Prove no isomorphism from rationals to reals

[issisoccer10]

Homework Statement

Prove that there is no isomorphism, \phi, from Q under addition to R under addition

Homework 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)

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.

Is there a better way to go about this proof, relying only on group theory?

 

[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.

 

[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.

Thanks in advance,
Doug

 

[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?

 

[issisoccer10]

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

Thanks a lot, I really appreciate it.