(a) We need to prove that is reflexive, symmetric and transitive. 2b + a = 3a + 3b (2a + b) = 3a + 3b 3k = 3(a + b k)

Transcription

1 MATH 111 Optional Exam 3 lutions 1. (0 pts) We define a relation on Z as follows: a b if a + b is divisible by 3. (a) (1 pts) Prove that is an equivalence relation. (b) (8 pts) Determine all equivalence classes. Prove your answer for the equivalence class of 1. Proof : (a) We need to prove that is reflexive, symmetric and transitive. reflexive: We need to prove: a Z : a a Let a Z. Then a + a = 3a is divisible by 3. a a. symmetric: We need to prove: a, b Z : a b = b a Let a, b Z with a b. a + b is divisible by 3. Thus a + b = 3k for some k Z. Then b + a = 3a + 3b (a + b) = 3a + 3b 3k = 3(a + b k) Since a + b k Z, we see that b + a is divisible by 3. b a. transitive: We need to prove: a, b, c Z : a b and b c = a c Let a, b, c Z with a b and b c. a + b and b + c are divisible by 3. Thus a + b = 3m and b + c = 3n for some m, n Z. Then a + c = (a + b) + (b + c) 3b = 3m + 3n 3b = 3(m + n b) Since m + n b Z, we see that a + c is divisible by 3. a c. (b) Recall that [a] = {b Z : b a} = {b Z : a b} and that the equivalence classes form a partition of Z (note that we do not know how many equivalence classes there are). We start with [0] = {b Z : 0 b} = {b Z : 0 + b is divisible by 3} = {b Z : b is divisible by 3} = {3m : m Z} = {..., 6, 3, 0, 3, 6,...} Next, we pick an integer that is not an element of [0], say 1. Then We prove that We use mutual inclusion. [1] = {b Z : 1 b} = {b Z : 1 + b is divisible by 3} = {b Z : b + is divisible by 3} {b Z : b + is divisible by 3} = {3m + 1 : m Z} 1

2 Let x {b Z : b + is divisible by 3}. x Z with x + divisible by 3. Thus x + = 3k for some k Z. Hence x = 3k = 3(k 1) + 1. Since k 1 Z, we get that x {3m + 1 : m Z}. Hence {b Z : b + is divisible by 3} {3m + 1 : m Z}. Let x {3m + 1 : m Z}. Then x = 3m + 1 for some m Z. x Z and x + = 3m + 3 = 3(m + 1). Since m + 1 Z, we get that x + is divisible by 3. x {b Z : b + is divisible by 3}. Hence {3m + 1 : m Z} {b Z : b + is divisible by 3}. Since {b Z : b + is divisible by 3} {3m + 1 : m Z} and {3m + 1 : m Z} {b Z : b+ is divisible by 3}, we get that {b Z : b+ is divisible by 3} = {3m+1 : m Z}. [1] = {b Z : b + is divisible by 3} = {3m + 1 : m Z} = {..., 5,, 1, 4, 7,...} Next, we pick an integer that is not an element of [0] [1]), say. Then [] = {b Z : b} = {b Z : + b is divisible by 3} = {b Z : b + 4 is divisible by 3} Similarly as in the case for [1], we can show that [] = {b Z : b + 4 is divisible by 3} = {3m + : m Z} = {..., 4, 1,, 5, 8,...} Since [0] [1] [] = Z we know we have found all the equivalence classes. There are three equivalence classes: {3m : m Z} {3m + 1 : m Z} {3m + : m Z} [1] = {3m + 1 : m Z}. (0 pts) What is the infimum of the interval ( 1, + )? Prove your answer! Proof : We prove that inf( 1, + ) = 1. First, we show that 1 is a lower bound for ( 1, + ). Let x ( 1, + ). Then 1 < x. 1 is a lower bound for ( 1, + ). Next, we show that 1 is the greatest lower bound for ( 1, + ). let m be a lower bound for ( 1, + ). Suppose that 1 < m. Let x R with 1 < x < m. Since x > 1, we have that x ( 1, + ). Since m is a lower bound for ( 1, + ) and x ( 1, + ), we have that m x, a contradiction since 1 < x < m. Hence 1 m. 1 is the greatest lower bound for ( 1, + ). Thus 1 = inf( 1, + ).

