*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.

Or f(x)=x/0 maybe alsođź™‚

I was however, thinking of maybe real numbers greater than zerođź™‚

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!