and problem sheet 6

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1 2-28 and 5-5 problem sheet 6 Solutions to the following seven exercises and optional bonus problem are to be submitted through gradescope by 2:0AM on Thursday 9th October 207. There are also some practice problems not to be turned in for those seeking more practice and also for review prior to the exam. Problem Let f : R \ {0} R be a function satisfying Determine f. f(x) 2f( x ) x2 x R \ {0}. Calculation. Let x R \ {0}. Note that x is a non-zero real number and substituting x the functional equation yields f( x ) 2f(x) ( x )2. for x in Multiplying by 2 and adding the result to the original functional equation yields Thus f(x) if it exists must be defined by 3f(x) x 2 + 2( x )2. 3 (x2 ) 2 3 ( x )2. A computation shows that a function defined in that way solves the original functional equation. Problem 2 Let g : R \ {0 } R be a function satisfying g(x) + g( ) x x R \ {0 }. x Determine g. You may assume that the function h : R \ {0 } R \ {0 } defined by h(x) x is well-defined. Hint: compute h(h(x)) and h(h(h(x))).

2 Calculation. Let x R\{0 }. We note that h(h(h(x))) x. Substituting h(x) and then h(h(x)) for x in the functional equation yields and g(h(x)) + g(h(h(x))) h(x) g(h(h(x))) + g(h(h(h(x)))) g(h(h(x))) + g(x) h(h(x)). Subtracting the first and then adding the second to the original functional equation yields Thus g(x) if it exists must be defined by 2g(x) x h(x) + h(h(x)). g(x) 2 (x h(x) + h(h(x))) 2 (x3 x + x 2 x ). A computation shows that a function defined in that way solves the original functional equation. Problem 3 Verify that the function f : (0 ) R defined by f(x) 2x 2x( x) for all x (0 ) is a bijection. Solution. First note that given x y R with y 0 we have f(x) y ()x 2 + 2( y)x 0 { y + y x 2 + y } y 2 + Call this result ( ). We prove that f is a bijection by finding an inverse for f. Define a function g : R (0 ) by letting for y R y + y 2 + g(y) 2 if y 0 if y 0 We verify that g is a well-defined function and that g is a left inverse and a right inverse for f. 2

3 g is well-defined. First note that the cases in the definition of g are mutually exclusive and cover all possibilities for y R. Also note that we may divide by y and take the square root in the first case since y 0 and y Finally note that g(y) (0 ) for all y R. Certainly if y 0 then g(y) 2 (0 ). If y 0 then y 2 > 0 so that y 2 + >. Moreover y 2 + < y y + ( y + ) 2 and so y 2 + < y + Hence 2 g(y) y (y + y 2 + ) y 2 + y y < y 2 y 2 by combining fractions by simplifying since y 2 + > 0 since y 2 + < y + by cancellation Since 2 g(y) < 2 it follows that 0 < g(y) < as required. g is a right inverse for f. To see this fix y R and let x g(y). We need to show that f(x) y. If y 0 then x 2 so that f(x) 0 y as required. If y 0 then x y + y 2 + so that f(x) y by ( ) above. g is a left inverse for f. To see this fix x (0 ) and let y f(x). We need to show that g(y) x. If x 2 then y 0 so that g(y) 2 x as required. If x 2 then y 0 so that x y + y 2 + or x y y 2 + by ( ) above. If x takes the first of these values then x g(y) so we re done. Hence it remains to show that 3

4 x y y 2 +. To see this note that if x y y 2 + then 2 x y (y y 2 + ) by combining fractions + y 2 + by simplifying + y 2 y 2 y + 2 > 2 so that 2 x > 2 and x (0 ). Hence f is a bijection and g f. since y 2 + > y by separating fractions since 2 y > 0 Problem 4 For all positive integers n define the function f n : R R via f n (x) xn + x 2. Determine all n > 0 for which f n is injective. You may assume that for any odd positive integer n and real numbers x > y x n > y n. You may not use calculus. Solution. We claim that the answer is all odd integers n larger than. First note that these are the only possible cases which can work. that f (2) ( ) f 2 4 and for n even one can note that f n () f n ( ). For n one can note Let n 3 be odd. Suppose that x and y are real numbers such that x y. WLOG we may assume that x > y. Since + x 2 and + y 2 are both positive x n + x 2 > yn + y 2 xn + x n y 2 > y n + y n x 2 (x n y n ) + (xy) 2 (x n 2 y n 2 ) > 0. 4

5 The last inequality is true since n and n 2 are odd positive integers and x > y (note that the first term is positive and the second term is non-negative). Thus f(x) > f(y) and in particular f(x) f(y). Thus f n is injective when n is odd and larger than. Problem 5 Given real numbers a b c d let f : R 2 R 2 be defined by f(x y) (ax + by cx + dy) for all (x y) R 2. Prove that f is an injection if and only if 0. Note: it is also true that f is a surjection if and only if 0 but you needn t show that. Solution. We show that both injectivity and surjectivity depend on 0 for completeness. Suppose 0. Define g(x y) ( ) dx by cx + ay for all (x y) R 2 Note that g is well-defined: 0 so we haven t divided by zero. Now g is a left inverse for f. To see this let (x y) R 2. Then g(f(x)) g(ax + by cx + dy) ( d(ax + by) b(cx + dy) ( (da bc)x + (db bd)y ( ()x ()y ) (x y) ) c(ax + by) + a(cx + dy) ( ca + ac)x + ( cb + ad)y ) 5

