Even and Odd Integers form Partition of Integers
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Theorem
The odd integers and even integers form a partition of the integers.
Proof
Let $n \in \Z$ be an integer.
Let $\Bbb O$ be the set of odd integers and $\Bbb E$ be the set of even integers.
By the Division Theorem:
- $\forall n \in \Z: \exists! q, r \in \Z: n = 2 q + r, 0 \le r < 2$
from which it follows that either:
- $n = 2 q \in \Bbb E$
or:
- $n = 2 q + 1 \in \Bbb O$
Thus:
- $(1): \quad$ each element of $\Z$ is in no more than one of $\Bbb E$ and $\Bbb O$
- $(2): \quad$ each element of $\Z$ is in at least one of $\Bbb E$ and $\Bbb O$
and:
- $(3): \quad$ setting $q = 0$ it is seen that $0 \in \Bbb E$ and $1 \in \Bbb O$ and so neither $\Bbb E$ or $\Bbb O$ is empty.
Thus $\set {\Bbb E \mid \Bbb O}$ is a partition of $\Z$ by definition.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $2$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2$
- 2008: David Joyner: Adventures in Group Theory (2nd ed.) ... (previous) ... (next): Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson