MATH 1701: Discrete Mathematics
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Module 2: Elementary Number Theory and
Methods of Proof
This Assignment is worth 5% of your final grade.
Total number of marks to be earned in this assignment: 32
Assignment 2, Version 11:
After completing Module 2, including the learning activities, you are asked to
complete the following written assignment in the space provided.
1)
Determine whether the following statements are true or false. Justify
your answer with a proof or a counterexample.
a. The product of any three consecutive integers is divisible by 6. (3 marks)
Let n, n+1, and n+2 be three consecutive integers.
Considering that odd and even integers alternate, at least one of these three
integers is even.
The even integer among them will be divisible by 2,
implying that the product n(n+1)(n+2) will include a factor of 2.
Since every third integer in a sequence is divisible by 3,
one of the three consecutive integers must be a multiple of 3.
Consequently,
the product n(n+1)(n+2) will also have a factor of 3.
By the principle of divisibility, if a∣c and b∣c, then ab∣c for coprime a and b.
Applying this principle with a=2 and b=3, it follows that the product
n(n+1)(n+2) is divisible by 6.
Therefore, by the laws of divisibility, the product of any three consecutive
integers is divisible by 6. ☐
1
Unless otherwise cited, all questions are created by the author. Any resemblance to existing
texts is purely coincidental.
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Module 2: Assignment 2, Version 1
b. The product of any two odd integers is odd. (3 marks)
Consider two odd integers, a and b.
By definition, odd integers can be represented in the form 2k+1, where k is an
integer.
Let a=2n+1 and b=2m+1, where n and m are integers.
The product a⋅b is given by:
a⋅b=(2n+1)(2m+1)
a⋅b=4nm+2n+2m+1
a⋅b=2(2nm+n+m)+1
Let r=2nm+n+m, which is an integer since it is a sum of products and sums of
integers.
Therefore, we can express a⋅b as:
a⋅b=2r+1
Since r is an integer, 2r is even, and 2r+1 is odd. The product of two odd
integers, a and b, is odd. ☐
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MATH 1701: Discrete Mathematics
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2)
a. Write the rational number 0.3234234234… as a ratio of integers. (2 marks)
0.3 + 234/10^3 + 234/10^6 +234/10^9 + ... + 234/10^n
a=234/10^3
r=1/10^3
234/1000 = 234/1000 = 234/1000 = 234
1-1/1000
1-1/1000
999/1000
999
1.
2.
3.
4.
5.
6.
7.
8.
9.
0.3 + 234/999
(0.3)(999/999)+(234/999)
(3)(9990/9990)+(2340/9990)
32310 /9990 = 3.23423423423
32310 /10 = 3213
(3231/9990)=0.3234234234…
gcd(3231/9990)=9
(3231/9990)/9 = 359/1110
359/1110 = 0.3234234234…
b. Prove that if
is rational, then
is rational. (2 marks)
Let x be a non-zero rational number.
A rational number can be expressed as the ratio of two integers.
Thus, x=a/b​for some integers a and b, where b≠0.
The reciprocal of x is given by 1/x​.
Substituting the expression for x:
1/x=1/(a/b)
To simplify this expression, we multiply the numerator and the denominator
by the reciprocal of a/b​, which is b/a:
1/x=1/(a/b)×(b/a)
Simplifying, we find that:
1/x=b/a
Since x is non-zero, a is also non-zero (because if a were zero, x would not be
defined).
Thus, 1/x​is expressed as a ratio of two integers b and a, with a non-zero. This
satisfies the definition of a rational number.
Therefore, 1/x​is a rational number. ☐
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Module 2: Assignment 2, Version 1
3) Determine whether the following statements are true or false. Justify your
answer with a proof or a counterexample.
a. For all integers
, if
then
or
. (3 marks)
Proof by counterexample:
Counterexample: Let a=6,b=2, and c=3.
It's true that 6 divides 2×3, but 6 does not divide 2 and 6 does not divide 3.
This contradicts the claim because a divides bc but a divides neither b nor c
individually.
Given the statement: 6|(2x3)→(6|2)∨(6|3),
6|(2x3) is true,
(6|2) is false,
(6|3) is false,
therefore (6|2)∨(6|3) is false.
Hence, the statement 6|(2x3)→(6|2)∨(6|3) is false because an argument
whose premises are true and whose conclusion is false is a false argument.
Thus, the statement is false as demonstrated by the counterexample where the
hypothesis is true (6 divides 2×3) but the conclusion is false (neither 6 divides
2 nor 6 divides 3), proving the claim incorrect. ☐
b. Determine the prime factorization of 44,574. (2 marks)
1)
2)
3)
4)
5)
6)
44,574 / 2 = 22287
22287 / 3 = 7429
5,7,11,13 ∤ 7429
7429 / 17 = 437
1061 is a prime number.
Therefore 44574 = 2x3x17x437
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MATH 1701: Discrete Mathematics
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7) Determine if the following statement is true and either prove it or provide a
counterexample: “For all integers
greater than 1, there exists a positive
integer
”. (3 marks)
such that
Proof by counterexample:
There does not exist a positive integer y such that x + y ≡ 0 (mod n).
