Use proof by contradiction to prove that cuberoot(2) is irrational. Use proof by
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Use proof by contradiction to prove that cuberoot(2) is irrational.
Use proof by contradiction to prove that cuberoot(2) is irrational. So I know I need to do proof by contradiction. And I can get to nearly the end: Please tell my mistake, my teacher says I'm just assuming b^3 is even and that's wrong: Assume cuberoot(2) is rational. Then, let a/b = cuberoot(2), where a, b epsilon mathbb{Z} and b != 0, in the lowest divisible form. Then, 2b^3 = a^3. Let a^3 =2k, since a^3 is an integer because of 2b^3 being an even integer, k is an even integer. Then, (2k)^3 = 2b^3. 8k^3 = 2b^3. Proof by contradiction, since 2 is a common factor.Explanation / Answer
The cube root of 2 is irrational. The proof that the square root of 2 is irrational may be used, with only slight modification.
Assume the cube root of 2 is rational. Then, it may be written as a/b, where a and b are integers with no common factors. (This is possible for all nonzero rational numbers). Since a/b is the cube root of 2, its cube must equal 2. That is,
(a/b)3 = 2
a3/b3 = 2
a3 = 2b3.
The right side is even, so the left side must be even also, that is, a3 is even. Since a3 is even, a is also even
2b^3 is even because when we multiply any number with 2, it become even number..
(because the cube of an odd number is always odd). Since a is even, there exists an integer c such that a = 2c.
Now,
(2c)3 = 2b3
8c3 = 2b3
4c3 = b3.
The left side is now even, so the right side must be even too. The product of two odd numbers is always odd, so b3 cannot be odd; it must be even. Therefore b is even as well.
Since a and b are both even, the fraction a/b is not in lowest terms, thus contradicting our initial assumption. Since the initial assumption cannot have been true, it must be false, and the cube root of 2 is irrational.
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