Math 111 Test Template

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MATH 111
Test 2 - Solutions
3
2
1. Let x ∈ Z. Prove that if x
+ x
+ x is odd, then x is odd. What type of proof (direct,
by contrapositive, or by contradiction) did you use?
We will prove this statement by contrapositive. If x is even, then x = 2k for some
3
2
3
2
3
2
3
2
k ∈ Z. Then x
+ x
+ x = (2k)
+ (2k)
+ 2k = 8k
+ 4k
+ 2k = 2(4k
+ 2k
+ k).
3
2
3
2
Since 4k
+ 2k
+ k ∈ Z, x
+ x
+ x is even.
2. Prove or disprove.
The sum of two irrational numbers is irrational.
The statement is false. Counterexample:
2 + (
2) = 0. We know that
2 is
0
irrational. Below we show that
2 is also irrational. However, 0 =
is rational.
1
We will show that
2 is irrational by contradiction. Suppose it is rational, then
m
m
m
2 =
for some m, n ∈ Z, n = 0. Then
2 =
=
. Since m, n ∈ Z and
n
n
n
n = 0,
2 is rational. However, this is not the case. Therefore
2 is irrational.
3. Prove or disprove.
The sum of a rational number and an irrational number is irrational.
The statement is true. We will prove it by contradiction, namely, suppose there exist a
k
rational number r and an irrational number x such that r + x is rational. Then r =
l
m
m
k
and r + x =
for some k, l, m, n ∈ Z, l = 0, n = 0. Then x = (r + x)
r =
=
n
n
l
ml
nk
. Since ml
nk ∈ Z and nl = 0, x is rational. We get a contradiction.
nl
4. Prove or disprove.
Let A, B, and C be sets. If A ∪ B = A ∪ C, then B = C.
The statement is false. Counterexample: A = {1, 2, 3}, B = {1}, C = {2}. Then
A ∪ B = {1, 2, 3} = A ∪ C, but B = C.
5. Let A = {1, 2, 3, 4}. Give an example of a relation on A that is symmetric but not
reflexive.
R = {(1, 2), (2, 1)} is symmetric (because for any (a, b) ∈ R, (b, a) ∈ R), but not reflexive
(because e.g. (1, 1) ∈ R).
6. Let R be the relation on Z defined by (a, b) ∈ R iff a + b < 5.
(a) Is R reflexive?
No, because e.g. (3, 3) ∈ R since 3 + 3 < 5.
1

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