- 29 May '14 19:00 / 1 editSo PI is irrational, and on a game show they said the 5th root of 64 is irrational. Is that root really irrational?

How can you tell if a number like that is going to be irrational or not?

Is there some kind of math test or function that can predict irrationality?

Last question:

Is there some function or procedure that can turn any irrational number into a rational one? - 30 May '14 13:49

The last one is easy:*Originally posted by sonhouse***So PI is irrational, and on a game show they said the 5th root of 64 is irrational. Is that root really irrational?**

How can you tell if a number like that is going to be irrational or not?

Is there some kind of math test or function that can predict irrationality?

Last question:

Is there some function or procedure that can turn any irrational number into a rational one?

Say that the number x is irrational, then the function f(x) = 0*x is not only rational but also natural. - 30 May '14 14:42 / 5 edits

Or f(x)=x/0 maybe also*Originally posted by FabianFnas***The last one is easy:**

Say that the number x is irrational, then the function f(x) = 0*x is not only rational but also natural.

I was however, thinking of maybe real numbers greater than zero

So I found this, proof that the square root of 2 is irrational.

The proof that square root of 2 is irrational:

Let's suppose sqr root of 2 were a rational number. Then we can write it sqr root 2 = a/b where a,b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to the lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in its simplest terms, both a and b must be not be even. One or both must be odd. Otherwise, you could simplify.

From the equality sqr root 2 = a/b it follows that 2 = a^2/b^2, or a^2 = 2 * b^2. So the square of a is an even number since it is two times something. From this we can know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a * a would be odd too. Odd number times odd number is always odd. Check if you don't believe that!

Okay, if a itself is an even number, then a is 2 times some other whole number, or a = 2k where k is this other number. We don't need to know exactly what k is; it won't matter. Soon is coming the contradiction:

If we substitute a = 2k into the original equation 2 = a^/b^2, this is what we get:

2 = (2k)^2/b^2

2 = 4k^2/b^2

2*b2 = 4k^2

b2 = 2k^2.

This means b^2 is even, from which follows again that b itself is an even number!!!

WHY is that a contradiction? Because we started the whole process saying that a/b is simplified to the lowest terms, and now it turns out that a and b would both be even. So the square root of 2 cannot be rational.

I'm still digesting this! - 23 Jul '14 19:49Proof that the 5th root of 64 is irrational.

Suppose not and that it is a/b where a and b are coprime ie have no factor in common. Now factorise both a and b into prime factors. As a and b are coprime then they must have no prime factor in common and hence a^5 and b^5 also have no prime factors in common and a^5=(2^6)*(b^5). Comparing powers of 2 says that 2 divides a and hence not b. But then the power of 2 appearing on the LHS must be a multiple of 5 but the power of 2 appearing on the RHS is 6. This is a contradiction.