Sum of Even Integers is Even
Theorem
The sum of any finite number of even integers is itself even.
Proof 1
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
- The sum of $n$ even integers is an even integer.
$\map P 1$ is trivially true, as this just says:
- The sum of $1$ even integer is an even integer.
The sum of $0$ even integers is understood, from the definition of a vacuous summation, to be $0$, which is even.
So $\map P 0$ is also true.
Basis for the Induction
$\map P 2$ is the case:
- The sum of any two even integers is itself even.
Consider two even integers $x$ and $y$.
Since they are even, they can be written as $x = 2 a$ and $y = 2 b$ respectively for integers $a$ and $b$.
Therefore, the sum is:
- $x + y = 2 a + 2 b = 2 \paren {a + b}$
Thus $x + y$ has $2$ as a divisor and is even by definition.
This is our basis for the induction.
$\Box$
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- The sum of any $k$ even integers is itself even.
Then we need to show:
- The sum of any $k+1$ even integers is itself even.
Induction Step
This is our induction step:
Consider the sum of any $k + 1$ even integers.
This is the sum of:
- $k$ even integers (which is even by the induction hypothesis)
and:
- another even integer.
That is, it is the sum of two even integers.
By the basis for the induction, this is also even.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
That is:
- The sum of any finite number of even integers is itself even.
$\blacksquare$
Proof 2
Let $S = \set {r_1, r_2, \ldots, r_n}$ be a set of $n$ even numbers.
By definition of even number, this can be expressed as:
- $S = \set {2 s_1, 2 s_2, \ldots, 2 s_n}$
where:
- $\forall k \in \closedint 1 n: r_k = 2 s_k$
Then:
\(\ds \sum_{k \mathop = 1}^n r_k\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n 2 s_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \sum_{k \mathop = 1}^n s_k\) |
Thus, by definition, $\ds \sum_{k \mathop = 1}^n r_k$ is even.
$\blacksquare$