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pracexam1sol

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Unformatted text preview: SOLUTION for Practice Exam, CMSC 250    These questions are intended to provide a reasonably complete overview of the  concepts we have covered so far this semester.  The actual exam will have  somewhat fewer questions, but drawn largely from this general type of question.  Since there was no homework set on the most recent material, these questions  include a bit more on that material than will be on the exam, in order to provide  extra practice. There may also be some True‐False or Multiple‐choice questions due  to lack of time for “problem‐questions” on all topics.    1. Is √2 < √2 ?   EXPLAIN.  The LHS is the sq root of 2, about 1.4; the RHS is the  ceiling of about 1.4, namely 2. So the ineq is correct.    2. Which real numbers x satisfy  x = x ?   (Find the maximum domain for x.)  The only way a (real) number can be its floor is if it is an integer. So the truth  set (or domain where the equation holds) is simply the subset Z of R.    3. Is the following argument form valid? If so, show it is valid; if not, show it is  not:  It is not valid. There is no way to get ~R from the premises. Here is a proof:  Look at a truth­table row in which R is true and  P is false; that will make all  three premises true, and the conclusion false. But that cannot happen for a  valid argument.  Step number  Statement form    1.   2.  3.  4.  P ‐> Q  Q ‐> R  ‐P  ‐R   Premise  Premise  Premise  Conclusion    4. Give a number domain containing the number 10, such that the following  sentence is true, with the usual interpretation of the mathematical symbols:                       ∃z  ∀x  ∀y [ x‐y ≤  z ] .   EXPLAIN. The domain {0,10} having 0 and  10 as its  only elements will do.  Just let z=10; any difference x­y between two of those  elements is either ­10, 0, or 10, so the absolute value is ≤ 10=z.    5. Give a number domain containing the number 10 such that the sentence in  problem 4 is false (again with the usual interpretations of the symbols).   EXPLAIN.  The domain R will do; there is no z in R such that all differences  between elements  are no larger than that.  In particular, for any real z, we  have that | (1+|z|) – 0 | > z, so letting x=1+|z| and y=0 makes |x­y|>z.   (The  reason this does not happen in the answer to prob 4 where z is 10, is that then  1+z  is 11, which is not in the domain {0,10} at all, so it is not a possible value  for x.)         6. Prove that for all natural numbers x, if  x^4 ≡3 0 , then x ≡3 0 .   (Recall  a ≡d b   means a‐b is divisible by d.)  Use the unique factorization theorem (UFT).   Suppose x^4 = 3m. So (x^2) (x^2) is divisible by 3, and since 3 is prime then (by  a Lemma) x^2 is divisible by 3.  We then have x^2 = x x divisible by 3, and by the  same Lemma, so is x. So,  x ≡3 0.  7. Prove the same result as in problem 6—but  do notuse UFT. Suppose x^4 =  3m; consider the remainders when dividing x by 3: 0,1,2. These give us three  cases, and we want to rule out the cases 1 and 2, to conclude that the  remainder is 0 and that x itself is divisible by 3.  Case 1: x = 3k+1. Then x^4 =  3^4 k^4  +  4( 3^3 k^3)  +  6(3^2 k^2)  +  4(3k)  + 1  =  3(…) + 1 ≡  1 (mod 3).   But this cannot be, since we know x^4 ≡ 0 (mod 3).  So case 1 cannot happen.  Now case 2: x=3k+2. Again we calculate x^4 = 3^4 k^4  +  4( 3^3 k^3) 2  +   6(3^2 k^2)2^2  +  4(3k)2^3  + 2^4  =  3(…) + 2^4 ≡ 2^4 = 16 ≡  1 (mod 3).  So  this case cannot happen either; and thus only case 0  (x≡3 0) can – and so must  – hold.          8. Show that 3^(1/4) is irrational, using the result in problems 6 and 7.   Suppose not: suppose 3^(1/4) is rational, hence 3^(1/4) = m/n for suitable  integers m and n. Then 3 = m^4 / n^4 so m^4 = 3n^4 and thus m^4 is divisible  by 3.  