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In some cases, zero is considered to be even.Is the zero even or odd? Could you explain it, please?
In other cases, it is counted as neither odd nor even.
Did you ask Google?
In some cases, zero is considered to be even.Is the zero even or odd? Could you explain it, please?
In some cases, zero is considered to be even.
In other cases, it is counted as neither odd nor even.
Did you ask Google?
I have to agree with lookagain on this one. I prefer to use multiplication and since 2*0 = 0 (and 0 is an integer), then 0 is even. Done, no debate.In some cases, zero is considered to be even.
In other cases, it is counted as neither odd nor even.
Did you ask Google?
However:I have to agree with lookagain on this one. I prefer to use multiplication and since 2*0 = 0 (and 0 is an integer), then 0 is even. Done, no debate.
However:
An "even" number with "even" exponent will result in "even" number.
2^0 = 1 ........ Now what??
You came up with a counterexample to 2integer is always even. However, an integer is even iff it on the two times table. This is a definition of an even number.However:
An "even" number with "even" exponent will result in "even" number.
2^0 = 1 ........ Now what??
Arguments about definitions are neverending. A common definition of a natural number is the set {1, 2, 3, 4, ... }. And a common definition of an even number isI have to agree with lookagain on this one. I prefer to use multiplication and since 2*0 = 0 (and 0 is an integer), then 0 is even. Done, no debate.
Arguments about definitions are neverending. A common definition of a natural number is the set {1, 2, 3, 4, ... }. And a common definition of an even number is
\(\displaystyle n \text { is an even number } \iff \exists\ k \in \mathbb N \text { such that } 2 * k = n.\)
Now, according to those common definitions, zero is not an even number. I myself prefer to define the natural numbers as the set {0, 1, 2, 3, 4, ...} so, according to my preferred definitions, zero is even. But my preferred definitions are not the most common so I should qualify my statements when they result from my less common definitions.
I certainly agree with you that zero is an even number if we work with this definition:
\(\displaystyle n \text { is an even number } \iff \exists \text { a number } k \text { such that } 2k = n.\)
Of course, under that definition \(\displaystyle \dfrac{1}{7}\) and \(\displaystyle e ^{\pi}\) are even numbers.
It makes sense to argue about which definitions are useful, but it makes little sense to argue when different conclusions arise from different definitions. A common definition of the natural numbers excludes zero, and a common definition of an even number is based on the natural numbers. Using those common definitions leads ineluctably to the conclusion that zero is not an even number. This is merely one reason why I find the alternative definition of natural numbers preferable. Or, as you have done, you can define even numbers over the integers, but that too is not the most common definition.
I think you misread me. Under my preferred definitions, zero is even.Well, under your definition, every number is "even", so it's not much of a definition. To make it a valid definition, you need to specify what kind of number k has to be, and that is the crux of the issue.
As I see it, if someone asks whether 0 is even, they are not thinking only of natural numbers (by the restricted definition, i.e. positive integers); they are working either with non-negative integers ("whole", by some definitions), or with integers, or with some larger set. And as soon as you allow 0 to be considered, the appropriate definition of "even" includes 0: \(\displaystyle n \text { is an even number } \iff \exists \text{ an }\mathbf {integer\ } k \text { such that } 2k = n.\) That is, even if you start with the most restricted definition, as you move to larger sets of numbers, the definition naturally expands to include them (though we don't move beyond integer multiples because to do so would trivialize the concept).
Of course, this is just a definition, not a proof, so I can't force anyone to agree. But do you have evidence that the most common definition of "even" actively excludes zero (as opposed to not mentioning it, just because zero is not under consideration)? My evidence is that it does not.
I think you misread me. Under my preferred definitions, zero is even.
But I gave commonly accepted definitions under which zero is not included in the set of even numbers.
I find it odd that people at a math site ask for proof that (1) different definitions reach different conclusions, (2) specifying definitions is important to do, and (3) even numbers are often defined in terms of natural numbers, which in turn are often defined to start with 1. So yes, any definition of "even" framed in terms of natural numbers, with those defined to start with 1, necessarily excludes 0. I am not aware of the difference between "actively excluding" and "passively excluding." If something is not in a set, please define the difference between actively not being in the set and passively not being in the set.
Now with that I disagree.I agree that arguing over definitions is silly.
Is zero a natural number? For most working mathematicians it is a natural number as you can see in that link. But \(\displaystyle 0\) is not positive.Natural numbers are positive,
Now with that I disagree.
A common definition of a prime number is a positive integer having itself & one as its only factors. Of course that makes \(\displaystyle 1\) a prime.
Whereas, a prime number is a positive integer having exactly two divisors does the job.
Before anyone questions the word two above, here is anecdote.
The great geometer E.H.Moore began an invited lecture at Princeton with "Let a be a point and b be a point." Solomon Lefschetz shouted out "But why don't you just say 'Let a & b be points' ?" Moore reply "Because a could be b". Lefschetz stormed out of the room. (p199, Mathematical Apocrypha)
Most working mathematicians do not come here for help.Is zero a natural number? For most working mathematicians it is a natural number as you can see in that link.
This commentary reminded me of discussion about multiplicity of roots.… A common definition of a prime number is a positive integer having itself & one as its only factors. Of course that makes \(\displaystyle 1\) a prime.
Whereas, a prime number is a positive integer having exactly two divisors does the job …
… "Let a be a point and b be a point." … "Because a could be b" …
Could you say they had a point in contention? (Ahem!)The great geometer E.H.Moore began an invited lecture at Princeton with "Let a be a point and b be a point." Solomon Lefschetz shouted out "But why don't you just say 'Let a & b be points' ?" Moore reply "Because a could be b". Lefschetz stormed out of the room. (p199, Mathematical Apocrypha)
No it was matter of the precision of language. It is also a matter of mathematical envy.Could you say they had a point in contention? (Ahem!)-Dan
I certainly agree with you that zero is an even number if we work with this definition:
\(\displaystyle n \text { is an even number } \iff \exists \text { a number } k \text { such that } 2k = n.\)
Of course, under that definition \(\displaystyle \dfrac{1}{7}\) and \(\displaystyle e ^{\pi}\) are even numbers.