Midterm — March 14th 2013
Mathematics 220-201
Page 1 of 7
This midterm has 6 questions on 7 pages, for a total of 50 points.
Duration: 75 minutes
• Read all the questions carefully before starting to work on them.
• With the exception of Q1, you should give complete arguments and explanations for
all your calculations; answers without justifications will not be marked.
• Continue on the back of the previous page if you run out of space.
• Please do not try to guess, but do not be afraid to think and communicate
your reasoning. Points will be taken off for statements that make no sense. Partial
credit may be given for attempts that make sense but do not lead to a complete solution.
• This is a closed-book examination. None of the following are allowed: documents,
cheat sheets or electronic devices of any kind (including calculators, cell phones, etc.)
Full Name (including all middle names):
Student-No:
Signature:
Question:
1
2
3
4
5
6
Total
Points:
9
7
8
7
7
12
50
Score:
Mathematics 220-201
9 marks
Midterm — March 14th 2013
Page 2 of 7
1. Short answers question:
(a) Does the set A = {1 −
1
n
| n ∈ N} have a minimal element?
Solution: Yes; it is 0. (note that 1 −
is minimal when n = 1).
1
n
is an increasing sequence, so the value
(b) Is the set B = {1, 2, 3} ∪ [4, 5] well-ordered?
Solution: No. For example, the subset (4, 5) does not have a minimal element.
(c) Does there exist a bijective function f : A → B, where
A = {n ∈ N | 1 ≤ n ≤ 10 and n is even},
B = {3m − 1 | 1 ≤ m ≤ 10, m ∈ N}?
Explain briefly why or why not.
Solution: No; the sets A and B have different cardinality: |A| = 5, |B| = 10.
(d) Is the following relation from [−1, 1] to [−1, 1] a function? Justify your answer.
R = {(x, y) ∈ [−1, 1] × [−1, 1] | x2 − y 2 = 1}.
Solution: No. In fact, for all x except for x = ±1, y is undefined, so this is not
a function on [−1, 1].
Mathematics 220-201
7 marks
Midterm — March 14th 2013
2. Find the domain and range of the function f (x) =
x−3
x
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(both should be subsets of R).
Solution: The function is defined for all x 6= 0, so its domain is R − {0}.
Claim: the range of f is the set R − {1}.
Proof: We need to prove that ran(f ) ⊇ R − {1} and ran(f ) ⊆ R − {1}.
The first inclusion: we need to show that for every y ∈ R−{1}, there exists x ∈ R−{0}
such that f (x) = y. We do some scratch work: let y = x−3
and solve for x. We get
x
3
x = 1−y , which is well-defined since y 6= 1. Plugging it back into f , we check that
3
f ( 1−y
) = y.
The second inclusion: we need to show that f (x) 6= 1 for any x ∈ R − {0}. Suppose
= 1, so we get x − 3 = x, which is impossible.
f (x) = 1. Then x−3
x
Mathematics 220-201
8 marks
Midterm — March 14th 2013
Page 4 of 7
3. Decide if the following statements are True or False. If True, give a proof. If False,
provide a counter-example. For any sets A, B, C,
(a) C − (A ∩ B) = (C − A) ∪ (C − B).
Solution: This is true.
Proof: x ∈ C − (A ∩ B) is equivalent to x ∈ C and x ∈
/ A ∩ B (by definition of
the set difference), which is equivalent to: (x ∈ C) ∧ ((x ∈
/ A) ∨ (x ∈
/ B)) (this is
DeMorgan’s law). By distributivity, this is equivalent to ((x ∈ C) ∧ (x ∈
/ A)) ∨
((x ∈ C) ∧ (x ∈
/ B)), which is exactly the statement that x ∈ (C − A) ∪ (C − B).
Note: this is the shortest form of writing this solution; it was also fine to check
that every element contained in the left-hand side is contained in the right-hand
side, and vice versa.
(b) (C − A) ∪ B = (C − B) ∪ A.
Solution: This is false. There were many good counterexamples. Here’s one of
the simplest: let C = ∅, A = ∅, B = {1}. Then (C − A) ∪ B = ∅ ∪ {1} = {1},
and (C − B) ∪ A = ∅ ∪ ∅ = ∅.
Mathematics 220-201
7 marks
Midterm — March 14th 2013
Page 5 of 7
4. Use induction to prove that 3|n3 + 2n for all natural numbers n.
Hint: (a + 1)3 = a3 + 3a2 + 3a + 1.
Solution: Base case: n = 1. Then n3 + 2n = 1 + 2 = 3, so the statement is true for
n = 1.
Induction step: Suppose the statement is true for some natural number n = k. We
need to prove that it is true for n = k + 1. That is, we are assuming that 3|k 3 + 2k,
and we need to prove that 3|(k + 1)3 + 2(k + 1).
Let us simplify the expression (k + 1)3 + 2(k + 1). We have
(k + 1)3 + 2(k + 1) = k 3 + 3k 2 + 3k + 1 + 2k + 2 = (k 3 + 2k) + (3k 2 + 3k + 3).
By the induction assumption, there exists an integer m such that k 3 + 2k = 3m.
Then we get:
(k +1)3 +2(k +1) = (k 3 +2k)+(3k 2 +3k +3) = 3m+3(k 2 +k +1) = 3(m+k 2 +k +1),
and thus we have shown that there exists an integer q = m + k 2 + k + 1 such that
(k + 1)3 + 2(k + 1) = 3q, which proves the induction step.
By the Principle of Mathematical induction, the statement holds for all n ∈ N.
Mathematics 220-201
7 marks
Midterm — March 14th 2013
Page 6 of 7
5. Prove that for all natural numbers n ≥ 6, 2n ≥ n2 + 10.
Solution: Base case: n = 6. Then 2n = 64 and n2 + 10 = 46, so the statement is
true.
Induction step: Assume the statement is true for an integer n = k ≥ 6, that is,
2k ≥ k 2 + 10. We need to show that 2k+1 ≥ (k + 1)2 + 10. We have: 2k+1 = 2 · 2k .
By the induction assumption, 2k ≥ k 2 + 10, and therefore 2k+1 ≥ 2(k 2 + 10).
Now, observe that for k ≥ 6, we have 2(k 2 + 10) > (k + 1)2 + 10. Indeed, subtracting
the right-hand side from the left-hand side, we get:
2(k 2 + 10) − ((k + 1)2 + 10) = k 2 − 2k + 9 = (k − 1)2 + 8 > 0.
Thus, we have:
2k+1 ≥ 2(k 2 + 10) > (k + 1)2 + 10,
and the induction step is proved.
By the Principle of Mathematical induction, the statement holds for all natural numbers n ≥ 6.
Midterm — March 14th 2013
Mathematics 220-201
12 marks
6. (a) Prove that
√
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10 is irrational.
Solution: Proof by contradiction.
√
Suppose 10 was rational.
Then there exist√integers a and b with no common
√
a
factors, such that 10 = b . 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.
(c) Prove that
√
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
√
10 =
7 − q2
.
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