COMP232 - Mathematics for Computer Science
Tutorial 5
Ali Moallemi
moa [email protected]
Iraj Hedayati
h [email protected]
Concordia University, Winter 2016
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
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Table of Contents
1
1.7 Introduction to Proofs
Exercise 5
Exercise 10
Exercise 17
Exercise 26
Exercise 31
Exercise 39
Exercise 40
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
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Exercise 5
Prove that if m + n and n + p are even integers, where m, n, and p are
integers, then m + p is even. What kind of proof did you use?
Answer: From the book
Direct proof: Suppose that m + n and n + p are even. Then m + n = 2s
for some integer s and n + p = 2t for some integer t. If we add these, we
get m + p + 2n = 2s + 2t. Subtracting 2n from both sides and factoring,
we have m + p = 2s + 2t − 2n = 2(s + t − n). Because we have written
m + p as 2 times an integer, we conclude that m + p is even.
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
3/9
Exercise 10
Use a direct proof to show that the product of two rational numbers is
rational.
Answer:
Direct proof: Suppose that m and n are rational numbers. Then
∃p, q (m =
p
p0
∧ q 6= 0) and ∃p0 , q 0 (n = 0 ∧ q 0 6= 0)
q
q
We can express product of m and n as:
mn =
p p0
pp0
× 0 = 0
q q
qq
Considering the fact that product of two integer numbers is also an integer
number as well as if q 6= 0 and q 0 6= 0 then qq 0 6= 0, we can conclude that
mn is a rational number.
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
4/9
Exercise 17
Show that if n is an integer and n3 + 5 is odd, then n is even using
a) a proof by contraposition.
Answer: (From the book) Assume that n is odd, so for some integer
k, n = 2k + 1. Then
n3 + 5 = 2(4k 3 + 6k 2 + 3k + 3)
Because n3 + 5 is two times some integer, it is even
b) a proof by contradiction.
Answer: (From the book) Suppose that n3 + 5 is odd and n is odd.
Because n is odd and the product of two odd numbers is odd, it
follows that n2 is odd and then that n3 is odd. But then
5 = (n3 + 5) − n3 would have to be even because it is the difference
of two odd numbers. Therefore, the supposition that n3 + 5 and n
were both odd is wrong
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
5/9
Exercise 26
Prove that if n is a positive integer, then n is even if and only if 7n + 4 is
even.
Remarks
p: n is even
q: 7n + 4 is even
p↔q
Answer:
p → q: First, assume that n is even, so that n = 2k for some integer
k. Then 7n + 4 = 7(2k) + 4 = 14k + 4 = 2(7k + 2). Hence, 7n + 4 is
even. (The method was direct proof)
q → p: To prove the converse, suppose that n is odd, so that
n = 2k + 1 for some integer k. Then
7n + 4 = 14k + 9 = 2(7k + 4) + 1, so 7n + 4 is odd. (The method
was contraposition)
Hence, n is even if and only if 7n + 4 is even
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
6/9
Exercise 31
Show that these statements about the integer x are equivalent:
(i) 3x + 2 is even
(ii) x + 5 is odd
(iii) x2 is even
Answer: We prove that all these are equivalent to x being even. If x is
even, then x = 2k for some integer k. Therefore
3x + 2 = 3 × 2k + 2 = 6k + 2 = 2(3k + 1), which is even
Similarly, x + 5 = 2k + 5 = 2k + 4 + 1 = 2(k + 2) + 1, so x + 5 is odd;
and x2 = (2k)2 = 2(2k 2 ), so x2 is even.
For the converses, we will use a proof by contraposition. So assume that x
is not even; thus x is odd and we can write x = 2k + 1 for some integer k.
Then
3x + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1, which is odd (i.e., not
even).
Similarly, x + 5 = 2k + 1 + 5 = 2(k + 3), so x + 5 is even (i.e., not odd).
and x2 = (2k + 1)2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 concludes x2 is odd.
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
7/9
Exercise 39
Prove that at least one of the real numbers a1 , a2 , . . . , an is greater than
or equal to the average of these numbers. What kind of proof did you use?
Answer:(From the book)
We will give a proof by contradiction. Suppose that a1 , a2 , . . . , an are all
less than A, where A is the average of these numbers. Then
a1 + a2 + . . . + an < nA. Dividing both sides by n shows that
A = (a1 + a2 + . . . + an )/n < A, which is a contradiction
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
8/9
Exercise 40
Use Exercise 39 to show that if the first 10 positive integers are placed
around a circle, in any order, there exist three integers in consecutive
locations around the circle that have a sum greater than or equal to 17.
Answer:
Let each a1 , a2 , . . . , a10 be one of the first 10 positive integers in any order
and
A1 = a1 + a2 + a3
A2 = a2 + a3 + a4
A3 = a3 + a4 + a5
...
A10 = a10 + a1 + a2 .
Then,
10
Σ10
i=1 Ai = 3Σi=1 ai = 3 × 55 = 165
The average of Ai -s would be 165
10 = 16.5. Using Exercise 39, some
Ai ≥ 16.5. Considering the fact that all ai -s are integer, Ai -s should be
also integer. Thus we can conclude that some Ai ≥ 17
Ali Moallemi, Iraj Hedayati
COMP232 - Mathematics for Computer Science
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