447 HOMEWORK SET 1 IAN FRANCIS

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1 7 HOMEWORK SET 1 IAN FRANCIS For each n N, let A n {(n 1)k : k N}. 1 (a) Determine the truth value of the statement: for all n N, A n N. Justify. This statement is false. Simply note that for 1 N, A 1 = {(1 1)k : k N} = {0 1, 0,...} = {0}., and since 0 is not a natural number, this serves as a counterexample to the claim. (b)find A A 3. A A 3 = {( 1)k : k N} {(3 1)k : k N} = {k : k N} {k : k N} = {k : k N} = N, the set of all even natural numbers. This is because N N. (c)find n N A n and n N A n. n N A n = N {0} n N A n = φ. Suppose that A = B := {x R : 0 x 1}. Consider the subset C := {(x, y) : x = y + 1} of A B. Is set C a function? Justify your answer. Solution I claim the answer is no. In order for C to be a function, all elements of A must map to a single element in B. To prove this is not the case, we simply point out a counterexample. Let x = 0, then if C is a function, y B such that 0 = y = y. Since this equation has no solution for y B (or in R for that matter), we can conclude that not all elements on the domain of C map to an element in B, hence C is not a function. Suppose f : A B is a function. 3 (a) If E and F are subsets of A, prove that: (i) f(e F ) = f(e) f(f ) Proof ( ) Let y f(e F ). Then x E F such that maps to y under f. Since x maps uniquely to y under f, and x E or x F, then we have y f(e) or y f(f ). Hence, f(e F ) f(e) f(f ). ( ) Let y f(e) f(f ). Then x, where x E or x F, such that f(x) = y. Hence, f(x) = y f(e) f(f ). (ii) f(e F ) f(e) f(f ). Proof Let y f(e F ). Then x E F, meaning x E and x F, such that f(x) = y. Thus, y f(e F ). Since y was arbitrary, this proves the claim. Date: September 17, 01. 1

2 IAN FRANCIS (b) If G and H are subsets of B, prove that: (i) f 1 (G H) = f 1 (G) f 1 (H). Proof ( ) Let x f 1 (G H). Then y G H, such that f(x) = y. Now since y G or y H (or both), x f 1 (G) or x f 1 (H) (or both). Since x was arbitrary, f 1 (G H) f 1 (G) f 1 (H). ( ) Let x f 1 (G) f 1 (H). Then y either in G or H (or both), such that f(x) = y. Since y G H, x f 1 (G H). This proves ( ) as x was arbitrary. Our original claim (=) now follows. (ii) f 1 (G H) = f 1 (G) f 1 (H). Proof ( ) Let x f 1 (G H). Then x f 1 (G) and x f 1 (H). So, y G H such that f 1 (y) = x, since f 1 (x) = y G and f 1 (x) = y H, we have x f 1 (G) f 1 (H). ( ) Let x f 1 (G) f 1 (H). Then y G H, such that y = f(x). since y G and y H, and f(x) = y, x f 1 (G) and x f 1 (h), which is what we wanted to show.. (c) Give an example to show that equality in 3(a)(ii) need not hold in general. Proof Let f : {1,, 3} {1,, 3} defined by {(1, 1), (, 1), (3, 3)}, and let E = {1}, F = {, 3}. Then f(e F ) = f(φ) = φ 1 = {1} {1, 3} = f(e) f(f ). (d) Give an example to show that it is NOT in general true that f(e \ F ) f(e) \ f(f ). Proof Let f : {1,, 3, } {1,, 3, } defined by {(1, 1), (, ), (3, ), (, )}, and let E = {1,, 3, }, F = {3}. Then f(e \ F ) = f({1,, }) = {1,, } {1, } = {1,, } \ {} = f(e) \ f(f ). 16 Let Y := {y R : 1 < y < 1}. Show that the function f defined by f := x x +1, is a bijection of R onto Y. 19 Proof We must show that f is both injective and surjective. First, assume that x, y R such that f(x) = x = y = f(y) x +1 y +1 (y + 1)x = y (x + 1) x = y. So either x = y, x or x = y. However, if x = y, = y x +1 x, but this is false for x nonzero, since these +1 two quantities are assumed to be equivalent, but one is negative and one is positive. If x = 0 the statement f(x) = 0 = = f( x) = f(y) x = y is trivial. Thus, x, y Y, +1 f(x) = f(y) x = y, and we have that f is injective. Now, to show that for every element in Y, there exists an element in R that maps to it under f, we let y Y and solve y = x y for x, in terms of y. This gives x =, 1 y x +1 whose right-hand side only makes sense (and thus, we only have a corresponding x R) for y ( 1, 1) = Y. Thus we have surjectivity, and together with injectivity, proves that f is a bijection as desired. (a) Show that if f : A B is injective and E A, then E = f 1 (f(e)). Give an example to show that equality need not hold if f is not injective. Proof ( ) Let e E. Then f(e) f(e). Since f maps e into f(e), e f 1 (f(e)). ( ) Now let e f 1 (f(e)). Then f(e) f(e). Since f is an injection, e is the unique element that is mapped to f(e) f(e), thus e E.

