prove that cubic root 6 is irrational
Answers
Answer:
6#3cube is an rational
6 can be written as 3 root2
root2 is an irrational so 6 is also an irrational
Step-by-step explanation:
hence proved
AnswEr:-
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=a³/b³
a³ = 6b³
a³ = 2(3b³)
Therefore, 2 divides a³ or a² * 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
8k³ = 6b³
b³ = 2{(2k³) / 3)}
Therefore, 2 divides b³. 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.
So, cube root 6 is irrational