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Mathematics 220
Midterm Practice problems from old exams
Page 1 of 8
1. (a) Write the converse, contrapositive and negation of the following statement:
For every integer n, if n is divisible by 3 then n2 is divisible by 3.
Solution:
• Converse: For every integer n, if n2 is divisible by 3, then n is divisible by
3.
• Contrapositive: For every integer n, if n2 is not divisible by 3 then n is
not divisible by 3.
• Negation: There exists an integer n such that n is divisible by 3 and n2 is
not divisible by 3.
(b) Let An be the interval [−n , 2 + n43 ) for n ∈ N. Find
\
n∈N
An and
[
An (no proof is
n∈N
required).
Solution:
T
• n∈N An = [−1, 2]
S
• n∈N An = (−∞, 6)
(c) Write R − N as a union of an indexed collection of sets where each set is an interval.
Solution: If we let I := N ∪ {0}, let A0 := (−∞, 1) and, for every n ∈ N, let
An := (n, n + 1), then:
R−N=
[
Ai .
i∈I
2. Consider the following two statements:
1. ∀n ∈ N, ∃ z ∈ Z such that z = n
2. ∃ z ∈ Z such that ∀n ∈ N, z = n
One of the statements is true, and the other is false. Determine which is which and prove
both of your answers.
Solution:
The first statement is true.
Proof. Since N ⊂ Z, given any n ∈ N we can simply choose z to be equal to n.
The second statement is false.
Mathematics 220
Midterm Practice problems from old exams
Page 2 of 8
Proof. We prove that its negation: ‘∀z ∈ Z, ∃n ∈ N such that z 6= n’ is true. Given
any z ∈ Z we can simply choose n := z 2 + 1. Since z 2 + 1 ∈ N and since z 6= z 2 + 1
for any integer z (why not?) this completes the proof.
3. (a) Give the definition of the power set P(A) of a set A.
Solution:
See textbook.
(b) Let A = {1, 2, {1, 2}}. Determine whether the following statements are True or
False (and provide a brief explanation why).
(a)
(b)
(c)
(d)
{1, 2} ⊆ A.
{1, 2} ⊆ P(A).
{1, 2} ∈ A.
{1, 2} ∈ P(A).
Solution:
(a) True because 1 ∈ A and 2 ∈ A.
(b) False because 1 ∈
/ P(A) and 2 ∈
/ P(A).
(c) True because it is listed as an element of A.
(d) True because (a) was true.
4. Let A := {n ∈ N : ∃ z ∈ Z such that n = 2z + 1} and
let B := {n ∈ N : ∃ k ∈ N such that n = 2k}. Determine the following:
1. A ∩ B,
2. A ∪ B,
3. A − B, and
4. B − A.
Solution: Notice that A coincides with the odd natural numbers while B coincides
with the even natural numbers. As such:
1. A ∩ B = ∅,
2. A ∪ B = N,
3. A − B = A, and
4. B − A = B.
Mathematics 220
Midterm Practice problems from old exams
Page 3 of 8
5. True or False: You do not have to give explanations.
(a) N ⊂ Z.
Solution: T
(b) [2, 3] ∈ P(R).
Solution: T
(c) 2 ∈ P(R).
Solution: F
(d) ∪n∈Z (n, n + 2) = R.
Solution: T
√
√
(e) Q ∩ ( 2, ∞) = Q ∩ [ 2, ∞).
Solution: T
6. Using any method you like, prove that the following statements are logically equivalent:
Statement 1: Q ⇒ (R ⇒ S)
Statement 2: (Q ∧ R) ⇒ S.
Solution:
Q ⇒ (R ⇒ S) ≡ (∼ Q) ∨ (R ⇒ S)
≡ (∼ Q) ∨ ((∼ R) ∨ S)
On the other hand
(Q ∧ R) ⇒ S ≡ (∼ (Q ∧ R)) ∨ S
≡ ((∼ Q) ∨ (∼ R)) ∨ S
The two right hand sides are logically equivalent by Associative law.
