CHAPTER 3 REVIEW QUESTIONS MATH 3034 Spring a 1 b 1

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1 . Let U = { A M (R) A = and b 6 }. CHAPTER 3 REVIEW QUESTIONS MATH 334 Spring 7 a b a and b are integers and a 6 (a) Let S = { A U det A = }. List the elements of S; that is S = {... }. (b) Let T = { A U det A is a prime integer }. List the elements of T ; that is T = {... }. (c) Let W = { p p is a prime integer and det A = p for some A U }. List the elements of W ; that is W = {... }. (d) V = { A U det A = 6 }. List the elements of V ; that is V = {... }.. Let S = { }. (a) Let A = { (a b) S S a + b = c for some integer c }. List the elements of A. (b) Let B = { a S (a b) A for some b S }. List the elements of B. (c) Let C = { c Z a + b = c for some (a b) A }. List the elements of C. 3. Let S = { A M (R) det A = 5 } and let a T = { B M (R) B = where a and b are integers }. List the elements in b S T. 4. The sets A = { x R 3 x < } and B = { x R x < 5 } can be expressed as intervals on the real line. Specifically A = 3 ) and B = 5). Express each of the sets A B A B A B and B A as an interval on the real line. 5. Let g be the function define on the reals by g(x) = e x + for all real numbers x. The range of g is the set R(g) = { y R there exists x R such that g(x) = y }. (a) Complete the following statement: R(g) provided.... (b) Prove that R(g). (c) Complete the following statement: R(g) provided.... (d) Prove by contradiction that R(g) Notation for Exercises 6 8: Recall that M x (R) denotes the set of all real matrices; that is x M x (R) = { x x x R }. Let A = 4

2 O = { the zero matrix. 6. Using the notation given above let S = { X M (R) AX = O }. (a) List three distinct elements of S. (b) The set S is said to be closed under addition provided for all X Y S we have X + Y S. Prove that S is closed under addition. (c) Prove that for every X S and for every k R we have kx S. 7. Using the notation given before Exercise 6 let T = { X M (R) AX = (a) List three distinct elements of T. (b) Prove that for all X Y T X + Y T. (c) Prove that for all X T and for every real number k if k then kx T. 8. With the matrix A defined above Exercise 6 let W = { X M (R) AX = Prove that W =. (HINT: Try a proof by contradiction.) 9. Let D(R) denote the set of all functions defined and differentiable on R. Set S = { f D(R) f (x) = 3f(x) for all x R }. (a) We say that the set S is closed under addition provided for all f g S we also have f + g S. Verify based solely on the definition of S that S is closed under addition. (b) Prove that for every f S and for every real number k we have kf S. (c) Exhibit three specific elements of S. Definition: Let f be a function defined on the reals and let A be a subset of R. The image of A under f denoted by f(a) is the set f(a) = { f(a) a A } = { y R there exists a A such that y = f(a) }. Example: Let f(x) = x and let A = { 3 3 }. Then f(a) = { f( 3) f( ) f( ) f() f() f() f(3) } = { 4 9 }.. Let f be a function defined on the reals and let A and B be subsets of R. (a) (b) Prove that f(a B) = f(a) f(b). Prove that f(a B) f(a) f(b). (c) With f(x) = x give examples of specific sets A and B such that f(a B) f(a) f(b). (d) Prove that f(a) f(b) f(a B). (e) With f(x) = x give examples of specific sets A and B such that f(a B) f(a) f(b). }. }.

