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My textbook defines even as such:

If $n$ is an integer, then $n$ is even $\iff \exists$ an integer k such that $n=2k$.

Question 1: which of the following is the correct formal restatement of the definition?

(1) $\forall n \in \mathbb{Z} (($$n$ is even$) \iff(\exists k \in \mathbb{Z}$ such that $n=2k))$.

(2) $\forall n ((n \in \mathbb{Z}) \implies ((n$ is even$) \iff (\exists k \in \mathbb{Z}$ such that $n=2k)))$.

Question 2: How does the definition guarantee that any element that is not an integer is not even?

This is the part I am mainly confused about. I know that any element that is not an integer obviously cannot be even, but I am not sure if the definition supports that. If the definition is an if statement as in statement (2), then isn't the conditional vacuously true for all non-integers? Then why would any non-integer not be even with this definition?

Subquestion: If the domain of a predicate variable is restricted to a set, then is the predicate false for all elements outside the domain?

For instance, does statement (1) imply that the biconditional is false for non-integers?

Cynicrom
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    The statement, as written, is silent on the question of non-integers. One could imagine some other definition applied to real numbers (say) which coincided with this one for integers, but allowed some other "even numbers". Typically, and somewhat confusingly, of course, some definitions are implicitly "if and only if" statements, though it's wise to clarify that point when you state any given definition. – lulu Jul 31 '23 at 14:44
  • Would you be able to justify that 3.14 (example) is not even from this definition? Is it valid to say that all elements of a set are not even if "even" is not defined for this set? – Cynicrom Jul 31 '23 at 15:14
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    As the definition is silent regarding non-integers, it simply does not apply to that number. Again, one could imagine a different definition which would apply, but that would be a different question. – lulu Jul 31 '23 at 15:15
  • Usually , "even" and odd" only apply to integers as well as "composite or prime" for which we usually even demand integers greater than $1$. Many concepts can be generalized , but it always the question whether it is worth the effort. – Peter Jul 31 '23 at 15:16
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    For instance, we say that a triangle is isosceles if at least two sides have equal length. Having made that definition, you can't then ask if some five sided figure is isosceles...the definition just doesn't apply. However, you could easily extend the definition to cover, say, other polygons. – lulu Jul 31 '23 at 15:19
  • @lulu Then would the statement "A five sided figure is isosceles" be true or false? Or would it not have any truth value at all? – Cynicrom Jul 31 '23 at 15:22
  • It would be "vacously true" – Peter Jul 31 '23 at 15:22
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    You keep asking the same question and I keep giving you the same answer. For the last time: if one has defined some property in that context, that definition is, a priori, silent outside that context. It may, or may not, be possible to generalize the definition to other contexts, but it isn't "automatic" in any sense. And different generalizations may not coincide apart from the original context. – lulu Jul 31 '23 at 15:23

3 Answers3

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We say that a natural number is 'perfect' if and only if it is the sum of its divisors, including $1$, but excluding itself.

Suppose we now change domains, and ask "Is this the perfect answer?"

Here is what you do not want to say:

'An answer is a not a natural number, and so yes, it is trivially perfect'

Here is what you also do not want to say:

'An answer is a not a natural number, and so no, it is trivially not perfect'

Here is what you do want to say:

'I know what it means for a natural number to be perfect, but I am not sure what it means for an answer to be perfect. I just don't know what 'perfect' means in this context.'

Bram28
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  • This misses a key point: the OP's definition of "even" applies in any ring (if we remove the (artificial) restriction to $\Bbb Z$), but your definition of "perfect number" does not apply to "answers". So the context is very different. – Bill Dubuque Jul 31 '23 at 18:36
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The purpose of the initial universal quantifier is to set the scope of the definition to the ring of integers $\Bbb Z$. Thus the odds (= not evens) are the complement of the evens within $\,\Bbb Z,\,$ i.e. $\,\Bbb Z\,\backslash\, 2\Bbb Z.\,$ It is a type error to attempt to apply the definition to some other ring without first explicitly extending the definition to the new context.

Extensions of the notion of parity arithmetic do in fact exist and often prove handy, e.g. as explained here parity extends uniquely to imaginary integers by defining $\,i\,$ to be odd.

A prototypical instance of your question is "is $\,i\,$ odd?" which is analogous to the question "is $\,i\,$ irrational?" which is discussed at length here.

Bill Dubuque
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The statement ∀((∈ℤ) says clearly n has to be an integer, the same for k. So real numbers or fractions can not be integer, for example 3/4=6/8 so one could not define an "even" fraction.

trula
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    This answer is misleading since this definition of even (and parity structure) can in fact be applied in many rings, e.g. to complex integers $,m+n,i,\ m,n\in\Bbb Z,,$ as explained here. – Bill Dubuque Jul 31 '23 at 14:58
  • I talked about fractions and real numbers, not any other rings. – trula Jul 31 '23 at 15:05
  • But parity arithmetic does in fact work fine for certain subrings of rationals, e.g. those rationals writable with odd denominator - see here – Bill Dubuque Jul 31 '23 at 15:55