Positive Rational Numbers under Addition not Isomorphic to Natural Numbers

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Theorem

The positive rational numbers $\Q_{\ge 0}$ under addition:

$\struct {\Q_{\ge 0}, +}$

is not isomorphic to the natural numbers under addition:

$\struct {\N, +}$


Proof

From:

Positive Rational Numbers under Addition form Commutative Monoid
Natural Numbers under Addition form Commutative Monoid

both $\struct {\Q_{\ge 0}, +}$ and $\struct {\N, +}$ form commutative monoids.


Aiming for a contradiction, suppose there exists an semigroup isomorphism $\phi$ from $\struct {\Q_{\ge 0}, +}$ to $\struct {\N, +}$.

By definition of isomorphism:

$\phi$ is a homomorphism
$\phi$ is a bijection.


Let $n \in \N$ be odd.

Let $q \in \Q_{\ge 0}$ such that $\map \phi q = n$.

Such a $q$ exists and is unique by definition of bijection.


But then we have:

\(\ds \exists m \in \N: \, \) \(\ds \map \phi {\dfrac q 2}\) \(=\) \(\ds m\) Definition of Bijection
\(\ds \leadsto \ \ \) \(\ds \map \phi q\) \(=\) \(\ds \map \phi {\dfrac q 2 + \dfrac q 2}\)
\(\ds \) \(=\) \(\ds \map \phi {\dfrac q 2} + \map \phi {\dfrac q 2}\) Definition of Semigroup Homomorphism
\(\ds \) \(=\) \(\ds m + m\)
\(\ds \) \(=\) \(\ds 2 m\)
\(\ds \) \(=\) \(\ds n\)

But this contradicts the assertion that $n$ is odd.

So by Proof by Contradiction there can be no such isomorphism from $\struct {\Q_{\ge 0}, +}$ to $\struct {\N, +}$.

Hence the result.

$\blacksquare$


Sources