Let x = 2 and n = 3.
For any positive integer y, the sum 2 + y will always leave a remainder of either 2
or 0 when divided by 3
2 mod 3 = 2
Since there is no positive integer y such that 2 + y ≡ 0 (mod 3), the statement is
false.
Therefore, the counterexample x = 2, n = 3 disproves the statement. ☐
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Module 2: Assignment 2, Version 1
8) Prove or disprove the following statement: “For all real numbers
.” (3 marks)
Consider the statement: ⌊xy⌋ = ⌊x⌋ ⌈y⌉ for real numbers x and y.
Consider three cases for the statement:
case 1: the real numbers x=1.5 and y=3.3.
⌊xy⌋ = ⌊4.95⌋ = 4
⌊x⌋⌈y⌉ = ⌊1.5⌋x⌈3.3⌉ = 1 x 4 = 4
4=4
case 2: the real numbers x=2 and y =3.5
⌊xy⌋ = ⌊2 x 3.5⌋ = 7
⌊x⌋⌈y⌉ = ⌊2⌋x⌈3.5⌉ = 2 x 4 = 8
7≠8
case 3: the real numbers x=3 and y = 5
⌊xy⌋ = 3x5 = 15
⌊x⌋⌈y⌉ = 3x5 = 15
3=5
Since there is a case (case 2) where the statement is not true for all real numbers
(7≠8), the statement is disproven with this counterexample. ☐
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MATH 1701: Discrete Mathematics
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9)
a. Use a proof by contraposition to show that if
real numbers, then
or
, where
and
are
. (Inspired by Rosen, 2007, p. 85). (3 marks)
If x + y ≥ 2, where x and y are real numbers, then x ≥ 1 or y ≥ 1, the
contraposition of the statement is:
If neither x ≥ 1 nor y ≥ 1, then x + y < 2.
Assume that neither x ≥ 1 nor y ≥ 1. This implies that both x < 1 and y < 1.
Now, consider the sum of x and y: x + y < 1 + 1 = 2
Since x + y is less than 2, we have demonstrated that when neither x ≥ 1 nor y
≥ 1, then x + y < 2.
By contraposition, the original statement,"If x + y ≥ 2, where x and y are real
numbers, then x ≥ 1 or y ≥ 1."is affirmed.☐
b. Use a proof by contradiction to show that the difference of any rational
number and any irrational number is irrational. (3 marks)
Proof by Contradiction:
Assume the opposite, that the difference of a rational number and an
irrational number is rational. Assume there exists a rational number p and an
irrational number q such that p−q is rational.
We can express q as the difference between p and (p−q) (assumed to be
rational):
q=p−(p−q)
Since p is rational, and the difference between two rational numbers is also
rational, we conclude that q must be rational as well. However, this
contradicts the initial assumption.
Since our assumption leads to a contradiction, we must conclude that the
original statement is true:
"The difference between any rational number and any irrational number is
irrational." ☐
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Module 2: Assignment 2, Version 1
10) Prove or disprove the following statements:
a.
is irrational. (3 marks)
Proof by Contradiction:
Assume that ³√3 is a rational number.
This means we can write ³√3 as a fraction in its simplest form, where both the
numerator and denominator are integers, and the denominator is not zero:
³√3 = a/b where a,b ∈Z and b ≠ 0. and gcd(a,b)=1 (simplest fractional form).
cube both sides:
3= (a^3) / (b^3)
cross multiply:
3 = (a^3)
1 (b^3) =
3(b^3) = (a^3)
This implies that (a^3) is a multiple of 3. Therefore a must also be a multiple
of 3 because the cube of any integer not divisible by 3 would not be a
multiple of 3.
We can now express a as 3k, where k is an integer:
3b^3 = (3k)^3
3b^3 = 27k
b^3 = 9k
This implies that b^3 is also a multiple of 3 and hence b as well.
However, this contradicts the initial assumption that a and b are integers in
their simplest form, where gcd(a, b) = 1.
If both a and b are multiples of 3, then they have a common factor greater
than 1, which contradicts the assumption that they are in their simplest
form, which leads to the conclusion that ³√3 is irrational. ☐
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MATH 1701: Discrete Mathematics
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b. There exists three consecutive prime numbers whose sum is also a prime
number. (2 marks)
There exists three consecutive prime numbers whose sum is also a prime
number.
Let p, p + 2, p + 4 represent the three consecutive prime numbers.
The sum of the three consecutive prime numbers can be summed to get (p +
p + 2 + p + 4), which simplifies to (3p + 6) or (3(p + 2)).
Because 3 is a factor of this sum, it cannot be one of the three primes (as the
three primes are consecutive odd integers).
Therefore, the sum of three consecutive prime numbers is not a prime
number for any three consecutive prime numbers. ☐
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Module 2: Assignment 2, Version 1
Bibliography
Epp, S. S. (2020). Discrete mathematics with applications (5th Edition). Boston: Cengage.
Rosen, K. H. (2007). Discrete mathematics and its applications. New York: McGraw-Hill.
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