Divisors of One

Theorem

The only divisors of $1$ are $1$ and $-1$.

Proof

From Every Integer Divides Itself‎ we have that $1 \backslash 1$.

From Every Integer Divides Its Negative we have that $-1 \backslash 1$.

Now suppose $\exists a \in \Z: a \backslash 1$.

Then $\exists c \in \Z: a c = 1$.

From Product of Absolute Values we have that:

$\left \vert {a}\right \vert \cdot \left \vert {c}\right \vert = \left \vert {1}\right \vert$

Neither $a$ nor $c$ can be zero, from Integers form Integral Domain.

So $\left \vert {a}\right \vert \ge 1$ and $\left \vert {c}\right \vert \ge 1$.

But if $\left \vert {a}\right \vert > 1$ then $\left \vert {a}\right \vert \cdot \left \vert {c}\right \vert > \left \vert {c}\right \vert$ and so $\left \vert {a}\right \vert \cdot \left \vert {c}\right \vert > 1$.

So:

$\left \vert {a}\right \vert = 1$

that is:

$a = 1$ or $a = -1$

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 3.10$: Theorem $17 \ \text{(i)}$