Exponential on Real Numbers is Group Isomorphism

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

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

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

$x \mapsto \map \exp x$

where $\exp$ is the exponential function.

Then $\exp$ is a group isomorphism.

Proof 1

From Exponential of Sum we have:

$\forall x, y \in \R: \map \exp {x + y} = \exp x \cdot \exp y$

That is, $\exp$ is a group homomorphism.

Then we have that Exponential is Strictly Increasing.

From Strictly Monotone Real Function is Bijective, it follows that $\exp$ is a bijection.

So $\exp$ is a bijective group homomorphism, and so a group isomorphism.


Proof 2

From Real Numbers under Addition form Abelian Group, $\struct {\R, +}$ is a group.

From Strictly Positive Real Numbers under Multiplication form Uncountable Abelian Group, $\struct {\R_{> 0}, \times}$ is a group.

We have that for all $y \in R_{> 0}$ there exists $x = \map \ln y \in R$ which satisfies $\map \exp x = y$.

Thus $\exp$ is a surjection.

Then from Exponential on Real Numbers is Injection:

$\exp$ is an injection.

Therefore, $\exp$ is a bijection.

Let $x, y \in R$.

From Exponential of Sum:

$\map \exp {x + y} = \map \exp x \, \map \exp y$

So $\exp$ is a homomorphism and a bijection.

It follows by definition that $\exp: \struct {\R, +} \to \struct {\R_{> 0}, \times}$ is an isomorphism.