Prove that the following numbers are irrational numbers:

Prove that the following numbers are irrational:

Solution:

(i) √8
If √8 is a rational number
Consider √8 = p/q where p and q are integers
q ˃ 0 and p and q have no common factor
By squaring on both sides
8 = p²/q²
So we get
p² = 8q²
We know that
8p² is divisible by 8
p² is also divisible by 8
p is divisible by 8
Consider p = 8k where k is an integer
By squaring on both sides
p² = (8k)²
p² = 64k²
We know that
64k² is divisible by 8
p² is divisible by 8
p is divisible by 8
Here p and q both are divisible by 8
So our supposition is wrong
Therefore, √8 is an irrational number.
(ii) √14
If √14 is a rational number
Consider √14 = p/q where p and q are integers
q ≠ 0 and p and q have no common factor
By squaring on both sides
14 = p²/q²
So we get
p² = 14q² ….. (1)
We know that
p² is also divisible by 2
p is divisible by 2
Consider p = 2m
Substitute the value of p in equation (1)
(2m)² = 13q²
So we get
4m² = 14q²
2m² = 7q²
We know that
q² is divisible by 2
q is divisible by 2
Here p and q have 2 as the common factor which is not possible
Therefore, √14 is an irrational number.
(iii)
If is a rational number
Consider = p/q where p and q are integers
q ˃ 0 and p and q have no common factor
By cubing on both sides
2 = p³/q³
So we get
p³ = 2q³ ….. (1)
We know that
2q³ is also divisible by 2
p³ is divisible by 2
p is divisible by 2
Consider p = 2k where k is an integer
By cubing both sides
p³ = (2k)³
p³ = 8k³
So we get
2q³ = 8k³
q³ = 4k³
We know that
4k³ is divisible by 2
q³ is divisible by 2
q is divisible by 2
Here p and q are divisible by 2
So our supposition is wrong
Therefore , is an irrational number.