Download Solutions to Discussion #14 Worksheet

Solutions to Discussion #14 Worksheet
1. Let functions f: R  R and g: R  R be defined by f(x) = x2 + 1 and g(x) = 2x + 3. Find
fg
f  g  2 x  32  1 = 4x2 + 6x + 10
2. Let S = {0, 1, 2, 3, 4, 5} and let f: S  S be defined by f(x) = 4x mod 5. Write out all
ordered pairs for the function. Prove or disprove that the function is one-to-one.
The ordered pairs are 0,0, 1,4, 2,3, 3,2, 4,1, 5,0 . Since f 0  f 5  0 , it follows that
f is not 1-1.
3. Check each property possessed by the following relation: {(a, b) | a, b  Z and ab  0}
Let the relation be denoted by R .
 Reflexive: Since (1, 1)  R, R is not reflexive.
 Irreflexive: Since 0,0  R , R is not irreflexive.
 Symmetric: If a, b  R then ab  0 hence ba  0 which implies that b, a   R , thus R is
symmetric.
 Transitive: 1,1  R and  1,2  R , but 1,2  R , hence R is not transitive.
 Antisymmetric: 1,1  R and  1,1  R but 1  1 , hence R is not antisymmetric.
 Asymmetric: Since R is nonempty and is not irreflexive, it follows that R cannot be
asymmetric.
4. Suppose |S| = 20, and R is a relation on S.
a. If R is reflexive, what is the minimum value for |R|? _______________
The only pairs that absolutely have to be in R are those of the form a, a , a  S .
Since S  20 it thus follows that R can be as small as 20.
b. If R is symmetric, what is the maximum value for |R|?
The largest symmetric relation on S is S  S . Thus if R is symmetric, its maximum
cardinality is S  S  20 2  400 .
c. The number of relations on S is _____________________
A relation on S is a subset of S  S . Each and every subset of S  S is a relation on
S . Since S  S  20 2  400 , it follows that there are 2
S S
 2400 relations on S .
5. A pair of (6-sided) dice is rolled. Find the probability:
a. that 6 comes up exactly three times in 10 rolls
There are 5 ways to get a sum of 6. This is a Bernoulli trials problem. Hence
the probability of getting a sum of 6 exactly three time in ten rolls is
3
10  5   31 
    
 3  36   36 
7
.
b. that one of the dice is a 2, given that the sum is 7
There are six ways to get a sum of 7, so the given information has reduced the
sample space from 36 possible outcomes to 6. In two of those outcomes one of the
dice is a 2, hence pone of the dice is 2 | sum is 7   . Of course this problem can
2
6
1
3
also be solved by applying the definition of conditional probability directly to find
pone of the dice is 2 | sum is 7 
pone of the dice is 2 and sum is 7 2 36 1

 .
psum is 7
16
3
c. of rolling a pair, given that the total is at least 9
There are ten ways to get a sum of 9 or more, and two of them are pairs, namely
5,5 and 6,6 . Hence proll a pair | sum is at least 9 
2
. Applying the definition of
10
conditional probability yields
proll a pair | sum at least 9 
proll a pair and sum is at least 9 2 36 1

 .
psum is at least 9
10 36 5
6. A die is rolled. E is the event that the number rolled is odd, and F is the event that the
number rolled is greater than 4. Are E and F independent? Be sure to show all
probabilities used in determining your answer.
p E  
1
1
1
and pF   . Also, E  F  roll is odd and greater than 4 so that pE  F   .
6
2
3
Since pE   pF   pE  F  , it follows that E and F are independent.
7. Show that if R is an antisymmetric relation on set S, then R-1 is also antisymmetric.
Proof Assume R is antisymmetric and suppose a, b, b, a   R 1 . We must show
that a  b .
a, b, b, a  R 1  b, a, a, b  R , definition of the inverse relation.
definition of antisymmetric
R is antisymmetric, so a = b
1
Therefore R is antisymmetric. definition of antisymmetric