7. Let n ∈ Z. Prove that n3 − 5n2 + 13 is odd.
Solution:
Proof. Let n ∈ Z and consider the following two (exhaustive) cases:
Mathematics 220
Midterm Practice problems from old exams
Page 4 of 8
• If n is even, then there is some k ∈ Z such that n = 2k. In this case,
n3 − 5n2 + 13 = (2k)3 − 5(2k) + 13 = 8k 3 − 10k + 12 + 1 = 2(4k 3 − 5k + 6) + 1
and since 4k 3 − 5k + 6 ∈ Z we conclude that n3 − 5n2 + 13 is odd.
• On the other hand, if n is odd, then there is some k ∈ Z such that n = 2k + 1.
In this case,
n3 − 5n2 + 13 = (2k + 1)3 − 5(2k + 1) + 13 = 8k 3 + 12k 2 + 6k + 1 − 10k − 5 + 13
= 8k 3 + 12k 2 − 4k + 8 + 1 = 2(4k 3 + 6k 2 − 2k + 4) + 1
and since 4k 3 + 6k 2 − 2k + 4 ∈ Z we conclude that n3 − 5n2 + 13 is odd.
8. (a) Write the negation of the following statement:
“For every positive there exists a positive δ such that if |x| < δ then |f (x)| < .”
Solution: We start negating this sentence, and get: “There exists a positive
such that for every positive δ, |x| < δ does not imply that |f (x)| < ”. If
we recall the rules for negating the implication, it is tempting to write: “There
exists a positive such that for every positive δ, x < δ and |f (x)| ≥ ”. Note,
however, that this doesn’t mean the same thing as the sentence just above (and
actually, unlike the above sentence, it is not a statement at all, but is an open
sentence with x being a variable). This happens because when we say “|x| < δ
implies |f (x)| < ”, we mean that this happens for all x that satisfy |x| < δ.
Thus, when we say “|x| < δ does not imply that |f (x)| < ”, we mean that there
exists x such that |x| < δ and |f (x)| ≥ . Putting it all together we get:
“There exists a positive such that for every positive δ, there exists
x such that |x| < δ and |f (x)| ≥ . ”
(b) Is the number 0.7(π − 3.1415) rational or irrational? (Include a short proof; you can
assume without proof that π is irrational).
Solution: It is irrational. Two ways to prove this: could refer to the results
from class: the sum/difference of a rational and irrational number is irrational;
the product of a non-zero rational number and an irrational number is irrational.
Alternatively, you could just use proof by contradiction (this is how we obtained
those results, anyway): Let a = 0.7(π − 3.1415). Suppose this number was
rational. Note that the numbers 0.7 and 3.1415 are both rational. Then we
have: π = a 10
+ 3.1415 would have been rational as well – a contradiction (we
7
are given that π is irrational).
9. Prove that if n and m are odd integers, then n2 − m2 ≡ 0 (mod 8) .
Mathematics 220
Midterm Practice problems from old exams
Page 5 of 8
Solution: The intended solution. The easiest way to solve this problem was by
cases: consider the possible remainders mod 8. If n is odd, it can be congruent to 1,
3, 5 or 7 mod 8. Then n2 is congruent mod 8 to, respectively, 12 = 1, or 32 = 9 ≡ 1
mod 8, or 52 = 25 ≡ 1 mod 8, or 72 = 49 ≡ 1 mod 8. Thus we see that in all the
cases, if n is odd, then n2 ≡ 1 mod 8. This also shows that m2 ≡ 1 mod 8, since m
is odd as well. Thus, n2 − m2 ≡ 1 − 1 = 0 mod 8.
Alternative solution (more popular): Since n and m are odd, we can write
n = 2k + 1, m = 2` + 1 for some integers k, `. By definition of congruence, we need
to prove that n2 − m2 is divisible by 8. Then
n2 − m2 = (2k + 1)2 − (2` + 1)2 = 4(k 2 − `2 + k − `).
Note that one cannot stop here: this clearly shows that n2 − m2 is divisible by 4, but
more work is needed to show that n2 − m2 is divisible by 8: namely, we need to prove
that the integer k 2 − `2 + k − ` is even. So let us prove the lemma:
Lemma. For any two integers k, `, the number k 2 − `2 + k − ` is even.
