Equivalence of Definitions of Odd Integer

Theorem

The following definitions of the concept of Odd Integer are equivalent:

Definition 1

An integer $n \in \Z$ is odd if and only if it is not divisible by $2$.

That is, if and only if it is not even.

Definition 2

An integer $n \in \Z$ is odd if and only if:

$\exists m \in \Z: n = 2 m + 1$

Definition 3

An integer $n \in \Z$ is odd if and only if:

$x \equiv 1 \pmod 2$

where the notation denotes congruence modulo $2$.

Proof

$(1)$ if and only if $(2)$

From the Division Theorem, setting $b = 2$:

$\forall a \in \Z: \exists! q, r \in \Z: a = 2 q + r, 0 \le r < 2$

Thus, either:

$\exists q \in \Z: n = 2 q$

or:

$\exists q \in \Z: n = 2 q + 1$

When $n = 2 q$, $n$ is even by definition.

When $n$ is not even, $n = 2 q + 1$

Likewise, when $n = 2 q + 1$ it follows that $n$ is not even.

Hence both definitions of odd integer are equivalent.

$\Box$

$(2)$ if and only if $(3)$

By definition of congruence modulo $2$:

$x \equiv y \pmod 2 \iff \exists r \in \Z: x - y = 2 r$

Setting $y = 1$:

$x \equiv 1 \pmod 2 \iff \exists r \in \Z: x - y = 2 r$

from which:

$x \equiv 1 \pmod 2 \iff \exists r \in \Z: x = 2 r + 1$

Thus definition 2 is logically equivalent to definition 3.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm