Prove the following are irrational numbers.

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Prove the following are irrational numbers.

(i) ∛2
(ii) ∛3
(iii) ∜5

Solution:

(i) ∛2
We know that ∛2 = 21/3
Let us consider 21/3 = p/q, where p, q are integers, q>0.
p and q have no common factors (except 1).
So,
21/3 = p/q
2 = p3/q3
p3 = 2q3 ….. (1)
We know that, 2 divides 2q3 then 2 divides p3
So, 2 divides p
Now, let us consider p = 2k, where k is an integer
Substitute the value of p in (1), we get
p3 = 2q3
(2k)3 = 2q3
8k3 = 2q3
4k3 = q3
We know that, 2 divides 4k3 then 2 divides q3
So, 2 divides q
Thus p and q have a common factor ‘2’.
This contradicts the statement, p and q have no common factor (except 1).
Hence, ∛2 is an irrational number.
(ii) ∛3
We know that ∛3 = 31/3
Let us consider 31/3 = p/q, where p, q are integers, q>0.
p and q have no common factors (except 1).
So,
31/3 = p/q
3 = p3/q3
p3 = 3q3 ….. (1)
We know that, 3 divides 3q3 then 3 divides p3
So, 3 divides p
Now, let us consider p = 3k, where k is an integer
Substitute the value of p in (1), we get
p3 = 3q3
(3k)3 = 3q3
9k3 = 3q3
3k3 = q3
We know that, 3 divides 9k3 then 3 divides q3
So, 3 divides q
Thus p and q have a common factor ‘3’.
This contradicts the statement, p and q have no common factor (except 1).
Hence, ∛3 is an irrational number.
(iii) ∜5
We know that ∜5 = 51/4
Let us consider 51/4 = p/q, where p, q are integers, q>0.
p and q have no common factors (except 1).
So,
51/4 = p/q
5 = p4/q4
P4 = 5q4 ….. (1)
We know that, 5 divides 5q4 then 5 divides p4
So, 5 divides p
Now, let us consider p = 5k, where k is an integer
Substitute the value of p in (1), we get
P4 = 5q4
(5k)4 = 5q4
625k4 = 5q4
125k4 = q4
We know that, 5 divides 125k4 then 5 divides q4
So, 5 divides q
Thus p and q have a common factor ‘5’.
This contradicts the statement, p and q have no common factor (except 1).
Hence, ∜5 is an irrational number.

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