Group Isomorphism/Examples/Real Power Function

From ProofWiki
Jump to navigation Jump to search

Example of Group Isomorphism

Let $\struct {\R, +}$ be the additive group of real numbers.

Let $\struct {\R_{>0}, \times}$ be the multiplicative group of positive real numbers.

Let $\alpha \in \R_{>0}$ be a strictly positive real number.


Let $f: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ be the mapping:

$\forall x \in \R: \map f x = \alpha^x$

where $\alpha^x$ denotes $\alpha$ to the power of $x$.


Then $f$ is a (group) automorphism if and only if $\alpha \ne 1$.


Proof

Let $\alpha \in \R_{>0}$.

We have by definition of power to a real number that:

$\alpha^x = e^{x \ln \alpha}$

From Natural Logarithm of 1 is 0:

$\ln \alpha = 0 \iff \alpha = 1$


First we note that:

\(\ds \forall x, y \in \R: \, \) \(\ds \map f {x + y}\) \(=\) \(\ds \alpha^{x + y}\) Definition of $f$
\(\ds \) \(=\) \(\ds \alpha^x \times \alpha^y\) Product of Powers
\(\ds \) \(=\) \(\ds \map f x \times \map f y\) Definition of $f$

Thus $f$ is a (group) homomorphism from $\struct {\R, +}$ to $\struct {\R_{> 0}, \times}$ for all $\alpha \in \R_{>0}$.


Sufficient Condition

Let $\alpha \in \R_{>0}$ such that $\alpha \ne 1$.

We have that:

\(\ds \forall x, y \in \R: \, \) \(\ds \map f x\) \(=\) \(\ds \map f y\)
\(\ds \leadsto \ \ \) \(\ds \alpha^x\) \(=\) \(\ds \alpha^y\) Definition of $f$
\(\ds \leadsto \ \ \) \(\ds e^{x \ln \alpha}\) \(=\) \(\ds e^{y \ln \alpha}\) Definition of Power to Real Number
\(\ds \leadsto \ \ \) \(\ds x \ln \alpha\) \(=\) \(\ds y \ln \alpha\) taking (natural) logarithm of both sides
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds y\) dividing both sides by $\ln \alpha$, as $\ln \alpha \ne 0$

and so $f$ is injective by definition.

Then we have:

\(\ds \forall x \in \R_{>0}: \, \) \(\ds x\) \(=\) \(\ds \alpha^y\)
\(\ds \leadsto \ \ \) \(\ds \log_\alpha x\) \(=\) \(\ds y\) Definition of General Logarithm
\(\ds \leadsto \ \ \) \(\ds \forall x \in \R_{>0}: \exists y \in \R: \, \) \(\ds x\) \(=\) \(\ds \map f y\) Definition of $f$

This demonstrates that $f$ is surjective.

So by definition $f$ is a bijection.

Thus $f$ is an (group) isomorphism from $\struct {\R, +}$ to $\struct {\R_{> 0}, \times}$.

$\Box$


Necessary Condition

Let $f$ be a (group) automorphism.

Aiming for a contradiction, suppose $\alpha = 1$.

Then we have:

$\forall x, y \in \R: \map f x = \map f y = 1$

and so $f$ is not injective.

Hence $f$ is not a bijection.

Therefore $f$ is not an isomorphism.

Hence by Proof by Contradiction $f$ is not an automorphism.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Group_Isomorphism/Examples/Real_Power_Function&oldid=562588"