Math 2534 Final Exam Review Answer Key 1. Know deþnitions for the various terms: a. Modus Ponens b. Modus Tollens c. divides d. rational e.

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1 Math 534 Fnal Exam Revew Answer Key 1. Know deþntons for the varous terms: a. Modus Ponens b. Modus Tollens c. dvdes d. ratonal e. Quotent-Remander Theorem f. unon g. ntersecton h. complement. DeMorgan s Laws j. power set k. one to one l. onto m. Pgeon Hole Prncple n. countable o. equvalence relaton p. tautology Look them up.. Truth tables: a. Construct the truth table for p (q r) p q r r q r p (q r) T T T F F T T T F T T T T F T F F T T F F T F T F T T F F F F T F T T T F F F F F F F F F T F F b. Be able to use truth tables to verfy the logcal equvelence of two statement forms. (Such as: Provng DeMorgan s Law or the Absorpton Law s true.) 3. Fnd the nverse, converse, contrapostve, and negaton of the followng: a. If I pass the Þnal exam, then I wll pass ths class. Inverse: If I do not pass the exam, then I wll not pass ths class. Converse: If I pass ths class, then I passed the exam. Contrapostve: If I do not pass ths class, then I dd not pass the Þnal. Negaton: I passed the exam and dd not pass the class. b. For every hour I lve there s a moment to be sezed such that f I don t spend my tme wsely, I wll be old before I know t. Inverse: For every hour I lve there s a moment to be sezed such that f I spend my tme wsely, I wll not be old before I know t. Converse: For every hour I lve there s a moment to be sezed such that f I am old before I know then I dd not spend my tme wsely. Contrapostve: For every hour I lve there s a moment to be sezed such that f I am not old before I know t then I dd spend my tme wsely. Negaton: There s an hour n whch I lve such that for every moment to be sezed, I don t have to spend my tme wsely and I wll not be old before I know t. 4. Rewrte n If... then form: a. p only f q c. p s a necessary condton for q b. p f q d. p s a suffcent condton for q. a. If qthen p c. If pthen q b.ifqthenp d.ifpthenq. 5. Be able to use truth tables to verfy the valdty of argument forms. Examples: 1

2 a. p q q p p q b. p p q q r r a. p q p q p q p q T T T T T T F F T F F T T F F F F T T F The crtcal rows are n bold face. Snce the concluson s not always True whenever the hypothess are True, ths argument form s not vald.. Use basc argument forms (secton 1.3) to justfy the concluson from the premses, gvng reasons for each step. p r s m s u p w u w m w 1) w dsjunctve ) u Modus 3) p r s Modus u w syllogsm u p Ponens p Ponens u 4) r s Conjunctve s SmplÞcaton p 5) s Modus m s Tollens m r s ) m Dsjunctve m w Addton 7. Gven the followng Major Premse, P(x) Q(x), gve the Mnor Premse and Concluson for the followng argument forms: a. Modus Ponens b. Modus Tollens c. Inverse Error d. Converse Error Modus Ponens Modus Tollens Inverse Error Converse Error Mnor Premse: P(x) Q(x) P(x) Q(x) Concluson: Q(x) P(x) Q(x) P(x) 8. Be able to use dagrams to determne the valdty of argument forms. Example: All polynomal functons are contnuous. f(x) s contnuous. f(x) s a polynomal. (of sorts...) Ths argument form s nvald. The crcle of polynomal functons s contaned n the crcle of contnuous functons. f(x) s n the crcle of contnuous functons, but t could be outsde the polynomal functons. Therefore we cannot conclude that f(x) s a polynomal. 9. Prove the followng: (be very explct) a. If a b anda c, then a (b+c) Proof: Let a,b,c Z such that a b anda c. Thus ntegers m, n such that b = am and c = an. Thus b+c = am+an = a(m+n). Snce m and n are ntegers, m+n s an nteger. Thus by the deþnton of dvdes, a (b+c). QED b. The square of any odd nteger has the form of 8m+1 for some m Z.

