Question

prove that cubic root 6 is irrational​

Answer 1

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 2

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

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