MATH 2001 MIDTERM EXAM 1 SOLUTION

Transcription

1 MATH 2001 MIDTERM EXAM 1 SOLUTION FALL MOON Do not abbreviate your answer. Write everything in full sentences. Except calculators, any electronic devices including laptops and cell phones are not allowed. (1) Quick survey. (a) (1 pt) This class is: Too easy Moderate Too difficult (b) (2 pts) Write any suggestion for improving this class. (For instance, give more examples in class, explain proofs in detail, give more homework, slow down the tempo,...) (2) Write the definition. (a) (2 pts) Tautology. A sentence is tautology if it is alway true. (b) (2 pts) For a set A, the power set P(A). The power set P(A) is the set of all subsets of A. In other words, P(A) = {B B A}. (c) (2 pts) Injective function. A function f : X Y is injective if for every a, b X, if f(a) = f(b), then a = b. Date: October 8,

2 (3) (a) (4 pts) Use a truth table to prove that the statement is always true. ((p q) p) q p q p q p (p q) p ((p q) p) q T T T F F T T F T F F T F T T T T T F F F T F T (b) (4 pts) Give its contrapositive and simplify it. q ((p q) q) contrapositive q (p q) q De Morgan s law q (p q) q double negation q ( p q) q De Morgan s law Taking the contrapositive q ((p q) q) correctly: 2 pts. Getting a simplified statement q ( p q) q: 4 pts. 2

3 (4) (5 pts) Write a proof sequence for the following assertion. } p q r r (p q) Below is a list of (some of) tautologies. x x double negation (x y) x y implication (x y) x y De Morgan s law (x y) } x y x y modus ponens x y } y x modus tollens x y statement reason 1 p q given 2 r (p q) given 3 p q implication and 1 4 (p q) De Morgan s law and 3 5 r modus tollens, 2 and 4 Applying implication and getting p q: 2 pts. Applying De Morgan s law and obtaining (p q): 4 pts. Getting r: 5 pts. 3

4 (5) In modern calculus, we say a function f : R R is continuous at 1 if the following statement holds: For every ɛ R +, there exists δ R + such that if x 1 < δ, then f(x) f(1) < ɛ. (a) (3 pts) Translate the statement in predicate logic. ( ɛ R + )( δ R + )( x R)( x 1 < δ f(x) f(1) < ɛ). Omitting x R: -2 pts. (b) (3 pts) Negate (a). (( ɛ R + )( δ R + )( x R)( x 1 < δ f(x) f(1) < ɛ)) ( ɛ R + ) (( δ R + )( x R)( x 1 < δ f(x) f(1) < ɛ)) ( ɛ R + )( δ R + ) (( x R)( x 1 < δ f(x) f(1) < ɛ)) ( ɛ R + )( δ R + )( x R) ( x 1 < δ f(x) f(1) < ɛ) ( ɛ R + )( δ R + )( x R) ( ( x 1 < δ) f(x) f(1) < ɛ) ( ɛ R + )( δ R + )( x R) ( x 1 < δ) ( f(x) f(1) < ɛ) ( ɛ R + )( δ R + )( x R)( x 1 < δ f(x) f(1) ɛ) If one omitted x R in (a), one could get 2 pts. (c) (3 pts) Translate (b) into ordinary English. There is a positive real number ɛ such that for every positive real number δ, there is a real number x such that the distance x 1 between x and 1 is less than δ but the distance f(x) f(1) between f(x) and f(1) is at least ɛ. 4

5 (6) (a) (5 pts) Let a, b Z. Prove that if ab is a multiple of 3, then a is a multiple of 3 or b is a multiple of 3. (Note that every integer can be written as precisely one of 3k, 3k + 1, or 3k + 2 for some integer k.) We prove the contrapositive of the statement. Suppose that it is not the case that a is a multiple of 3 or b is a multiple of 3. Then both a and b are not a multiple of 3. There are three cases we must consider: 1) both a and b are of the form 3k + 1, 2) both a and b are of the form 3k + 2, 3) one is of the form 3k + 1 and the other is of the form 3k + 2. For the first case, suppose that a = 3k + 1, b = 3l + 1 for some k, l Z. Then ab = (3k + 1)(3l + 1) = 9kl + 3k + 3l + 1 = 3(3kl + k + l) + 1. Therefore ab is not a multiple of 3. For the second case, suppose that a = 3k + 2, b = 3l + 2 for some k, l Z. Then ab = (3k + 2)(3l + 2) = 9kl + 6k + 6l + 4 = 3(3kl + 2k + 2l + 1) + 1. Therefore ab is not a multiple of 3. Finally, suppose that a = 3k +1 and b = 3l+2 for some k, l Z. Then ab = (3k +1)(3l+2) = 9kl + 6k + 3l + 2 = 3(kl + 2k + l) + 2. Therefore ab is also not a multiple of 3. In any case, ab is not a multiple of 3. Therefore we obtain the desired result if we take its contrapositive. (b) (4 pts) Consider the following statement, which is a generalization of (a) (if a = 3, it is exactly (a)). If a bc, then a b or a c. Show that this is not true. Let a = 4, b = c = 2. Then a bc since 4 4, but a b and a c because

6 (7) Let X = {n N n 10} and Y = {m Z m 2 5}. (a) (2 pts) Find Y \ X. Y = { 2, 1, 0, 1, 2}, X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Y \ X = { 2, 1, 0} (b) (3 pts) Compute P(X Y ). X Y = {1, 2} P(X Y ) = {, {1}, {2}, {1, 2}} (c) (2 pts) Calculate X Y. X Y = X Y = 10 5 = 50 (8) Let f : X Y, g : Y X be two functions. Suppose that g f = id X. (a) (5 pts) Prove that f is injective. Let a, b X. Suppose that f(a) = f(b). Then g(f(a)) = g(f(b)). Thus a = id X (a) = (g f)(a) = g(f(a)) = g(f(b)) = (g f)(b) = id X (b) = b. Therefore f is injective. (b) (5 pts) Show that g is surjective. Let c Y. Then c = id X (c) = (g f)(c) = g(f(c)). So c is the image of f(c) X. Therefore g is surjective. (c) (3 pts) Give an example of f and g such that g f = id X but f is not surjective. Here is one of the simplest examples I can think of. Let X = {a, b} and Y = {1, 2, 3}. Define f : X Y as f(a) = 1, f(b) = 2. And define g : Y X as g(1) = a, g(2) = b, and g(3) = b. Then clearly f is injective, g is surjective, and g f = id X. But f is not surjective because there is no x X such that f(x) = 3. (And g is not injective because g(2) = b = g(3).) 6