Download MAT 102 Solutions – Take-Home Exam 2 Problem 1 a) The set of

MAT 102
Solutions – Take-Home Exam 2
Problem 1
a)
The set of whole numbers under subtraction is not a group. As checked below, it
violates all four group properties.

CLOSURE
The system is not closed. For example: 1  3  2 but 2 is not a whole
number.

IDENTITY & INVERSES
The system has no identity and, therefore, none of its elements have
inverses. To see that there is no identity here, argue by contradiction. Assume
Then, for any whole number a , we would have
i is the identity.
a  i  i  a  a . But that would imply that i  0  a  2a … Clearly this is not a
possibility whenever a is positive. Therefore, the identity doesn’t exist.

ASSOCIATIVITY
The operation of subtraction is not associative. In general, for any three
a,
c,
whole
numbers
and
we
have
b,
For
example:
 a  b   c  a  b  c  a  b  c   a  b  c .
 5  3 1  2 1  1  5  3 1  5  2  3 .
b)
The set of positive fractions under multiplication is an Abelian group. As checked
below, it satisfies all five conditions for commutative groups.

CLOSURE
The system is closed since the product of any two positive fractions is also a
positive fraction.

IDENTITY
The system has the usual multiplicative identity 1 since
positive fraction

a
a a
1  1  for every
b
b b
a
.
b
INVERSE
Every positive fraction
a
b
a b
has its reciprocal
as its inverse since   1 .
b
a
b a

ASSOCIATIVITY & COMMUTATIVITY
The operation of multiplication is clearly associative since for any three positive
e
a c
 a c  e a  c e  ace
fractions , , and
, we have          
. For example,
b d
f
 b d  f b  d f  bdf
 1 2 3 1  2 3 1
It is also clearly
        .
 2 3 5 2  3 5 5
a c c a ac
2 3 3 2 2
. For example,     .
   
b d d b bd
3 5 5 3 5
c)
d)
commutative
since
The set of even integers Zeven  ..., 4, 2,0, 2, 4,  under addition is an Abelian
group. As checked below, it satisfies all five condition for commutative groups.

CLOSURE
The system is closed since adding two even integers always yields another even
integer. For example: 6  4  2 .

IDENTITY
The system has the usual additive identity 0 since 2a  0  0  2a  2a for any
even integer 2a (where a is any integer).

INVERSE
Every even integer 2a has its opposite 2a as its inverse since 2a   2a   0 .
For example, 6  6  0 .

ASSOCIATIVITY & COMMUTATIVITY
Since the operation of addition is associative and commutative for integers, it
must also have the same properties for even integers.
For example:
2   4  10    2   4    10  4 and 2  4  4  2  2 .
The set of real numbers (i.e. all the numbers that can be placed on a number line) under
division is not a group. As checked below, it fails 3 out of the 4 group conditions.

CLOSURE
The system is closed since the quotient of any two real numbers is also a real
number. Note that the operation of division does not include cases where the
denominator is zero since dividing any number by zero is not defined as a valid
operation in arithmetic.
e)

IDENTITY & INVERSES
The system has no identity and so none of its elements have inverses.
Arguing by contradiction, if the system had an identity i , then it would satisfy the
x i
following equations for any non-zero real number x :   x . But this would
i x
2
imply that i  1  x . This is clearly not true for any real number x  1 . Thus,
the identity doesn’t exist.

ASSOCIATIVITY
The operation of division is not associative. Here’s a counterexample:
1
1 1
1
1
1  2  10  2     1   2  10   2  5 .
10 2 10 20
10
The set of positive fractions/rationals under the “averaging” operation is not a group.
As checked below, it fails 3 out of the 4 group conditions.

CLOSURE
The system is closed since averaging any two positive fractions clearly yields
another positive fraction.

IDENTITY & INVERSES
The system has no identity and so none of its elements have inverses.
Arguing by contradiction, if the system had an identity i , then you can check that
it would satisfy the following equation for any positive fraction a :
ai
 a  a  i  2a  a  i . But this would imply then that the inverse is not
2
uniquely determined. Thus, the identity doesn’t exist for this system.

ASSOCIATIVITY
The operation of “averaging” is not associative. Here’s a counterexample:
1 3
5
consider the positive fractions , , and . Suppose you average the fractions
4 4
4
1
3
1
and
first. Then you will get . Averaging this number to the third
4
4
2
5
7
3
5
fraction
yields . However, the average of
and
is 1 and averaging this
4
8
4
4
1
5 7
number now to the first fraction
returns  .
4
8 8
f)
The set of truth values under the logical operation of disjunction is not a group. As
checked below, it fails 1 out of the 4 group conditions.

