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MTH 200 – Graded Assignment 3
Material from Module 5 (The Foundations of Algebra) and Module 6 (More than You
Ever Wanted to Know about Addition)
P1: We have discussed four properties that relations on sets may have in the lecture.
What are they? State both the name and the formal definition. I’ll start to give you an idea
of where this is coming from…
1. Reflexive. A relation R on a set A is reflexive if for every x  A , xRx .
2. ?
3. ?
4. ?
P2: Which three of the four properties define an equivalence relation?
P3: Which three of the four properties define a partial order?
P4: For the relation  (“greater than or equal to”) on the set ¡ , which of the four
properties hold, and which fail? Support your answers (see suggested problems in the
section on equivalence relations as a model).
P5: Is the relation  on ¡ an (1) equivalence relation, (2) partial order, (3) both, (4)
neither?
P6: A set A is given: A  {1, a, Z }
Is the relation defined on set A by
R  {(1, a), (a,1), (1, Z ), ( Z ,1), ( a, Z ), ( Z , a)}
an equivalence relation? Support your answer by referring to the properties of
equivalence relations – which hold, which fail, and why?
P7: Let A be the set of integers ¢ , and let R be defined by
x y
R  {( x, y )|
¢}
2
a) Verify that R is an equivalence relation.
Note: Reflexive and symmetric are straightforward, but transitive is a little tricky,
x y
yz
 ¢ and
 ¢ . You
so here’s some hints: To show transitive, assume
2
2
xz
x  z x  z  2z
 ¢ . Use the facts that

need to get to
, and that
2
2
2
xz  x y  yz


.
2
 2   2 
b) Give some examples of integers that are equivalent to 1 under this relation.
[1]  {...???...}
1 y
is an integer?)
2
c) Give some examples of integers that are equivalent to 2 under this relation.
[2]  {...???...}
(What are some values of y for which
2 y
is an integer?)
2
d) This relation sorts the integers into two familiar categories – how could you
describe [1] and [2] in general?
(What are some values of y for which
P8: For which of the sets given below does the Well-Ordering principle guarantee a
smallest member? Explain.
a) The set of all integers in the form 22 x  13 y .
b) The closed interval of real numbers [5, 272] .
c) The set {x | x  ¥ , x  25} .
P9: Prove by induction: For each natural number n  2 , n3  n is divisible by 3 .
Basis step: Show P (2) :
Induction step:
1. Assume P ( n ) :
2. We wish to show P ( n  1) :
3. Consider:
4. And prove it:
Formal conclusion: Since P(n)  P(n  1) for all n  2 , and P (2) holds, it must be that
P ( n ) is true for all n  2 .  For each natural number n  2 , n3  n is divisible by 3 .
P10: Prove by induction: For each natural number
1·2  2·3  3·4  ...  n(n  1) 
n(n  1)(n  2)
.
3
Basis step: Show P (1) :
Induction step:
1. Assume P ( n ) :
2. We wish to show P ( n  1) :
3. Consider:
4. And prove it:
Formal conclusion: Since P(n)  P(n  1) for all n  1, and P (1) holds, it must be that
P ( n ) is true for all n  1.  For each natural number n  1,
n(n  1)(n  2)
1·2  2·3  3·4  ...  n(n  1) 
3
P11: What can we define the natural numbers in terms of? How is addition of natural
numbers defined in this context?
P12: How are the integers defined in terms of natural numbers?
P13: Take the integers 5 and 14 .
a) Give two representations for each.
b) Compute the sum 5  (14) using the definition of addition of integers as
equivalence classes of ordered pairs (use either of your examples in part (a)), and
verify that your answer is a member of the equivalence class for 9 .
c) Compute the product 5·(14) using the definition of multiplication of integers as
equivalence classes of ordered pairs (use either of your examples in part (a)), and
verify that your answer is a member of the equivalence class for 70 .
P14: Use the arithmetic properties (or any of the previously established properties such as
Proposition 1 on p. 12) of the integers to prove the following:
Let a and b be integers. Prove that (a)·b  (ab) .
This is #4 in the section 1.2 textbook exercises, p. 13.
Hints: You need to show that ab  (a )·b  0 ; that implies that (a)·b must be the
additive inverse of ab . I started with 0·b  0 (Proposition 1) and got there in two steps.