Prove that there is an infinite number of even numbers:
Assume there is a largest even number, $E$.
$E + 2$ would also be even, as $E$ must be divisible by 2, so $E + 2$ is divisible by $2$, and clearly greater than $E$.
If $E$ is the largest even number, then $-E$ is the largest negative even number.
$-E – 2$ would also be even, as $-E$ must be divisible by 2, so $-E – 2$ is divisible by $2$, and clearly -E – 2 is a greater negative even number than E
Therefore, this contradicts the original assumption, so it must be incorrect.
Therefore, there is an infinite number of even numbers.
Is this proof sufficient, or is something missing? As the textbook's answer was much longer.
EDIT: Textbook's answer is:
Suppose that there is a finte number N of even numbers
This finite list can be ordered so that E1 < E2 < E3 < …
Then the largest even number is En
But 2En would also be even and clearly greater than En, so is not in the list.
Therefore, there are more than N even numbers.
This contradicts the initial proposition.
Therefore, there are infinitely many even numbers
Best Answer
Consider mapping even numbers to odd numbers. $$\varphi : A\to B\quad:\quad2n\mapsto2n-1$$ where $A$ is the set of even numbers and $B$ is the set of odd numbers.
This will give you a bijection. As $\mathbb{N}=A\cup B$ shows the set of natural numbers $\mathbb{N}$ is finite giving you a contradiction.
$\Big($ Assuming you can show every even and odd number is of the form $2n$ and $2n-1$ for $n\in\mathbb{N}$ respectively. $\Big)$