Prove by Contradiction and Contrapositive Methods

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.