3 . (a) Give an elementwise proof that for all sets A B and C we have A (B C) (A B) (C A). (b) Disprove: For all sets A B and C we have A (B C) = (A B) (C A).. (a) Give an elementwise proof that for all sets A and B we have (A B) (B A) = (A B) (A B). (b) Use known set equalities to prove that for all sets A and B we have (A B) (B A) = (A B) (A B). 3. Let A B and C be sets. Use known set equalities to prove that if A C then C (B A) = A (C B). 4. Prove that for all sets A B and C we have (A C) (B C) = if and only if A B C. Note: This is an equivalence so two proofs are required. In each case a contrapositive proof works well. SOLUTIONS Solutions for : (a) S = { } (b) T = { } 3 4 (c) W = { 3 5 }. (d) V = Solution for : (a) A = { ( 4 3) ( 4 3) ( 3 4) ( 3 4) (3 4) (3 4) (4 3) (4 3) }. (b) B = { }. (c) C = { 5 5 } Solution for 3: S 5 T = { Solution to 4: A B = 3 5) A B = ) A B = 3 ) and B A = 5). Solutions for 5: (a) R(g) provided there exists x R such that g(x) =. (b) Set x = ln. Note that > e so ln > ln e =. Therefore ln > and x = ln exists. Further }. 3

4 g(x) = e x + = e ln + = e ln + = e ln = =. This proves that R(g). (c) R(g) provided for every x R g(x). (d) The proof that R(g) is by contradiction. Thus assume that R(g). This means there exists a real number x such that g(x) =. Therefore e x + = so e x + =. Taking the natural log of both sides yields ln(e x + ) = ln ; that is x + =. Therefore x =. This is a contradiction to the fact that x for every real number x. We conclude that R(g). Solutions to 6: (a) Construction: We first solve the system of equations with matrix equation AX = O. The augmented matrix for that system is. Using elementary 4 / row operations to reduce this matrix yields. This means that if x X = then AX = O x if and only if x x = ; that is if and only if X has the form X = x. Since x is an arbitrary real number there are infinitely many such matrices. x Solution to (a): The matrices and 3 6 are examples of matrices in S. (b) Proof: To prove that S is closed under addition let X Y S. Then AX = O and AY = O. It follows that A(X + Y ) = AX + AY = O + O = O. Therefore X + Y S. (c) Proof: Let X S and let k be a real number. Then AX = O so A(kX) = k(ax) = ko = O. Therefore kx S. Solutions to 7: (a) Construction: We first solve the system of equations with matrix equation AX =. The augmented matrix for that system is. Using elementary 4 / / row operations to reduce this matrix yields. This means that if x X = then AX = if and only if x x x = ; that is if and only if X has the form X = x +. Since x is an arbitrary real number there are infinitely many such matrices. x Solution to (a): The matrices and 3 are examples of matrices in T. 4

5 Solution to (b): Let X Y T. Then AX = and AY = A(X + Y ) = AX + AY = + = 4. Therefore X + Y T.. Therefore Solution to (c): Let X T and let k be a real number k. Then AX = k A(kX) = k(ax) = k = since k. Therefore kx T. k Solution to 8: The proof is by contradiction so assume W. Then there exists a real matrix X x such that X W. Therefore by definition of W AX =. If X = then the x augmented matrix for this matrix equation is. Using elementary row 4 / / operations to reduce this matrix yields. The second row of this matrix implies that x + x = which is quite impossible for any choice of real numbers x and x. This contradiction leads us to conclude that for all real matrices X we have AX so X W. Therefore W = Solution to 9: (a) Suppose that f and g are in S. Then for every real number x we have f (x) = 3f(x) and g (x) = 3g(x). Therefore (f + g) (x) = f (x) + g (x) = 3f(x) + 3g(x) = 3(f + g)(x). Consequently f + g S. (b) Suppose that f S and let k be a real number. Then for every real number x we have f (x) = 3f(x). Thus (kf) (x) = k(f (x)) = k(3f(x)) = 3(kf(x)). It follows that kf S. (c) If θ is the zero function defined by θ(x) = for all x R then θ S. Otherwise functions in S have the form f(x) = ce 3x where c R. So for instance f(x) = e 3x and g(x) = e 3x are in S. Solution to : (a) To see that f(a B) f(a) f(b) let y f(a B). Thus there exists x A B such that f(x) = y. Since x A B we know that either x A or (fork) x B. Case : Suppose that x A. Then f(x) f(a) so f(x) f(a) f(b). Case : Suppose that x B. Then f(x) f(b) so f(x) f(a) f(b). This proves that f(a B) f(a) f(b). To prove that f(a) f(b) f(a B) let y f(a) f(b). Thus either y f(a) or (fork) y f(b). Case : Suppose that y f(a). Then there exists a A such that f(a) = y. Now a A implies that a A B so y = f(a) f(a B). 5 so

