Question

Prove that ³√6 is an irrational number​

Answer 1

• We shall start by assuming (cube root 6) as rational. In other words, we need to find integers x and y such that
• (cube root 6) = x/y.
• Let x and y have a common factor other than 1, and so we can divide by that common factor and assume that x and y are coprime. So, y(cube root 6) = x.
• Cubing both side, we get, 6 y^3 = x^3.
• Thus, x^3 is divisible by 6, and by theorem we can say that x is also divisible by 6.
• Hence, x = 6z for some integer z.
• Substituting x, we get, 6 y^3 = 216 z^3 i.e. y^3 = 36 z^3; which means y^3 is divisible by 6, and so y will also be divisible by 6.
• Now, from theorem, x and y will have 6 as a common factor. But, it is opposite to fact that x and y are co-prime.
• Hence, we can conclude (cube root 6) is irrational.

Answer 2

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• 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......