Is $\mathbb(Q, +)$ isomorphic to $\mathbb(R, +)$? (without using uncountability)

Question

I came across this problem:
Prove that $\mathbb(Q, +)$ is not isomorphic to $\mathbb(R, +)$ without using the argument of uncountability.

The alternative argument given was the following:

$\mathbb(Q, +)$ has a property allowing it to act as pseudo-cyclic which $\mathbb(R, +)$ does not:

$Φ(\frac{m}{n}) = (\frac{m}{n})Φ(1)$

(A) Is this argument valid?

(B) I understand that if the arrival set were $Q$ under addition , it would have made more sense, but could one argue that $K$ is isomorphic to $H$ if and only if $H$ is isomorphic to $K$?

(C) If we are to find an argument relating to surjectivity, would that not be referring to the uncountability of $\mathbb R$?

Answer 1

When working with $\mathbb{Q}$ and $\mathbb{R}$ as abelian groups, saying that $\Phi(\tfrac{m}{n}) = \tfrac{m}{n}\Phi(1)$ isn't technically well-defined since $\tfrac{m}{n}$ is not an integer (although, this will make sense when you view them as vector spaces over $\mathbb{Q}$). You are correct about the fact that $\mathbb{Q}$ has a certain notion of pseudo-cyclicity that $\mathbb{R}$ does not share with it.

An (additive) group $G$ is "pseudo-cyclic" if for all non-zero $x, y \in G$, there exist integers $m$ and $n$ satisfying the equality $mx = ny$.

Prove that $\mathbb{Q}$ is pseudo-cyclic and $\mathbb{R}$ is not. Conclude that they are non-isomorphic.

Answer 2

When the Criteria is "Do not use Countability" , I am not entirely sure what we can use & what we can not use , though I would use "limits" to show the "Non-Isomorphism" here.
Proof given here comes with that Disclaimer.

When we have 2 rational numbers $a$ & $b$ , we know that $a+b$ is also rational.
When we have 2 real numbers $a$ & $b$ , we know that $a+b$ is also real.
We can use the Isomorphism to get $\phi(a)+\phi(b)=\phi(a+b)$ which converts rationals to reals.

[[ looks Isomorphic !! ]]

When we add $n$ rational numbers , we should still get rational number.
$D_n \in Q \implies [D=\sum_{i=1}^{n}D_n] \in Q$

When we add $n$ real numbers , we should still get real number.
$D_n \in R \implies [D=\sum_{i=1}^{n}D_n] \in R$

We can use the Isomorphism to get $\phi(D)=\phi(\sum_{i=1}^{n}D_n)=(\sum_{i=1}^{n}\phi(D_n))$ which converts rationals to reals.

[[ looks Isomorphic here too !! ]]

In the limit , when $n \rightarrow \infty$ , the Isomorphism will break down :
We can choose the $D_n$ such that $D$ is irrational. Eg $D_n=10^{-(n^2)}$
In the limit , that Summation is closed under addition , when we consider reals.
In the limit , that Summation is NOT closed under addition , when we consider rationals.

In other words , that Summation shows that $\phi$ [ which should convert rational to real ] has converted at least 1 irrational number to 1 real number.
The Inverse of that 1 real number should have been 1 rational number.

The Isomorphism has broken down !!

Answer 3

I assume that we can use the fact that there are irrational numbers and the set of real numbers is nothing but union of rational and irrational number sets and also that irrationals are those numbers which cannot be written in the form $\frac{m}{n}$, where $m$ and $n$ are integers (with $n\neq0$).

If this is allowed, if $\varphi(1)$ is rational, then the image of $\varphi$ is a subset of rational numbers and hence the map is not onto. Suppose $\varphi(1)$ is irrational, say, $x$. Simply conisder $x^2$. If $x^2$ has a pre image, say $q$, then $$x^2=\varphi(q) = q\varphi(1) = qx $$ This would imply that, $x=q$, a contradiction. Thus, in any case, the map cannot be onto.

This nowhere assumes uncountability of $\mathbb{R}$.

Answer 4

Suppose we have an isomorphism $\phi:(\mathbb R,+)\to(\mathbb Q,+)$. Then $\phi(1)=q$ is a non-zero rational, $\phi(m)=mq$ for an integer $m$, and so for $t=m/n$ we have $n\phi(t)=\phi(nt)=\phi(m)=mq$, so that $\phi(t)=mq/n=qt$. Since multiplication by $q$ on the rationals is onto, we see that for every irrational $x$ there is a rational $t$ such that $\phi(x)=\phi(t)$, a contradiction.