3 Proof: Let x be any odd nteger. Usng the Quotent Remander Theorem, we know that x = 4q+1 or x=4q+3forsomeq Z. Case 1: x = 4q+1. x =(4q+1) = 1q + 8q + 1 = 8(q + q) + 1. Snce q s an nteger, q +qs an nteger. Thus x = 8m+1 for some nteger m. Case : x = 4q+3. x =((4q+3) = 1q + 4q + 9 = 1q +4q+8+1=8(q +3q+1)+1. Snce q s an nteger, q + 3q+ 1 s an nteger. Thus x = 8m+1 for some nteger m. Therefore f x s any odd nteger, x = 8m+1 for some nteger m. QED c. The product of a ratonal number and an rratonal number s rratonal. (Hnt: Use contradcton) Proof: (by contradcton) Assume r Q and s/ Q such that rs Q. By the deþnton of ratonal numbers, a,b,c,d Z such that r = a b and rs = c rs d, wth a,b,c,d=0. Snce s = r By deþnton, ths means that s Q, but ths s a contradcton for we earler assumed s / Q. Thusour assumpton was false, and hence the product of a nonzero ratonal and rratonal number s rratonal. QED d. For all n Z, fn s odd, then n s odd. (Hnt: Use the contrapostve) We wll prove: for all n Z, fnseven,thenn s even. Proof: Let n be an even nteger. Thus n = k for some nteger k. Thus n =4k =(k ). Thus by deþnton of even n s even. QED 10. Fnd formulas for the followng sequences (c and d are recursvely deþned): a. 0,,4,,8,... b. 0,,,1,0,30,4,5,... c. 0,1,1,3,5,11,1,43,... d. 1,1,,,4,10,... a. s n =s, n 0 b. s n = n(n +1),n 0 c. s 0 =0,s 1 =1,s n =s n + s n 1, n d. s 0 =1,s n = n(s n 1 ), n Prove va Mathematcal Inducton: Xn a. n(n + 1)(n +1) =,forn 1. Proof: Base case: It works. Now assume for k 1 that Pk+1 k+1 = (k+1)(k+)((k+1)+1). P Pk = (k+1)(k(k+1)+(k+1)) = (k+1)(k +7k+) = (k+1)(k+)(k+3) n P Pk,s= cb da,whereda=0. = k(k+1)(k+1). We want to show that +(k+1) = (by the nductve hypothess) = k(k+1)(k+1) +(k+1) = k(k+1)(k+1)+(k+1) = = n(n+1)(n+1),forn 1. QED,whchswhatwewantedtoshow. b. n! > n for n 4 Proof: Base case: It works. Now assume for some k 4, that k! > k. We want to show that (k +1)!> k+1. (k +1)! = (k +1)k! > (by the nductve hypothess) > (k +1) k > (snce k 4, k>) > k = k+1.. Thus (k +1)!> k+1,whchswhatwewantedtoshow. n! > n for n 4. QED. 3