6 Case : Suppose that y f(b). Then there exists b B such that f(b) = y. Now b B implies that b A B so y = f(b) f(a B). This proves that f(a) f(b) f(a B) so we conclude that f(a) f(b) = f(a B). (b) To see that f(a B) f(a) f(b) let y f(a B). Then there exists x A B such that y = f(x). Since x A B x A and x B. Since x A it follows that y = f(x) f(a). Likewise since x B it follows that y = f(x) f(b). Therefore y f(a) f(b) and this proves that f(a B) f(a) f(b). (c) As one example let A = { } and let B = { }. then A B = { } so f(a B) = { f() } = { }. But f(a) = f(b) = { 4 }. Consequently f(a) f(b) = { 4 } f(a B). (d) To see that f(a) f(b) f(a B) let y f(a) f(b). Thus y f(a) but y f(b). Therefore there exists a A such that f(a) = y. But a B since y = f(a) f(b). Thus a A B and it follows that y = f(a) f(a B). This proves that f(a) f(b) f(a B). (e) Let A and B be as in (c) above. Then A B = { } so f(a B) = { 4 }. But f(a) = f(b) = { 4 } so f(a) f(b) =. Certainly then f(a) f(b) f(a B) Solution to : (a) Let A B and C be sets. To prove that A (B C) (A B) (C A) suppose that x A (B C). Then x A but x B C. It follows that x A x B and x C. Since x A and x B we get that x A B. Since x C it follows that x C A. Consequently x (A B) (C A) and this proves that A (B C) (A B) (C A). (b) Set A = { 3 4 } B = { } and C = { 5 7 }. Then A (B C) = { 3 4 } { } = { 3 }. But (A B) (C A) = { 3 } { 5 7 } = { 3 }. Thus A (B C) (A B) (C A) Solution to : (a) First to show that (A B) (B A) (A B) (A B) let x (A B) (B A). Therefore either x A B or (fork) x B A. Case : Suppose that x A B. Then x A but x B. Since x A it follows that x A B. But x B so x A B. Therefore x (A B) (A B). Case : Suppose that x B A. Then x B but x A. Since x B it follows that x A B. But x A so x A B. Therefore x (A B) (A B). This proves that (A B) (B A) (A B) (A B). To prove that (A B) (A B) (A B) (B A) let x (A B) (A B). Thus x A B but x A B. Since x A B either x A or (fork) x B. Case : Suppose that x A. Since x A B it follows that x B. Thus x A B which implies that x (A B) (B A). Case : Suppose that x B. Since x A B it follows that x A. Thus x B A which implies that x (A B) (B A). 6

7 This proves that (A B) (A B) (A B) (B A) so we now conclude that (A B) (A B) = (A B) (B A). Proof of (b): For all sets A and B we have (A B) (A B) = (A B) (A B) = (A B) (A B ) = (A B) A (A B) B = (A A ) (B A ) (A B ) (B B ) = (B A ) (A B ) = (A B ) (B A ) = (A B) (B A). Proof of 3: Let A B and C be sets and assume that A C. It follows that A C = A. Thus C (B A) = C (B A ) = C (B A ) = C (B A) = (C B ) (C A) = (C B ) A = A (C B). Proof of 4: We will first prove that for all sets A B and C if (A C) (B C) = then A B C. The proof is by contrapositive so assume that A B C. This means that there exists x such that x A B but x C. Thus x A and x B but x C. It follows that x A C and x B C; that is x (A C) (B C). Therefore (A C) (B C). Now we will prove that if A B C then (A C) (B C) =. The proof is by contrapositive so assume that (A C) (B C). Then there exists x (A C) (B C); that is x A C and x B C. It follows that x A and x B but x C; that is x A B but x C. This proves that A B C. 7