3 3. (0 pts) Consider the function f : Z Z where f(x) = 5x. (a) (10 pts) Is f one-to-one? Prove your answer! (b) (10 pts) Is f onto? Prove your answer! lution : (a) We prove that f is one-to-one. Let x 1, x Z with f(x 1 ) = f(x ). Then 5x 1 = 5x Adding to both sides, we get that 5x 1 = 5x Dividing both sides by 5, we find that x 1 = x Hence f is one-to-one. (b) We prove that f is not onto. Suppose that f is onto. Since 0 Z and f is onto, there exists x Z with f(x) = 0. 5x = 0 Thus 5x = Since x Z, this implies that is divisible by 5, a contradiction. Hence f is not onto. Remark: Note that the function g : R R where g(x) = 5x is onto. ( ) n 1 4. (0 pts) What is the limit of the sequence? Prove your answer! 3n + n + n=1 ( ) n 1 Proof : We prove that the limit of the sequence is 0. we have to 3n + n + n=1 prove: ɛ > 0 : N N : n N : n 1 3n + n + 0 < ɛ Let ɛ > 0. Let N N with N > (note that N exists by the Archimedes Axiom). Let 3ɛ n N. Then n 1 3n + n + 0 = n 1 3n + n + < n 3n = 3n 3N < ɛ 3

4 Explanations: n 1 < n 3n + n + > 3n > 0 so 1 3n + n + < 1 3n n N > 0 so 1 n 1 N N > 3ɛ so 3N > 1 ɛ > 0 and thus 3N < ɛ. 5. (0 pts) Let f : A B and g : B C be functions such that g f is bijective. Prove that f is injective and g is surjective. Proof : First, we prove that f is injective. Let x 1, x A with f(x 1 ) = f(x ). Then and so g(f(x 1 )) = g(f(x )) (g f)(x 1 ) = (g f)(x ) Since g f is bijective, we have that g f is injective and thus Hence f is injective. Next, we prove that g is surjective. x 1 = x Let c C. Since g f is bijective, we know that g f has an inverse: there exists a function h : C A such that h (g f) = id A and (g f) h = id C. Hence ((g f) h)(c) = id C (c) g(f(h(c))) = c Put b = f(h(c)). Then b B and g(b) = g(f(h(c))) = c. g is surjective. Extra Credit (15 pts) Let S be a non-empty subset of R that is bounded above. Put T = { s : s S}. (a) Prove that T is a non-empty subset of R that is bounded below. (b) What is inf T in terms of bounds related to S? (c) Suppose S = ( 3, 5). What is T? 4

5 Proof : (a) First, we show that T is non-empty. Since S, we can pick s S. Then s T. T. Next, we show that T is bounded below. Since S is bounded above, we can pick an upper bound m for S. We show that m is a lower bound for T. let t T. Then t = s for some s S. Since m is an upper bound for S, we get that s m. Hence m s. m t. Since this is true for all t T, we have that m is a lower bound for T. This means that T is bounded below. (b) Note that inf T exists since T is non-empty and bounded below. We prove that inf T = sup S. Since sup S is an upper bound for S, it follows from the proof of (b) that sup S is a lower bound for T. Now we show that sup S is the greatest lower bound for T. let m be a lower bound for T. We show that 1 m is an upper bound for S. Indeed, let s S. Then s T. Since m is a lower bound for T and s T, we have that m s. Hence 1 m s. Since this is true for all s S, we have that 1 m is an upper bound for S. Since sup S is the least upper bound for S and 1 m is an upper bound for S, we must have that sup S 1 m. Thus sup S m. Since this is true for any lower bound m for T and sup S is a lower bound for T, we finally get that sup S is the greatest lower bound for T. Thus inf T = sup S. (c) We need to multiply every element in S by. Note that x R : 3 < x < 5 10 < x < 6 T = { s : s S} = { s : 3 < s < 5} = ( 10, 6) 5