Question

Show that cube root of 6 is irrational.

Answer 1

Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0)
 Cubing both sides : 6=a^3/b^3
 a^3 = 6b^3
 a^3 = 2(3b^3)
Therefore, 2 divides a^3 or a^2 * a . By Euclid's Lemma if a prime number
 divides the product of two integers then it must divide one of the two integers
 Since all the terms here are the same we conclude that 2 divides a
.
 Now there exists an integer k such that a=2k
 Substituting 2k in the above equation
 8k^3 = 6b^3
 b^3 = 2{(2k^3) / 3)}
 Therefore, 2 divides b^3. Using the same logic as above. 2 divides b.
 Hence 2 is common factor of both a & b. But this is a contradiction of the fact that a & b are co-prime. Therefore, the initial assumption is wrong.
 
cube root 6 is irrational

Answer 2

Answer:

Step-by-step explanation:

Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0)

 Cubing both sides : 6=a^3/b^3

 a^3 = 6b^3

 a^3 = 2(3b^3)

Therefore, 2 divides a^3 or a^2 * a . By Euclid's Lemma if a prime number

 divides the product of two integers then it must divide one of the two integers

 Since all the terms here are the same we conclude that 2 divides a

.

 Now there exists an integer k such that a=2k

 Substituting 2k in the above equation

 8k^3 = 6b^3

 b^3 = 2{(2k^3) / 3)}

 Therefore, 2 divides b^3. Using the same logic as above. 2 divides b.

 Hence 2 is common factor of both a & b. But this is a contradiction of the fact that a & b are co-prime. Therefore, the initial assumption is wrong.

 

cube root 6 is irrational