Exponential of Sum/Real Numbers/Proof 2

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Theorem

Let $x, y \in \R$ be real numbers.

Let $\exp x$ be the exponential of $x$.


Then:

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


Proof

Lemma

Let $x, y \in \R$.

Let $n \in \N_{> 0}$ such that $n > -\paren {x + y}$.


Then:

$1 + \dfrac {x + y} n + \dfrac {x y} {n^2} = \paren {1 + \dfrac {x + y} n} \paren {1 + \dfrac {\paren {\frac {x y} {n + x + y} } } n}$

$\Box$


This proof assumes the definition of $\exp$ as defined by a limit of a sequence:

$\exp x = \displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$


From Powers of Group Elements we can presuppose the Exponent Combination Laws for natural number indices.


First we introduce a lemma:


By definition:

\(\displaystyle \paren {\exp x} \paren {\exp y}\) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n \lim_{n \mathop \to +\infty} \paren {1 + \frac y n}^n\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {\paren {1 + \frac x n} \paren {1 + \frac y n} }^n\) Combination Theorem for Sequences
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n + \frac {x y} {n^2} }^n\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {\paren {1 + \frac {x + y} n} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n} }^n\) Lemma: Without loss of generality let $n > - x - y$: therefore $n + x + y > 0$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n}^n \lim_{n \mathop \to +\infty} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n}^n\) Combination Theorem for Sequences
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n}^n\) Null Sequence in Exponential Sequence
\(\displaystyle \) \(=\) \(\displaystyle \map \exp {x + y}\)

$\blacksquare$