Proove that:
a)2^(1/2) is not rational.
b) prove that 2^(1/3) is not rational.
I have managed to prove the first part, but i am stuck on the second can someone give me an hint please?
assume 2^(1/2) is a rational number in the form a/b, where a/b is reduced to lowest terms
2^(1/3) = a/b
cube both sides
2= a^3/b^3
2b^2 = a^3
the left side of this equation is even
then the right side of the equation must be even
Properties of integers:
if you cube an even number the result is even
so a must be even.
Which means we could write a = 2k
and back in 2b^3 = a^3 , we can say
2b^3 = 8k^3
b^3= 4k^3
Now the right side is even, which means the LS is even and b itself must be even
so our original assumption that 2^(1/3) = a/b , where a/b is in lowest terms, led us to conclude that both a and b were even.
This is a contradiction, since a/b can be further reduced.
Thus our assumption that 2^(1/3) is rational MUST BE FALSE
Therefore 2^(1/3) is NOT rational
Note that this follows almost exactly what you must have done for the first question
To prove that 2^(1/3) is not rational, you can use a similar approach to the one you used to prove that 2^(1/2) is not rational. Here's how you can do it:
Assume that 2^(1/3) is rational, which means it can be expressed as a fraction in the form p/q, where p and q are integers with no common factors other than 1, and q is not equal to 0.
Now, let's raise both sides of this equation to the power of 3:
(2^(1/3))^3 = (p/q)^3
Simplifying the left side, we get:
2 = p^3/q^3
Rearranging the equation, we have:
2q^3 = p^3
This implies that p^3 is an even number since 2q^3 is even. Now, recall that any number raised to an odd power will always be odd. Therefore, p must also be an even number because if p were odd, p^3 would also be odd.
Knowing that p is even, we can rewrite it as p = 2k, where k is an integer. Plugging this back into the equation, we have:
2q^3 = (2k)^3
Simplifying, we get:
2q^3 = 8k^3
Dividing both sides by 2, we have:
q^3 = 4k^3
This implies that q^3 is even because 4k^3 is even.
Now, similar to what we did before, we can conclude that q is also even. If q were odd, then q^3 would be odd.
However, we initially assumed that p and q had no common factors other than 1, but now we know that both p and q are even, so they have 2 as a common factor. This contradicts our assumption.
Hence, our initial assumption that 2^(1/3) is rational must be false. Therefore, we can conclude that 2^(1/3) is indeed irrational.