# 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

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.1$