show that cube root of 6 is irrational
Answers
Answered by
3
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......
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......
Similar questions