Solutions To Mathematics Textbooks/Proofs and Fundamentals/Chapter 2

Exercise 2.1.1 1 If is a real number, then the area of a circle of radius is . 2 If there is a line and a point not on , then there is exactly one line containing that is parallel to . 3 If is a triangle with sides of length and then 4 If e is raised to the power of x+y, then it is equivalent to the product of e, each one rised to the power of x and y, respectively. 5 If is a continuous function on the interval [a, b] and is any function such that , then the integral of f(x) on [a,b] equals F(b) - F(a). Exercise 2.2.2 1 If , then there is an integer q such that . Let q = n. 2 If , then there is an integer q such that . Let q = 1. 3 If , then there is an integer q such that . This implies , and so , and thus . Exercise 2.2.3 1 If n is an even integer, then for some integer k, . Let . Then . 2 If n is an odd integer, then for some integer k, . Let . Then . Exercise 2.2.4 If n is even, then . For integers j and k, let . , so is even. If n is odd, then . For integers j and k, let . , so is odd. Exercise 2.2.6 If a|b, and b|bm then a|bm, implying aj = bm for some integer j. Also, if a|c, and c|cn then a|cn, implying ai = cn for some integer i. We let x = (j+i). ax = aj+ai ax = bm+cn Which implies a|(bm+cn). Another proof: Suppose that and . Hence there are integers and such that and . Define the integer by . Then Because , it follows Exercise 2.2.7 implies that for some integer, x. implies that for some integer, y. Therefore, for some integer, j. Let , hence . Exercise 2.2.8 Suppose that . Hence there is an integer such that . if is a positive integer, define the integer by . Then Because , it follows Exercise 2.3.2 Proof by contrapositive: Assume is not even, then and . Let then , it follows that if is not even then is not even. Exercise 2.3.3 It is true that does not divide . Suppose that . This means there is an integer such that . Then, we have: We may consider the integer . Therefore, we have that . Then , Contradiction! Exercise 2.3.4 Let a non-zero rational number, hence there are integers and both different from zero, such that . Let an irrational number. Suppose that the product is a rational number, hence there are integers and different from zero such that , this is , it follows that . The last equality means that is a rational number, which is a contradiction because we supposed that was irrational. By contradiction, it follows that the product must be irrational. Exercise 2.3.5 Suppose that and , but does not divide . Hence there are integers and such that and . Suppose that the equation has a solution such that and are integers, then Let , then , it follows that , which is a contradiction.