Proof of the Lemma: either by cases (you need to consider 3 cases: k, ` both even,
both odd, and one is even one is odd), or better, note that for any k, k 2 − k is even
(this can be proved by cases: k even/odd, or by considering congruence mod 2).
Then k 2 − `2 + k − ` = (k 2 − k) + (`2 − `) is even.
Now, by the Lemma, we have that there exists an integer q such that k 2 − `2 + k − ` =
2q, and putting it all together, we get that n2 − m2 = 4 · 2q = 8q.
10. (a) Prove that
√
10 is irrational.
Solution: Proof by contradiction.
√
Suppose 10 was rational.
Then there exist√integers a and b with no common
√
factors, such that 10 = ab . Then a = b 10, and therefore a2 = 10b2 ; in
particular, a2 is even. We proved in class that a2 is even if and only if a is even,
so we get that a is even, that is, a = 2k for some integer k. Let us plug it in. We
get: 4k 2 = 10b2 , and thus 2k 2 = 5b2 . Then 5b2 is even, and therefore b2 is even,
since the product of two odd numbers is odd. Then b is even, and we arrive at
a contradiction, since we were assuming that a and b had no common factors,
but we arrived at the conclusion that they are both even.
Note that the same argument with using divisibility by 10 rather than by 2 does
not work so well, because it is harder to prove the lemma that 10|a2 if and only
if 10|a.
(b) Prove that the following statement is False:
If x, y are both irrational, then x − y is irrational.
Solution: To show that this√statement is false, all we need is a counterexample.
Take, for example, x = y = 2. Then x − y = 0 is rational, while x and y are
both irrational.
Mathematics 220
(c) Prove that
Midterm Practice problems from old exams
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√
√
5 − 2 is irrational.
√
√
Solution: We prove this by contradiction. Suppose √ 5 − 2 = q is a rational
number. Then, squaring both sides, we get 5 + 2 − 2 10 = q 2 , and therefore
√
7 − q2
10 =
.
2
Now, note that if q is a rational number, then q 2 is rational, 7 − q 2 is rational,
2
and 7−q
is rational. We get a contradiction with part (a).
2
11. (a) Write the negation of the following statement:
“For every (a, b) ∈ N × N, if a > b then (a + b)2 ≥ (a − b)2 .”
Solution: ∃(a, b) ∈ N × N s.t. (a > b) and ((a + b)2 < (a − b)2 ).
(b) Write the converse and contrapositive of the following statement:
“If it is raining outside then this is Vancouver.”
Solution: Converse — “If this is Vancouver then it is raining outside.”
Contrapositive — “If it is not Vancouver then it is not raining outside.”
(c) Give a precise mathematical definitions of the following sets
\
[
Sα
Sα
C=
B=
α∈I
α∈I
Solution:
S
• α∈I Sα = {x|∃α ∈ I s.t. x ∈ Sα }
T
• α∈I Sα = {x|∀α ∈ I, x ∈ Sα }
(d) For any n ∈ N, let An =
B=
1
n
− 1, 3 +
[
n∈N
An
1
n2
. Simplify the following sets
\
C=
An
n∈N
Solution: Union is B = (−1, 4]. Intersection is C = (0, 3].
(e) Let A, B be sets in some universal set U . Suppose that Ā = {3, 8, 9}, A−B = {1, 2},
B − A = {8} and A ∩ B = {5, 7} Determine A, B, U .
Solution:
• It really helps to note that A = (A − B) ∪ (A ∩ B).
Mathematics 220
Midterm Practice problems from old exams
Page 7 of 8
• Then A = {1, 2} ∪ {5, 7} = {1, 2, 5, 7}
• Similarly B = {8} ∪ {5, 7} = {5, 7, 8}.
• Finally U = A ∪ Ā = {1, 2, 5, 7} ∪ {3, 8, 9} = {1, 2, 3, 5, 7, 8, 9}.
12. (a) Prove or disprove the following statement
Let a, b, c, d ∈ R. If ab ≥ cd then a ≥ c and b ≥ d.
