Proof of even numbers

Question

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

Answer 1

The cardinality of the set $2\Bbb Z$ equals the cardinality of the set $\Bbb Z$. We can consider the map $x\mapsto 2x$ from $\Bbb Z\rightarrow 2\Bbb Z$. Since we know that $|\Bbb Z|=\infty$, we also know that $|2\Bbb Z|=\infty$.

Answer 2

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. $\Big)$

Answer 3

To answer this question we need a definition of "finite" and "infinite".

"Not having a largest element" isn't a good definition. Take the set: $\{$ all the rational numbers that are equal or greater than $0$ and equal or less than $1\}$. It does have a largest element-- $1$-- but it is not finite.

Also $0, -1, -2, -3,.....$ has a largest element: $0$.

What the textbook is doing is something subtler. I'm not sure the exact wording, but it is probably something like this. A set is finite if there is a set of natural numbers $1$ through $n$ so that each number corresponds to precisely one element of the set. In other words you can list them in some order and there will be a first, second, and so on and there will be a last item.

And if that is impossible, then then set is infinite.

What the book is doing is subtle. It is saying: If we take a list of elements in some order so that there is a last element, I can show that the list is not completely. That means such a list can not exist. And so the set is infinite.

And that is why you can say: If the numbers were even we could list them in order from smallest to largest. The largest and last item on the list would be some even number $E$. If we add $2$ to $E$ to get $E+2$ that number is even and larger than the largest on the list. So it isn't on the list. So it is impossible make such a list. So there are an infinite number of even integers.

My only complaint would be, I'd like an explanation why $E + 2$ is even. (You could say. $E$ is even so there is some integer $N$ so that $E = 2N$ so $E +2 = 2N + 2 = 2(N+1)$. So $E+2$ is $2$ times the integer $N+1$. So $E$ is even.)

The textbook did something similar but it used the direct definition of even. It said if there were a finite number of even integers we could list them in order from smallest to largest. There'd be a largest $E$. Then $2E$ would be even. And $2E$ would be larger than the largest and not on the list. So a list is impossible.

My complaint with that is the book assumed $2E > E$. That would not be the case if $E \le 0$. The book, I think, should point out that $2$ is an even number so $E \ge 2 > 0$ so $2E > E$.