Contradiction and Contrapositive Proofs in Number Theory Analysis

This assignment demonstrates proofs using the contradiction and contrapositive methods. It includes proving the irrationality of the cube root of 2, showing that the Diophantine equation x^2 - y^2 = 10 has no positive integer solutions, demonstrating the absence of rational number solutions for the equation x^5 + x^4 + x^3 + x^2 + 1 = 0, and proving that the sum of a rational and an irrational number is irrational. Additionally, it covers proofs by contrapositive, such as showing that if the product of two integers is even, at least one must be even, and if the product is odd, both must be odd. It also demonstrates that if n is a positive integer such that n (mod 3) = 2, then n is not a perfect square, and if the product of two real numbers is irrational, then either a or b must be irrational. Desklib offers a range of solved assignments and study tools for students.

Prove each of the following by the contradiction method.
1. The cube root of 2 is irrational.
Step 1: By contradiction, let 3
√ 2 be rational
Step 2: Then 3
√ 2 = a
b where, a
b is the simplest possible fraction form and a, b ∈ Z, b ≠ 0.
Step 3: Implying 2b3 = a3.
Step 4: a3 is even because any number contained in Z multiplied by 2 is divisible by 2 without
a remainder and thus even.
Step 5: Also, 2
a3 and 2
a are even by step 4.
Step 6: Therefore, there exists k ∈ Z, such that a = 2k.
Step 7: Substitute a = 2k in step 3: 2b3 = (2 k )3.
Step 8: Then, b3 = 4k3, therefore, 2/b. Thus, by contradiction a and b have a common factor of
two. Therefore, 3
√2 is irrational by contradiction method.
2. There are no positive integer solutions to the Diophantine equation x2− y2=10.
Step 1: Let us assume by contradiction that there is a solution (x, y) where x, y ∈ Z+¿ ¿
Step 2: Also, x2− y2=10 can be written as (x - y) (x + y) = 10 expanded form.
Step 3: Since x, y ∈ Z+¿ ¿ it follows that either (x - y) = 10 and (x + y) = 10 or (x - y) = -10
and (x + y) = -10.
Step 4: In the first case we can simultaneously solve for x and y to obtain x = 10 and y = 0.
Thus, contradicting assumption of x and y being positive integers.
Step 5: In the second case we can simultaneously solve for x and y to obtain x = -10 and y =
0. Thus, contradicting assumption of x and y being positive integers. Therefore, by
contradiction there are no positive integer solution to the Diophantine equation x2− y2=10.

3. There is no rational number solution to the equation x5 + x4 + x3+x2+ 1 = 0.
Step 1: The roots of this polynomial are algebraic integers. If an algebraic integer is a real
number, then it is either an integer or irrational.
Step 2: Thus, we only need to show that there are no integer solutions to x5 + x4 + x3+x2+ 1 =
0.
Step 3: Clearly the function f(x) = x5 + x4 + x3+x2+ 1 is a continuous polynomial with the
roots x = -0.73, x = -0.035 – 0.74i and x = -0.035 + 0.74i. Since the roots are not integers
there does not exists rational number solution to the equation.
4. If a is a rational number and b is an irrational number, then a + b is an irrational
number.
Step 1: since a is rational it can be represented as m
n
Step 2: We leave b unchanged and a + b can also be represented as ratio of two integers j
k
Step 3: writing the two assumption we obtain
m
n +b= j
k
⟹ b= j
k −m
n
⟹ b=km−n j
k n
The last equation in step 3 says that b equals the product of two integers (km) minus the
product of two other integers (nj), all divided by another integer (kn). This implies that b is
rational. However, from the definition b is irrational so the assumption that;
rational + irrational = rational does not hold.

Prove each of the following by the contrapositive method.
5. If x and y are two integers whose product is even, then at least one of the two must
be even.
Step 1: Let x be even and y be odd. Then by definition x = 2n and y = 2m + 1 for n, m ∈ Z
Step 2: Also, consider xy = (2n) (2m + 1) = 4nm + 2n = 2(2nm + n) + 2k where k = 2nm + n
is an integer. Therefore, by definition the product xy is even.
Step 3: Let both x and y be even, then x = 2n and y = 2m for n, m ∈ Z. Consider the product
xy = (2n) (2m) =2(2nm) = 2k where k = 2nm is an integer.
Step 4: Therefore, the product xy is even by definition and both cases verify the truth of the
hypothesis.
6. If x and y are two integers whose product is odd, then both must be odd.
Step 1: By definition x = 2n + 1 and y = 2m + 1 for n, m ∈ Z.
Step 2: Now consider the product xy
xy = (2n + 1) (2m + 1) = 4nm +2n +2m +1= 2(2nm +n + m) + 1 = 2k.
Where k = 2nm + n + m is an integer. Therefore, the product xy is odd by definition of odd.
7. If n is a positive integer such that n (mod3) = 2, then n is not a perfect square.
Step 1: Contrapositive version – If n is perfect square then n(mod3) must be 0 or 1.
Step 2: Suppose n = k2 the four cases are;
1. If k mod (3) = 0, then k = 3q, for some integer q. Then, n = k2 = 9 q2, q2 = 3(3 q2).
2. If k mod (3) = 1, then k = 3q + 1, for some integer q.
Then, n = k2 = 9 q2 + 6q + 1 = 3(3 q2 + 2q) + 1.
Step 3: Therefore, n is not a perfect square.
8. If a and b are real numbers such that the product ab is an irrational number, then
either a or b must be an irrational number.
Step 1: Contrapositive if both a and b are rational, then ab is rational.

Step 2: Since a and b are rational, we have a = m
n and b = p
q
Where; m, n, p and q are integers.
Step 3: It follows that ab = a = m p
n q which is clearly a quotient of two integers. Thus, proved.