But by the result in probs 6 and 7, this means m is divisible by 3: m=3k for  some integer k. So m^4 = (3k)^4 = 3n^4, and then 3^3 k^4 = n^4; and this  means also n^4 is divisible by 3, and so then is n. Thus, m and n must both  divisible by 3.  But this cannot be since m/n is simply a rational number of some  particular value (namely 3^(1/4)) and that cannot force both numerator and  denominator to have the factor 3. So the supposition that 3^(1/4) is rational  must be wrong, so 3^(1/4) is not rational, hence it is irrational.  9. The only way a conditional statement  P ‐> Q  can be false is if:    P is true and  Q is false.  10. Describe in English how truth tables, disjunctive normal form, circuit design,  and propositional statement forms are related.  Be brief: 20‐30 words only.  A  circuit has a description as a logic statement, which has a unique truth­table,  that gives a unique disjunctive normal form, which leads to  a circuit with the  same input­output as the original one.    11. State the Quotient‐Remainder Theorem (QRT).  Given integers n and d with  d>0, there are unique integers q and r such that n = qd + r where ) 0 ≤ r < d.          12. Prove the QRT using floors.  Let q = n/d and r = n –qd.  Then from the  definition of floor,  q ≤ n/d < q+1. So qd ≤ n < qd + d (since d>0)., and then 0 ≤ n­ qd < d, i.e., 0 ≤ r < d.  This shows q and r exist.  (Uniqueness follows from the fact  that if q’ were larger than q, using it as quotient would force r to be negative;  and if q’ were smaller than q, using it as quotient would force r to be larger  than d.  This is easy to check.)  13. True or false:  ∃x ε R   3x   =  3  x  .   JUSTIFY.  True: let x = 0, for instance, or  any integer at all, since then the floor of x is just x, and the floor of 3x is just 3x.      14. True or false:  ∀x ε R   3x   =  3 x  .  JUSTIFY.  False; this does not hold when  x = 1/3, for instance. The floor of 1/3 is 0, so the RHS is 0; and the floor of 3x = 1  is then the floor of 1, so the LHS is 1.  15. One of the following is always true and one is not always true. Which is  which?    EXPLAIN.      The first one is always true: if one fixed x=x0 works  (makes P true) for all the different y values, then for any y there is an x that  works, namely that same x0.       But just because each y has some x that works  with it, does not guarantee that one fixed x works for all y.  (Intuitively: If one  person loves everybody, then everybody is loved by somebody.     But if  everybody is loved by somebody, it does not follow that everybody is loved by  one and the same person.                                                                                                     ∃ x ∀ y P(x,y)   ∀ y ∃ x P(x,y)                                                                                          ∀ y ∃ x P(x,y)   ∃ x ∀ y P(x,y)        16. Give the definition of logical equivalence of two statement forms:   S1  ≡  S2      means  “in each row of a common truth‐table for the two forms, both forms  have the same truth­value”.    17. Are these two forms logically equivalent”  If so, prove it; if not prove not.          (∼ P  & ~R)  and  ~(~P  R).  EXPLAIN your proof.  Yes, equivalent. The  second says that ~P  R is false, which means that ~P must hold and R must  not, i.e.,  ~P & ~R.      18. Prove or disprove:  if a ≡m b, and c ≡m d, then a‐c ≡m b‐d.  This is true. Let the  first two congruences hold. Then m | a­b and m | c­d, so a­b = km and c­d = rm.  Subtracting gives  a­c – b + d = (a­c) – (b­d) = (k­r)m. So (a­c) ≡m (b­d).  19. USE UFT to show that   70,000  =/=  49 x 25 x 64.  (Do not multiply the RHS.)   70,000’s prime factorization has only the prime powers 7^1, 2^4, and 5^4  (since it is 7x10^4). But the RHS has two factors of 7, so they cannot be equal.  ...
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