# Prove no isomorphism from rationals to reals

1. Jan 25, 2010

### issisoccer10

1. The problem statement, all variables and given/known data
Prove that there is no isomorphism, $$\phi$$, from Q under addition to R under addition

2. Relevant equations
An isomorphism $$\phi$$:Q to R is a bijection such that $$\phi$$(x + y) = $$\phi$$(x) + $$\phi$$(y), where x,y are elements of Q

$$\phi$$(0) = 0.

$$\phi$$(-x) = -$$\phi$$(x)

3. The attempt at a solution
My inclination is to attempt to attempt to show a contradiction from two equal rationals p/q and p'/q' occurs when,

$$\phi$$(p/q - p'/q') = $$\phi$$(0) = 0, for p,q integers.

So, $$\phi$$(p/q) - $$\phi$$(p'/q') = 0,

However, I cannot arrive at an algebraic contradiction.

2. Jan 25, 2010

### Dick

The rationals have the property that for every nonzero x and y there are integers n and m such that n*x+m*y=0 where n and m are nonzero integers. Can you prove that? phi(n*x)=n*phi(x), right? Now find two reals that don't have that property.

Last edited: Jan 25, 2010
3. Jan 25, 2010

### issisoccer10

Hi Dick,

Thank you very much for your help. I feel that I can prove the property of rationals that your described.

However, while I know that there are reals that don't have that property, such as $$\pi$$ and the sqrt(1), I do not know how to prove that those two numbers do not have that property.

Doug

4. Jan 25, 2010

### Dick

They don't. But let's pick an easier example, suppose phi(x)=1 and phi(y)=sqrt(2). Can you have n*1+m*sqrt(2)=0?

5. Jan 26, 2010

### issisoccer10

No, because that would imply $$\sqrt{2}$$ = -n/m, i.e. that $$\sqrt{2}$$ is rational..

Thanks a lot, I really appreciate it.