CLOSURE
The system is closed since the truth table for disjunctions only contains the
truth values T and F.

IDENTITY & INVERSES
The system has the identity F since T or F = T and F or F = F. Checking for
inverses, we see that the inverse of F is F since F or F = F. However, T fails
to have an inverse since any disjunction with a true component is necessarily
true!

ASSOCIATIVITY
It is easy to check that the operation of disjunction is associative.
Problem 2
a)
 is closed under  since for any two integers a and b , the group operation generates
the number a  b  a  b  5 , which is also an integer.
In other words, the group operation  could never generate a non-integer (like a
fraction, or an irrational number).
b)
Suppose the identity is denoted by i . Then to find the value of i , we solve the equation
a  i  a for any integer a (we could also solve i  a  a ):
a  i  a  i  5  a  i  5  0  i  5 .
So the identity is the number -5. You can check this for any integer a :
a  5  a  (5)  5  a   5  5  a  0  a .
c)
By definition, applying the group operation to an integer and its inverse must always
produce the identity -5.
Now, for any integer a we have:
a    a  10  a    a  10   5  a   a  10  5   a  a   10  5  5 .
We then conclude that every integer a has the inverse (a  10) under  .
d)
For any three integers a, b, c we have:
and
Therefore
 a  b   c   a  b  5  c  a  b  5  c  5  a  b  c  10
a   b  c   a   b  c  5  a  b  c  5  5  a  b  c  10 .
 a  b  c
 a   b  c  and the associative property holds for this group
operation. It is easy to check that the commutative property also holds for this group
operation.
Based on part a, b, c, and d, we conclude that  is an Abelian group under  .
e)
Problem 3
Let  denote the operation of adding in this “clock 7” arithmetic.
a)
Answer = 2
In clock 7 arithmetic, the inverse of 5 is 2 since 5  2  7 .
b)
Answer = 7
1 2  3  4  5  6  6  2  6  6 1  7 .
c)
Answer = 3 since 3  2017  6051  7  864  3 .
Problem 4
To check closure, consider the cases when a  b is odd and a  b is even.

If a, b are digits in D  1, 2,3,...,8,9 and a  b is even, then a  b is either 2, 4, 6, 8, ...,
16, or 18 and so ab returns either 1, 2, 3, …, 8, or 9.

If a, b are digits in D  1, 2,3,...,8,9 and a  b is odd, then a  b  1 is either 4, 6, 8, ...,
16, or 18 and so ab returns either 2, 3, …, 8, or 9.
In both cases, the operation only yields digits in D , so D is closed under  .
Problem 5
Exercise # 66 on page 582.
a)
The elements of the set in this system are the numbers 3, 4, 5, and 8.
b)
The binary operation is the “turtle” symbol given in the table (the table itself actually
defines the operation.)
c)
The system is closed since only the numbers 3, 4, 5, and 8 appear in the table.
d)
Yes, the identity is the number 4.
e)
Yes, inv(3) = 8, inv(4) = 4, inv(5) = 5, and inv(8) = 3.
f)
Let * denote the “turtle” operation defined by this table. Then
(3*5)*8 = 8*8 = 5
and
3*(5*8) =3*3 = 5.
This example illustrates the associative property of *.
g)
Yes, the system is also commutative since there is symmetry with respect to the main
diagonal. For example, 5*8 = 8*5 = 3.
h)
Yes, it satisfies all 5 conditions for a commutative/Abelian group.
Problem 6
Exercise # 72 on page 583.
The mathematical system given by the table is not associative and it is also not
commutative.
For example, if we let * denote the “happy face” operation considered here, then the following
counterexample shows why the associative property fails here:
(delta*0)*pi = b*pi = pi,
but
delta*(0*pi) = delta*delta = a.
The following counterexample shows why commutativity also fails:
delta*pi = pi, but
pi*delta = a.
Problem 7
Exercise # 76 on page 583.
a)
b)
x
E
O
E
E
E
O
E
O
The set {E, O} is not a group under multiplication since E does not have a
multiplicative inverse! To see why, first note that O is the identity in this system.
Then, the table shows that E fails to have an inverse since E x E = E x O = E.
Bonus Problem
Exercise # 79 on page 584.
See answer in A-35 (back of the book).