6 g is a right inverse for f. To see this let (x y) R 2. Then ( ) dx by cx + ay f(g(x)) f ( ) dx by + ay by + ay a + b cx cdx + d cx ( ) a(dx by) + b( cx + ay) c(dx by) + d( cx + ay) ( ) ()x + ( ab + ba)y (cd dc)x + ( cb + da)y ( ()x ()y ) (x y) Now suppose 0. Note that f(0 0) f( b a) f( d c) (0 0). This provides an example to show that f is not injective unless a b c d 0. If a b c d 0 we note that f(0 0) f( ) (0 0) which shows that f is not injective in this case. Problem 6 Let f : A B and g : B C be functions and define h g f. Determine which of the following statements are true giving proofs for the true statements and counterexamples for the false statements: (a) If h is injective then f is injective. (b) If h is injective then g is injective. (c) If h is surjective then f is surjective. (d) If h is surjective then g is surjective. Solution. (a) True. If f is not injective then f(a) f(b) for some a b; but then g(f(a)) g(f(b)) so h(a) h(b) and h is not injective. (b) False. For example let A C {} and B { 2}. Let f() and g() g(2). Then h : {} {} is bijective since it is its own inverse (so h is definitely injective) but g is not since 2 but g() g(2). 6

7 (c) False. The same counterexample as in (b) works: h is surjective since it is bijective but f is not surjective since 2 f(x) for any x {}. (d) True. Given c C there exists a A with h(a) c since h is surjective. Let b f(a). Then g(b) g(f(a)) h(a) c. So g is surjective. Problem 7 Consider functions f : A B and g : B A. Prove that (a) If f g is the identity function on B then f is surjective. (b) If g f is the identity function on A then f is injective. To remind you: given a set X the identity function on X is the function id X : X X defined by id X (x) x for all x X. Solution. (a) Suppose f g id B and let y B. Let x g(y). Since f g id B we have f(x) f(g(y)) y. Hence f is surjective. (b) Suppose g f id A. Let x x A with f(x) f(x ). Then so f is injective. x g(f(x)) g(f(x )) x Bonus Problem - 2 points If the common difference d of an arithmetic progression starting from is relatively prime to 0 show that the sequence + d + 2d... contains infinitely many powers of 0. Is this still true if the arithmetic progression starts from 2? Solution. For each k we have from Euler s Totient theorem that 0 kϕ(d) (mod d). Equivalently there exists an integer n k (depending on the choice of k) such that 0 kϕ(d) +n k d. Therefore this term will be in the sequence and since k can be an arbitrary positive integer the result follows. For the second question the answer is no. 3 is relatively prime to 0 while 0 (mod 3) and so no power of 0 will be 2 (mod 3). 7

8 Extra Problem Consider a function f : A A. Prove that if f f is injective then f is injective. Solution. This is immediate from Q6(a) letting g f. Extra Problem 2 Let f : A B be a function. (a) Prove that there exists a set X and functions p : A X and i : X B with p surjective and i injective such that f i p. (b) Prove that there exists a set Y and functions j : A Y and q : Y B with j injective and q surjective such that f q j. Solution. (a) Define p : A f (A) by p(a) f(a) for all a A and define i : f (A) B by i(b) b for all b f (A). Then p is well-defined since if a A then f(a) f (A) and i is well-defined since f (A) B. Now If a A then i(p(a)) i(f(a)) f(a) so that f i p. Given b f (A) we have b f(a) p(a) for some a A so that p is surjective. Given b b f (A) if i(b) i(b ) then b b simply by definition of i so that i is injective. (b) If A then let Y A B and define for all a A and b B. Then j(a) (a f(a)) and q(a b) b j is injective. If a a A and j(a) j(a ) then (a f(a)) (a f(a )). Since the first components must be equal we have a a. q is surjective. Given b B let a A be any element (note: this is where we use the fact that A is non-empty). Then b q(a b). f q j since if a A then q(j(a)) q(a f(a)) f(a). On the other hand if A then f is vacuously injective and moreover the identity function id B : B B is surjective so we can define j f and q id B. 8

9 Extra Problem 3 Recall that if A B R a function f : A B is increasing if for all x y A if x < y then f(x) < f(y). Let A and B be subsets of R and let f : A B be a bijection. Prove that if f is increasing then f is increasing. Solution. Suppose f is increasing and let x y B with x < y. We prove that f (x) < f (y) by contradiction. Indeed if f (x) f (y) then one of the following must be true: f (x) f (y). In this case applying f to both sides yields x y contradicting x < y. f (x) > f (y). In this case since f is increasing applying f to both side yields x > y also contradicting x < y. It follows that f (x) < f (y). Hence f is increasing. 9