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Theorem:

The negative of any even integer is even.

proof:

Suppose $n$ is an even integer. By definition of even, $n$ is even $\iff \exists k \in\mathbb{Z}$ such that $n = 2k$. We must show that the negative of $n$ is even. Then $n = 2k$ $= -(n) = -(2k)$ $= -n = -2k$ $= -n = 2(-k)$. Let $-k = r$, notice that $r$ is an integer because it's the product of integers $-1$ and $k$. Since $-n = 2r$ follows the definition of even, the negative of n is even.

How can we abbreviate this to take less space? When quoting definitions, I have a tendency to quote its entirety.

Omar
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    "Suppose that $n$ is even. Then $n=2k$ for some $k\in \Bbb Z$, and hence $-n=-2k=2(-k)$ is even." – Dietrich Burde May 22 '23 at 17:10
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    I prefer good prose to a forest of symbols. Don’t trade space for clarity – lhf May 22 '23 at 17:11
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    Yes, most mathematicians avoid writing seas of symbols. Use words. – Ted Shifrin May 22 '23 at 17:19
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    Written math is about communication. Some times you want it to be dense and economical. Most of the time that isn't easier to read (although there is such a thing as too verbose, but you're nowhere close at the moment). – Arthur May 22 '23 at 17:19
  • Thanks for the help. I think I will edit the question to include the entire proof. – Omar May 22 '23 at 17:28
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    I only like to introduce notation for calculations or readability. Otherwise I prefer natural language. – CyclotomicField May 22 '23 at 17:33

2 Answers2

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The negative of any even integer is even.

Correctly applying a definition does not necessarily involve quoting the thing. For example, here's a sample proof:

  • Let $n$ be an arbitrary even integer. Then for some integer $k,$ we have $n=2k.$

    Let $m$ denote $-k.$ So, $m$ is an integer and $-n=-2k=2m;$ that is, $-n$ is even.

    Hence, every even integer's negation is even.

(For example, given the even integer -24, its negation 24 is also even.)

By the way, the given theorem is not well-worded: does "the negative of the even integer $-6$ (i.e., minus six)" equal $6$ (but this is not negative) or $-6$ (but this is not the negation of $-6$)?

Addendum corresponding to the OP's edit

Suppose $n$ is an even integer. By definition of even, $n$ is even $\iff \exists k \in\mathbb{Z}$ such that $n = 2k$. We must show that the negative of $n$ is even. Then $n = 2k = -(n) = -(2k) = -n = -2k = -n = 2(-k).$ Let $-k = r$, notice that $r$ is an integer because it's the product of integers $-1$ and $k$. Since $-n = 2r$ follows the definition of even, the negative of n is even.

This is a nicely-written proof, except that $$n = 2k = -(n) = -(2k) = -n = -2k = -n = 2(-k)$$ is incoherent. Try to write the string of equalities such that they read more linearly.

ryang
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  • This theorem is from the book: Discrete Mathematics with Applications 4th Edition. – Omar May 22 '23 at 17:55
  • When you wrote: "Then for some integer k,n=2k." does the comma mean "such that"? – Omar May 22 '23 at 17:57
  • @Omar Don't you think that the full sentence "For some $x$ such that $x$ is prime." is incoherent or non-meaningful? Anyhow, I've just added a "we have" there, to improve readability. – ryang May 22 '23 at 18:03
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$n$ is even $\iff$ $(\forall n)(n\in Z\implies (\exists k)(n=2k))$

This is a proper definition according to definition theory of Patrick Suppes, "Introduction to Logic".

a definition should satisfy criterion of eliminability, i.e. eliminate definiendum by axioms and previous theorems of the theory on the definiens.

And satisfy criterion of non-creativity as it does not provide anything that can not be proven without it, using axioms and previous theorems of the theory.

In Mathematics this is called a conditional definition.

The OP has changed the question from what is an even natural number, to proving that an even integer (may as well be negative) is even.

That is not fair play.

Yet i will answer.

Let $N=2k$ (already defined by me). Then $-N=(-1)*2k$ assuming the axioms of arithmetic.

Thus the number is divisible by $2$.

What I don't like is foul play of people asking for someone else to do their homework!

ryaron
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