Math 163 - Discrete Math - Spring 2014

Prof. Mark Janeba

Answers to selected even-numbered problems for chapter 2.4  3.2  5.2  5.3 7.1 10.1 April 11  April 14  11.4


Chapter 2.4

#16a) There is a student in Prof. Lenhart's C++ class that is not majoring in computer science and not majoring in mathematics.
b) Every student in Prof. Lenhart's C++ class is a non-history major.

#20 a) If a number is divisible by 21, then it is divisible by 7 (which is true, though you didn't have to answer that part)
   Converse:  If a number is divisible by 7, then it is divisible by 21 (which is false)
    Contrapositive: If a number is not divisible by 7, then it is not divisible by 21.  (What does the truth value have to be, given that the original is true?)

b) If a snake is a cobra then it is dangerous.  Converse:  If a snake is dangerous then it is a cobra.   Contrapositive:  If a snake is not dangerous then it is not a cobra.

c) For every complex number z, If z2 is real, then z is real. (if you got this one backwards, go back to 2.1 p. 48 item 2c and review)  (If you are curious, this one is false.  What does that tell us about the truth values of the converse and contrapositive?)
    Converse:  For every complex number z, If z is real, then z2 is real.  Contrapositive:  For every complex number z, If z2 is not real, then z is not real.

Chapter 3.2

#2) Recall that [a,b] means the set (or interval) of real numbers from a to b inclusive, while (a,b) means the set (or interval) of real numbers from a to b exclusive.
Drawings of the sets on the real line (number line) will help.
a) [2,3]    b) [0,7)    c) (-[infinity],0)[union](3,[infinity])       d) [0,2)[union](3,7)      e) [0,2)      f) (3,7)

#4a)
i) T     ii) F  (A contains 2 but C doesn't, also C contains 4 but E doesn't)
iii) F (consider the element 3)        iv) T     v) T       vi) F  (For example, the complement of D contains 4)
#4b)
i) C[intersect]E = E     ii) B[union]D=B     iii) A[intersect]B=D
iv) B[intersect]D=D     v) the complement of A is the set of all odd integers, which include -1,-3,-5, and so on, as well as the more familiar odd integers.
vi) A[intersect]E = E

Chapter 5.2 supplementary exercises

If A={6,7,8,9,10} and B={p,q,r,s,t,u},
tell whether each of the following is (a) not a function from A to B, (b) a function from A to B, but not a one-to-one function, or (c) a one-to-one function from A to B.
  1. (a) not function from A to B - no ordered pair begins with 10, one of the elements of A.
  2. (c) one-to-one function; the extra ordered pair with 8 is just a repeat and can be ignored, it is not a distinct pair
  3. (a) not a function, two distinct ordered pairs beginning with 8
  4. (b) function but not one-to-one; both 6 and 9 are paired with p.
  5. (a); This is a function from B to A, but that's not the same as a function from A to B.  Note this function is not 1-1, nor can it possibly be (why?)
  6. (a) not a function from A to B - no ordered pair begins with 9.  There's a repeat for 6, but it can be ignored.
  7. (b) a function, a very boring function, but most definitely not 1-1.
There are 65 functions from A to B; each entry of A can be paired with any of the 6 entries of B.
There are 6P5 one-to-one functions from A to B; there are 6 choices to pair with the first element of A, once that is done, there are 5 choices to pair with the next element of A, and so on.
There are 56 functions from B to A, and zero (!) one-to-one functions from B to A, because we run out of choices for the last element of B.

Chapter 5.3

2a) one-to-one AND onto      2b) one-to-one, NOT onto. The range is the odd integers.      2c) one-to-one AND onto

4a) 64 functions, 6P4 = 360 of them are one-to-one, none are onto b/c |B| is smaller than |A|.
4b) 46 functions, none of them are one-to-one, 46-(4C3)36+(4C2)26-(4C1)16=1560 are onto (use formula atop p. 262)

8) We need to count functions that send the 9 compounds to the 5 assistants, but only functions that are onto b/c each assistant gets at least one. 
         That's counting onto functions from a set of 9 to a set of 5,
         so 59-(5C4)49+(5C3)39-(5C2)29+(5C1)19=834120

Miscellaneous problems from March 14

Chapter 7.1

6) (5a) is a partial order, (5c) and (5f) are equivalence relations.
10a) 212.    10b) 210.   10c) 26.
Misc problems for March 31: 
R1 is none of these: reflextive, symmetric, transitive.
R4 is symmetric and transitive, under the usual definition of (full) sibling - though not transitive for half-siblings.  Not reflexive for any usual definition of "sibling."
R5 is reflextive, symmetric, and transitive, under most common definitions of "family."
R6 is transitive.

Chapter 10.1

2a)   This rearranges to an+1 = 1.5an , a geometric recursion. Our formula tells us an=Adn, where A=a0 and d=1.5 (1.5 from the rearranged equation).
            Since we aren't given a0 this time, all we can say is that an = a0(1.5)n.

2b)  Rearrange to an = (5/4)an-1, then continue as in 2a above to get an = a0(5/4)n=a0(1.25)n.


2c)  Rearrange to an+1 = (4/3)an , then use an=Adn to get an = a0(4/3)n, but notice they gave us a1 instead of a0, so use the formula we just found:
          5=a1=a0(4/3)1, now solve this new equation for a0 to get a0 = 15/4.  Put it all together:  an = (15/4)(4/3)n.

2d)  Using the methods outlined above, an = 16(3/2)n.

April 11 homework

    1. an+1 = (1.004)an , ao = 550.
      Answer:  an = 550(1.004)n.
    2. an+1 = an + 2.5, ao = 12.
      Answer:  an = 12 + 2.5n

April 14 homework

  1. an+1 = (1.0015)an + 50, a0 = 50.
  2. an(h) = a0(1.0015)n = 50(1.0015)n, an(p) = -33,333 so
    an = 33,383(1.0015)n - 33,333
  3. The balance after 10 years is $6628 (rounded).  This is only $628 more than the amount deposited (current savings interest rates are NOT good).
  4.  For the recurrence relation an+1 = (1.0125)an - 119, a0 = 5000:
    1. This models a loan of $5000 charging an interest rate of 1.25% per month or 15% per year, which is currently a typical credit-card rate. (which is also awful).
      The monthly payment being made is $119.
    2. an(h) = a0(1.0125)n = 5000(1.0125)n, an(p) = 9520 so
      an = -4520(1.0125)n + 9520
    3. It takes from 59 to 60 months (after 60 months, the balance is only -$4, i.e. only a tiny overpayment).
    4. 60 payments of $119 is $7140 paid back on a $5000 loan.  Credit cards are a bad way to borrow money.

Chapter 11.4

    8) a) 2   b) 6   c)  14   d) 2m, but only because m<n, so n must be at least one more than m.
   10) No.  Using the notation of Definition 11.18 on p. 541, every edge connects a point in V1 to V2; on a path, we alternate between vertices in V1 and V2.  To end up where we started, we need an even number of steps or edges.

Last Modified April 23, 2014.
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