3 7 HOMEWORK SET 1 3 Therefore, we have E = f 1 (f(e)). Example Let f : {1,, 3} {1,, 3} defined by {(1, 1), (, ), (3, )}, and let H = {1, 3}. Then E = {1, 3} {, 3} = f 1 ({}) = f(f 1 ({1, 3})) = f(f 1 (E)). (b) Show that if f : A B is surjective and H B, then H = f(f 1 (H)). Give an example to show that equality need not hold if f is not surjective. Proof ( ) Let h H. Then, by surjectivity, a A such that f(a) = h. Hence, a f 1 (H), so a maps uniquely to f(f 1 (h)) which implies that f(a) = h f(f 1 (H)). ( ) Now let h f(f 1 (H)). Then a f 1 (H) such that f(a) = h, which tells us that h H (since a is in the preimage of H and it must map uniquely to h). Example Let f : {1,, 3} {1,, 3} defined by {(1, 1), (, ), (3, )}, and let H = {3}. Then H = {3} φ = f(φ) = f(f 1 ({3})) = f(f 1 (H)). 0 (a) Suppose that f is an injection. Show that (f 1 f)(x) = x, x D(f) and that (f f 1 )(y) = y, y R(f). Proof ((f 1 f)(x) = x) Let x D(f), Then, y R(f) such that x y. Also, since f is an injection, the element x is unique (There is no other element in the domain that maps to y) Because of this, f 1 must map y uniquely back to x. Thus, (f 1 f)(x) = x for all x D(f). ((f f 1 )(y) = y) Let y R(f), then by definition of the range, there exists an x D(f), such that x y. Also note that this x is unique in D(f) by definition of injectivity. Since x is the unique element in the preimage of y, and f(x)=y, we have that (f f 1 )(y) = y. Therefore, (f f 1 )(y) = y, y R(f). (b) If f is a bijection of A onto B, show that f 1 is a bijection of B onto A. Proof (injective) Let x, y B such that f 1 (x) = f 1 (y). Then, since f is a function, f(f 1 (x)) = f(f 1 (y)) x = y (since f is bijective, this is a property we proved above). (surjective) Let x A, then y B such that f(x) = y, and applying f 1 to both sides, and using the fact that f is injective (and the result in 0a), f 1 (f(x)) = f 1 (y) x = f 1 (y). Therefore, f is injective and surjective, and thus, bijective. 1 Prove that if f : A B is bijective and g : B C is bijective, then the composite g f is a bijective map of A onto C. Proof (injective) Let x, y A such that g(f(x)) = g(f(y)). Since g is injective, we have f(x) = f(y), which implies x = y since f is also injective. Thus g f is injective. (surjective) Let z C. Since g is surjective, y B such that g(y) = z, and since f is surjective, x A such that f(x) = y, and thus g(f(x)) = g(y) = z. Thus g f is surjective. Therefore, g f is bijective. Let f : A B and g : B C be functions. (a) Show that if g f is injective, then f is injective. Proof Let x, y B such that f(x) = f(y). Since g is a function, we have g(f(x)) = g(f(y)), and since g f is injective, this implies x = y. Thus f(x)) = f(y) x = y, i.e. f is injective.

4 IAN FRANCIS (b) Show that if g f is surjective, then g is surjective. Proof Let y C. Since g f is surjective, x A such that g(f(x)) = y. Since f(x) B is in the domain of g, we have that g is surjective. Let f, g be functions such that (g f)(x) = x for all x D(f) and (f g)(y) = y for all y D(g). Prove that g = f 1. Proof By the previous problem, we have that both f, g are bijective (since the identity function is trivially a bijection), and thus, both have inverse functions. Now, for arbitrary y D(g) we have both f(g(y)) = y and f(f 1 (y)) = y, thus f(f 1 (y)) = f(g(y)). Since f is injective, this implies f 1 (y) = g(y), where y was arbitrary. This, together with the fact that g has a unique inverse function, tells us that f 1 = g. 5. Use Induction to prove the following ( ) (i) For all n N, n 3 = n(n+1). ( ) For the base case, we note that 1 3 = 1 = 1(). Now assume that the formula holds for some n N. Then, we have n 3 + (n + 1) 3 = ( n 3 ) + (n + 1) 3 ( ) n (n + 1) = + (n + 1) 3 = n (n + 1) (n + 1)3 + = n (n + 1) + (n + 1)(n + 1) = (n + n + ) (n + 1) = (n + ) (n + 1) ( ) (n + 1) (n + ) =, Thus, if the formula holds for some n, then it holds for n+1. Therefore, by mathematical induction, the formula holds for all natural numbers. (ii) For all n N, n k=1 ( 1)k+1 k = ( 1) n+1 n(n+1). For the base case, note that ( 1) 1 = 1 = ( 1) 1(). Now assume that the statement holds for some n N. Then, n+1 ( 1) k+1 k n+1 n(n + 1) = ( 1) + ( ( 1) n+ (n + 1) ) k=1 = ( 1) n+1 n(n + 1) [ (n + 1) ] = ( 1) n+ [ n + n + n + n + ] ( n = ( 1) n+ ) + 3n + n+ (n + 1)(n + ) = ( 1). Thus, by the principle of mathematical induction, the formula is valid. (iii) For all n N and n, n < n!.

5 7 HOMEWORK SET 1 5 It s clear that = 16 < =!. So the base case holds. Now, assume the inequality holds for some n N. Then we have n+1 = ( n ) < (n!) < (n + 1)n! = (n + 1)!. Thus, by the principle of mathematical induction, The inequality is valid for all natural numbers greater than 3.

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