4 1. Be able to prove two sets are equal usng the Element Method. Examples: a. A (A B) = A Let x A. Thus x A B, hence by deþnton of ntersecton x A (A B). Thus A A (A B). Let x A (A B). Thus by deþnton of ntersecton x A. Thus A (A B) A. A (A B) = A. QED. b. B-A=B A c Let x B-A. Then by deþnton x Bandx/ A. Thus x A c.thusbydeþnton of ntersecton x B A c. Let x B A c.thusx Bandx A c, hence x/ A. Thus by deþnton of set mnus, x B-A. x B A c =B-A.QED c. sets A, B, and C, f A C=B C, then A = B. Not true. Counterexample: Let A = {1,,3}, B ={1,4,5}, C ={1}. 13. Be able to prove thngs concernng the empty set Ø. Examples: a. A Ø=Ø b. A A c =Ø c. For all sets A and B, f A B, then A B c =Ø Assume false; thus sets A and B such that A B c =Ø. Letx A B c. Thus x Aandx B c. Snce x AandA B, x B. But thus x Bandx B c, whch s a contradcton. Therefore our assumpton was false, and the orgnal statement s true. QED. 14. State whch of the followng are true: a. {1,,3} Z b. π Q c. Ø Q d. Z Q = Z e. Z Q = R f. Ø Ø g. Ø Ø h. 1 {{1} } a. F (ts a subset, not an element) b. F (π s rratonal) c. F (a subset, not an element) d. T e. F (= Q, see d) f. F (subset, not an element) g. True h. F. 15. Let A = {1,,3,4}, B ={5,,7}. Usng arrow dagrams, do each of the followng f possble. If not possble, state why. a. Draw f:a B such that f s one to one. b. Draw f:a Bsuchthatfsonto. c. Draw f:a B such that f s a one to one correspondence. a. Not possble, for A has more elements than B. b. Many optons exst. c. Not possble, see (a) 1. Let A,B,and C be sets, and let f,and g be functons so that f:a B, g:b C. a. If g f s one to one, must f be one to one? Must g? Prove or gve a counterexample. f must be one to one, g does not. Play around wth dagrams to get a counterexample. Proof: Let x 1,x A such that f(x 1 )=f(x ). Snce g s a functon, ths means that g(f(x 1 )) = g(f(x )), whch s the same as sayng that (g f)(x 1 )=(g f)(x ). But snce g f s one to one, ths means that x 1 = x.thusfsonetoone. b. If g fsonto,mustfbeonto?mustgbe?proveorgveacounterexample. g must be onto, f does not. Play around wth dagrams to get a counterexample. 4

5 Proof: Let x C. Snce g fsonto,weknow y Asuchthat(g f)(y)=x.letz=f(y) B. g(z) =g(f(y))=x.thus x C, y Bsuchthatg(y)=x,andthusgsonto. 17. Dscrete Dorothy deþnes f:z Z by f(n) = n. Jve Junkn Jerry clams ths s not well deþned snce t s not one to one or onto. Dscrete Dorothy dsagrees. Who s correct and why? Dscrete Dorothy. A functon need not be one to one or onto to be well deþned. 18. Prove the set of odd ntegers are countable. (Do so by Þndng an approprate functon from a known countable set to ths set.) We know from prevous work that Z s countable. DeÞne f:z {odd ntegers} by f(n) = n+1. Prove ths s functon s 1-1 and onto ntegers are pcked. What s the maxmum number of ntegers that we can be sure of that are congruent mod 5? DeÞne a bnary relaton R on Z by xry m n(mod4). ProveR s an equvalence relaton. Lst the equvalence classes, and descrbe the elements n each. Reßexve: Snce x x (mod 4) for all x, xrx for all x. Symmetrc: Snce f m n(mod4),thenn m(mod4),fmrn, then nrm. Transtve: Let xry and yrz. Thus x y(mod 4)and y z(mod 4). Hence 4 (x-y) and 4 (y-z). Thus 4 ((x-y)+(y-z)) = (x-z), hence x z (mod 4), and hence xrz. The equvelance classes are [0], [1], [], [3], where the class [a] = all ntegers of the form 4q+a, for some q Z. 1. Let A be a collecton of sets. DeÞne a bnary relaton R on A by U R V U V. Prove R s a partal order relaton. Reßexve: for all sets U A, U UsoURU for all sets U. Transtve: for all sets U, V, W A, f URV andvrw, then U VandV WsoU W, hence URW. Antsymmetrc: If URV andvru, then U VandV U, hence U = V. Now let A = P({a,b,c}) (the power set of the set{a,b,c}.) Usng the relaton R above, draw the Hasse Dagram deþned by ths relaton. Hnt: It looks lke a cube.. You have 4 dfferent knds of food: Chocolate, Steak, Apples, and Beets (unlmted supply). You have 5 frends to whch you would lke to gve 1 knd of food. How many dfferent ways can you do ths? On planet Namtrah, the Namtrahanans use a 0 letter alphabet. They are gven a personalzed ID that conssts of ether 3,4, or 5 letters. How many dfferent ID s are possble? How many numbers between 1 and 100 are even or dvsble by three? (Use ncluson/excluson) = 7 5