# Algebra and number theory : an integrated approach by Martyn R Dixon; Leonid A Kurdachenko; Igor Ya Subbotin

By Martyn R Dixon; Leonid A Kurdachenko; Igor Ya Subbotin

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**Sample text**

2 In each of the following questions explain your reasoning, either by giving a proof of your assertion or a counterexample. 1. Let * = {(x, y) E N x N I 3x = y}. Is a functional correspondence? 2. Let = {(x, y) EN x N I 3x = Sy}. Is a functional correspondence? 3. Let = {(x, y) E N x N I x = 3y}. Is a functional correspondence? 4. Let = {(x, y) E N x N I x 2 = i}. 5. Let = {(x, y) EN x NIx= y 4 }. a functional correspondence? Is a functional correspondence? 6. Let f: Z ~No be the mapping defined by f(n) = Is f injective? *

3. Let * = {(x, y) E N x N I x = 3y}. Is a functional correspondence? 4. Let = {(x, y) E N x N I x 2 = i}. 5. Let = {(x, y) EN x NIx= y 4 }. a functional correspondence? Is a functional correspondence? 6. Let f: Z ~No be the mapping defined by f(n) = Is f injective? Is f surjective? 7. Let f: N ~ {x E Q I x > 0} be the mapping defined by f(n) = where n E N. Is f injective? Is f surjective? 8. Let f: N ~ N be the mapping defined by f(n) = (n n E N. Is f injective? Is f surjective? *

A left inverse to f exists if and only iff is injective. A right inverse to f exists if and only iff is surjective. Proof. Suppose first that g is a left inverse of f. 3 shows that f is injective. Conversely, suppose that f is injective. We choose and fix the element u in the set A. If b E Im f, then the element b has a unique preimage a, since f is injective. Put g(b) = a, where f(a) = b whenever bE Im f, Iu, if b ¢ Im f. By the definition of g we have, for every element a E A, go f(a) = g(f(a)) =a= eA(a), which shows that g is a left inverse to f.