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 $16.1$