Solution: The statement is false. Let a = b = −1 and c = d = 0. Then
ab = 1 > 0 = cd, but a < c and b < d.
(b) Prove or disprove the following
Let a, b, c, n ∈ Z so that n ≥ 3. If a + b ≡ 1 mod n and b + c ≡ 1 mod n
then a + c ≡ 2 mod n.
Solution: This is false. The negation of the statement is
There are some a, b, c, n ∈ Z so that n ≥ 3, so that a + b ≡ 1 mod n
and b + c ≡ 1 mod n but a + c 6≡ 2 mod n.
Consider n = 3, a = 2, b = 2, c = 2. Then a + b = b + c = 4 ≡ 1 mod 3, but
a + c = 4 ≡ 1 mod 3, not 2.
(c) Repeat part (b), but when n = 2.
Solution: This is now true.
Proof. Assume a + b ≡ 1 mod 2 and b + c ≡ 1 mod 2. Then there exist k, l ∈ Z
so that
a + b − 1 = 2k
a + c − 2 = 2(k + l − b)
b + c = 2l
hencea + 2b + c − 2 = 2k + 2l
Since k + l − b ∈ Z, it follows that a + c ≡ 2 mod 2 as required.
13. Let n ∈ Z. Prove that n2 + 1 is odd if and only if 7n + 3 is odd.
Solution:
Proof. The number n is either even or odd and so we consider the two cases.
• Assume n is even, so n = 2k for some k ∈ Z. Then n2 +1 = 4k 2 +1 = 2(2k 2 )+1
and so is odd. Similarly 7n + 3 = 14k + 3 = 2(7k + 1) + 1 and so is odd.
Mathematics 220
Midterm Practice problems from old exams
Page 8 of 8
• Now assume n is odd, so n = 2k+1 for some k ∈ Z. Then n2 +1 = 4k 2 +4k+2 =
2(2k 2 + 2k + 1) and so is even. Similarly 7n + 3 = 14k + 10 = 2(7k + 5) and so
is even.
Thus we see that either n2 + 1 and 7n + 3 are both even or both odd.
Note that you can also solve this using a couple of lemmas. Namely
“If n2 + 1 is odd then n is even.” and
“If 7n + 3 is odd then n is even.”
Both of which are easy to prove using their contrapositives. The proof of the main
result then becomes
Proof. We must prove both the forward and backward implications.
• Assume that n2 + 1 is odd, then by the previous lemma n is even and hence
n = 2k for some k ∈ Z. Then 7n+3 = 14k +3 = 2(7k +1)+1. Since 7k +1 ∈ Z,
it follows that 7n + 3 is odd.
• Now assume that 7n + 3 is odd, then by the above lemma, n is even and hence
n = 2`. Then n2 + 1 = 4`2 + 1 = 2(2`2 ) + 1. Since 2`2 ∈ Z, it follows that
n2 + 1 is odd.
14. Determine whether the following four statements are true or false — explain your answers
(“true” or “false” is not sufficient).
(i) ∀x ∈ R, ∀y ∈ R, if (xy ≥ 0) then (x + y ≥ 0).
(ii) ∀x ∈ R, ∃y ∈ R s.t. if (xy ≥ 0) then (x + y ≥ 0).
(iii) ∃x ∈ R s.t. ∃y ∈ R s.t. if (xy ≥ 0) then (x + y ≥ 0).
Solution:
(i) False — negation is ∃x s.t. ∃y s.t. (xy > 0) and (x + y ≤ 0). This is true,
Consider x = y = −2, then xy = 4 > 0 and x + y = −4 < 0. Since negation is
true, original is false.
(ii) True — Pick any x and set y = −x. Now if x = 0 then y = 0 and xy = x+y = 0
and so all is true. On the other hand, if x 6= 0 then xy = −x2 < 0 and so the
hypothesis is false, and the implication is therefore true. In either case the
implication is true.
(iii) True — pick (say) x = 11, y = 3 then xy = 3 > 0 and x + y = 4 > 0 as required.