Question

prove 3 cube root and 2√6 are irrational​

Answer 1

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