisfying the given conditions exists, then it is unique. Let fg: N —' S be two such mappings. Then f(l)=a=g(l). Further, iff(n)=g(n) forany nEN, thenf(n+ l)= cb(n,f(n)) = = g(n + 1). Hence, by the induction principle, f(n) = g(n) for all n E N. To prove the existence off let M be the set of all subsets X of N X S satisfying the following conditions: (l,a)EX,andif(n,x)EX,then(n+ N.

1, (n + I)! = (n + 1)'n! for all n N. Now the question is whether this is a valid procedure for describing a mapping. In other words, does there exist a unique mapping f: N —, S satisfying the conditions stated earlier? Intuitively, the answer seems to be yes. We prove it formally in the following theorem. 7 Theorem (recursion theorem). Let S be a set and a S. : N X S —. S. there exists a unique mappingf N —' S such that f(l)=a and f(n+ foralln€N. Integers, real numbers, and complex numbers 34 Proof We first show that if a mapping satisfying the given conditions exists, then it is unique.

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A radical approach to algebra by Mary W Gray


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