Identity Element of Natural Number Multiplication is One

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Theorem

Let $\N$ be the natural numbers.

Let $1$ be the element one of $\N$.


Then $1$ is the identity element of multiplication:

$\forall n \in \N: n \times 1 = n = 1 \times n$


Proof

Firstly, by definition of multiplication:

\(\displaystyle n \times 1\) \(=\) \(\displaystyle \left({n \times 0}\right) + n\)
\(\displaystyle \) \(=\) \(\displaystyle n\)

Next, recall that multiplication is recursively defined as:

$\forall m, n \in \N: \begin{cases} m \times 0 & = 0 \\ m \times \left({n + 1}\right) & = m \times n + m \end{cases}$

From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition for $m = 1$; that is, such that:

$\forall n \in \N: \begin{cases} f \left({0}\right) = 0 \\ f \left({n + 1}\right) = f \left({n}\right) + 1 \end{cases}$

Consider now $f'$ defined as $f' \left({n}\right) = n$.

Then evidently $f' \left({0}\right) = 0$.

Also:

\(\displaystyle f' \left({n + 1}\right)\) \(=\) \(\displaystyle n + 1\)
\(\displaystyle \) \(=\) \(\displaystyle f' \left({n}\right) + 1\)

showing that $f'$ also satisfies the definition for $1 \times n$.

Hence $n \times 1 = 1 \times n = n$ for all $n \in \N$.

That is, $1$ is the identity element for multiplication.

$\blacksquare$


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