# Show Rationals and Integers are not Isomorphic

1. Oct 28, 2009

### miqbal

1. The problem statement, all variables and given/known data
Show that the groups $$\textbf{Q}$$ and $$\textbf{R}$$ are not isomorphic (both under addition).

This was already answered before https://www.physicsforums.com/showthread.php?t=294687", but using different theory (generators and cyclic groups). We haven't covered that stuff in class yet. See below.

2. Relevant equations
Two groups, G and H, are NOT isomorphic if:
1) |G| $$\neq$$ |H|
2) Let $$\varphi$$ be a map from G to H. Then G, H are not isomorphic if for some $$x \in \textbf{G}, |x| \neq |\varphi(x)|$$
3) G is abelian, H is not abelian or vice versa.

Two groups are isomorphic iff there exists $$\varphi:\textbf{G} \rightarrow \textbf{H}$$ s.t:
1) $$\varphi$$ is a homomorphism.
2) $$\varphi$$ is a bijection.

Homomorphism:
$$\varphi(xy) = \varphi(x)\varphi(y)$$.

3. The attempt at a solution
I know it is possible to set up a bijection from Z to Q because Q is countable (isomorphic to N) and so is Z. Then the solution must be that no homomorphism exists from Z to Q.

Proof
p,q are rational. p < q. a,b are integers. $$\varphi:{\bftext{Q} \rightarrow \bftext{R}$$, assume it is a homomorphism.
If $$\varphi(p)=a, \varphi(q) = b$$, then $$\varphi(p + q)=\varphi(p)+\varphi(q) = a + b$$. But consider p + (q - p)/2. This must map to an integer in between a and b. Why? (I'm not sure, i'll have to investigate this rigorously). Since there are finitely many integers between a and b, but infinitely many rationals between p and q, $$\varphi$$ cannot be bijective. Which means that there is no homomorphic map from the rationals to the integers that is also bijective. So Q cannot be isomorphic to R.

Is this sufficient?

Last edited by a moderator: Apr 24, 2017
2. Oct 28, 2009

### Dick

It's a little sketchy. Let f be an isomorphism Q->Z. Let p=f^(-1)(1). Then what is f(p/2)? It must be an integer that when added to itself gives 1.

3. Oct 28, 2009

### miqbal

Yeah got it.

If $$f^{-1}(1) = p$$, then $$\varphi(\frac{p}{2}+\frac{p}{2})=1$$ but that implies $$0<\varphi(\frac{p}{2})<1$$ which is impossible since no integer exists between 0 and 1.

Thanks man.