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1 S 202, Sing 2005 Midtem 1: 24 eb 2005 Page 1/8 Name: KEY Pledge: Signatue: Thee ae 75 minutes fo this exam and 100 oints on the test; don t send too long on any one uestion! The 12 shot answe uestions euie only a sentence o two fo full cedit; the thee long answe uestions have thei own age, and obviously euie moe. The uestions ae oganized by toic, so the long answe uestions ae scatteed thoughout the exam uestions 5, 14, and 15. The long answe uestions ae woth about half of the test scoe; the shot answe uestions ae all woth 4 oints each, and constitute the othe half. The efeence sheet is on age 2. ll wok must be on these exam ages. Good luck! Pat : Logic / 33 Pat : Stuctues / 16 Pat : Poofs / 51 Total / 100

2 S 202, Sing 2005 Midtem 1: 24 eb 2005 Page 2/8 S 202 Exam 1 Refeence Sheet Set and logical identities Sets Rosen,. 89 Name oolean logic Rosen,. 24 dentity T Domination T T demotent omlementation law ommutative ssociative Distibutive DeMogan s bsotion omlement T Rules of infeence Rosen,. 58 Rule of nfeence Tautology Name ddition Simlification onjunction [ ] Modus onens [ ] Modus tollens [ ] Hyothetical syllogism [ ] Disjunctive syllogism [ ] Resolution

3 S 202, Sing 2005 Midtem 1: 24 eb 2005 Page 3/8 Pat : Logic Question 1 4 oints: What is the convese of? nswe: The convese is Question 2 4 oints: Given the oolean oosition, wite an euivalent comound oosition using only the oeatos,, and. nswe: Question 3 4 oints: State the negation of the following uantified statement: x y P x Q y nswe: x y P x Q y Question 4 4 oints: What ae the two ways to convet a oositional function into a oosition? nswe: Eithe by sulying it with a constant, o by adding a uantifie fo each vaiable

4 S 202, Sing 2005 Midtem 1: 24 eb 2005 Page 4/8 Question 5 17 oints: Pove that using logical euivalences. You must clealy label each ste of the logical euivalence. nswe: Oiginal statement ssociativity of ND emoving of aenthesis demotent law Distibutive law ssociativity of ND emoving of aenthesis Negation law Domination law dentity law demotent law Gimaldi, uestion 7a, section 2.2, age 66.

5 S 202, Sing 2005 Midtem 1: 24 eb 2005 Page 5/8 Pat : Stuctues sets and functions Question 6 4 oints: What is the diffeence between a subset and a oe subset? nswe: subset can be eual to the oiginal set; a oe subset cannot be it must have fewe elements than the oiginal set Question 7 4 oints: Why must a function f be 1-to-1 and onto if the function f is invetible i.e. you can find an invese function of f? nswe: The function must be 1-to-1, as othewise the invese function has multile answes fo a single inut. The function must be onto, as othewise the invese function is not defined fo cetain values. Question 8 4 oints: What is the cadinality of a owe set of a set of n elements? nswe: 2 n Question 9 4 oints: Let f x 5x + 2 and g x 2x + 3. What is f o g x? nswe: f o g x 52x x + 17

6 S 202, Sing 2005 Midtem 1: 24 eb 2005 Page 6/8 Pat : Poofs Question 10 4 oints: What two oeties must be shown fo a uniueness oof? nswe: Existence and uniueness Question 11 4 oints: Wite an existential genealization of the uantified statement xpx. f you intoduce new vaiables, etc., clealy descibe what they eesent. nswe: Py, whee y is some unknown constant, not a vaiable Question 12 4 oints: What is the diffeence between a vacuous oof and a tivial oof? nswe: vacuous oof is when the antecedent of the conditional is always false, and thus the conditional is always tue. tivial oof is when the conseuence of the conditional is always tue, and thus the conditional is always tue. Question 13 4 oints: What movie did Pofesso loomfield show a eview of duing class? nswe: aody taile of Sta Was: Eisode

7 S 202, Sing 2005 Midtem 1: 24 eb 2005 Page 7/8 Question oints: o this uestion, you will have to ove that if m is an even intege, then m+7 is an odd intege. You need to ove it two diffeent ways: by diect oof, indiect oof, o oof by contadiction. Each oof method is woth the same amount. nswe: Poof method 1 cicle one: Diect oof ndiect oof Poof by contadiction The statement tanslates into, whee means m is even, and means m+7 is odd. o the diect oof, we assume is tue, and show that must always be tue. f m is even, then m2k, whee k is some intege this is the definition of even numbes. Thus, m+7 2k+7 2k+3+1, whee k+3 is an intege as k was an intege. This is the definition of odd numbes two times an intege lus one. Thus, m+7 must be odd. Poof method 2 cicle one: Diect oof ndiect oof Poof by contadiction The statement tanslates into, whee means m is even, and means m+7 is odd. o the indiect oof, we ove the contaositive. The contaositive is, which tanslate to if m+7 is even, then m is odd. f m+7 is even, then m+7 2k+1, whee k is some intege definition of odd numbes. Solving fo m, we get m 2k-6 2k-3, whee k-3 is an intege as k was an intege. This is the definition of even numbes two times anothe intege. Thus, m must be even. Poof method 2 cicle one: Diect oof ndiect oof Poof by contadiction The statement tanslates into, whee means m is even, and means m+7 is odd. o the oof by contadiction, we assume that the statement is false, and aive at a contadiction. o the statement to be false, we assume that must be tue, and must be false. Thus, we assume that m is even, and m+7 is even. f m is even, then m2k, whee k is some intege this is the definition of even numbes. Thus, m+7 2k+7 2k+3+1. This, howeve, is the definition of odd numbes, and theefoe m+7 must be odd. This contadicts ou assumtion that m+7 is even. Thus, the statement must be tue, as the one case whee it is false i.e. when m is even, and m+7 is even cannot occu. Gimaldi, section 2.5, theoem 2.4, ages

8 S 202, Sing 2005 Midtem 1: 24 eb 2005 Page 8/8 Question oints: onside the following statements. 1. f Dominic goes to the acetack, then Helen will be mad. 2. f Ralh lays cads all night, then amela will be mad. 3. f eithe Helen o amela gets mad, then Veonica thei attoney will be notified. 4. Veonica has not head fom eithe of these two clients. om these, can we conclude the following? Dominic didn t make it to the acetack and Ralh didn t lay cads all night. Wite each of these statements in symbolic fom. lealy label what you oolean vaiables eesent! Then establish the validity of the conclusion. You must clealy label which ule of infeence is used fo each ste. nswe: Dominic goes to the acetack h Helen gets mad Ralh lays cads all night c amela gets mad v Veonica is notified 1. h 2. c 3. h c v 4. v 1. h c v 3 d hyothesis 2. v 4 th hyothesis 3. h c Modus tollens on stes 1 and 2 4. h c DeMogan s law on ste 3 5. h Simlification of ste 4 6. h 1 st hyothesis 7. Modus tollens on stes 5 and 6 8. c Simlification of ste 4 9. c 2 nd hyothesis 10. Modus tollens on stes 8 and onjunction on stes 7 and 10 Gimaldi, uestion 12b, section 2.3, age 86.

arxiv: v1 [math.nt] 12 May 2017

arxiv: v1 [math.nt] 12 May 2017 SEQUENCES OF CONSECUTIVE HAPPY NUMBERS IN NEGATIVE BASES HELEN G. GRUNDMAN AND PAMELA E. HARRIS axiv:1705.04648v1 [math.nt] 12 May 2017 ABSTRACT. Fo b 2 and e 2, let S e,b : Z Z 0 be the function taking