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:

\(\ds n \times 1\) \(=\) \(\ds \paren {n \times 0} + n\)
\(\ds \) \(=\) \(\ds n\)

Next, recall that multiplication is recursively defined as:

$\forall m, n \in \N: \begin{cases}

m \times 0 & = 0 \\ m \times \paren {n + 1} & = 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}

\map f 0 = 0 \\ \map f {n + 1} = \map f n + 1 \end{cases}$

Consider now $f'$ defined as $\map {f'} n = n$.

Then evidently $\map {f'} 0 = 0$.

Also:

\(\ds \map {f'} {n + 1}\) \(=\) \(\ds n + 1\)
\(\ds \) \(=\) \(\ds \map {f'} n + 1\)

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.

$\blacksquare$


Sources

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