D
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In some cases, zero is considered to be even.
In other cases, it is counted as neither odd nor even.
Did you ask Google?
In other cases, it is counted as neither odd nor even.
Did you ask Google?
No, zero is even. When zero, an integer, is divided by two, it leaves a remainder of zero. That is all you need to show.
There are no "in some cases."
Steven G
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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.
D
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However:
An "even" number with "even" exponent will result in "even" number.
2^0 = 1 ........ Now what??
Dr.Peterson
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How is that relevant? You seem to have made up a nonexistent "rule", and then shown a counterexample.
An even number is defined as 2 times any integer.
You could say that an even number multiplied by any integer is even; or that an even number raised to a positive integer power is even; but even exponents have nothing to do with it.
Steven G
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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.
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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.
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Dr.Peterson
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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.
Dr.Peterson
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I didn't say "your preferred definition", but referred to the definition you offered, which clearly was not intended seriously.
As for "actively excluding", the contrast is with "merely failing to include".
What I'm saying is that the fact that you often find a definition that only applies to positive integers does not mean that the authors of those definitions would tell you that zero is not even; it just isn't in view. (Maybe their audience doesn't even know what zero is.) They don't happen to include zero as an even number, because the context doesn't cover zero; but that doesn't explicitly exclude zero.
Can you show me any reasonable authority who says that zero is actually not an even number?
I agree that arguing over definitions is silly. The question here is, what do you say when someone asks? You say yes. You don't say "some people say it is neither even nor odd".
pka
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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)
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pka
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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.
Granted historically counting numbers ,\(\displaystyle \{1,2,3,\cdots\}\) are postive. In fact it was late in history before we even had the notion zero. Merchants was a sand table and smooth stones to add and subtract. When a stone was removed it left a nice \(\displaystyle {O}\) shape. So the zero was addopted.
Dr.Peterson
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Huh? I don't think I disagree with you at all.
I said that arguing over definitions is silly. Of course intelligent discussion of definitions is not, nor is careful consideration of wording.
The silliness is when two people say "I'm right, nyah, nyah, nyah" just because they're starting with different contexts, and each definition is valid in its context.
Careful dissection of a definition to make sure it is worded adequately (and, indeed, deciding whether 1 should be considered a prime number at all) is worth talking about.
But this is digressing a little too far.
mmm4444bot
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mmm4444bot
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pka
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No it was matter of the precision of language. It is also a matter of mathematical envy.
E.H. Moore was R.L. Moore's PhD advisor.(no relation)
R.L was a high school teacher is East Texas who proved that Hilbert's geometric axioms were not independent.
RLMoore's time at Chicago was just long enough to justify awarding the degree.
RLMoore became the most important American topologist of the last century.
Devlin described him as the greatest mathematics teacher ever.
Five of his students have been presidents of the American Mathematical Society.
If we add the Mathematical Association of America(MAA) that number more than doubles.
RLMoore passion for language came from EHMoore. But RLMoore passed it on to his students, then to their students, of which I am one.
k could not be either of those values above, because "n" is used for integers. That definition is faulty, because it contains a necessary condition,
but not a sufficient condition. So, that definition cannot even be considered.
Also, if k were to belong to the set {... ,-3/2, -1/2, 1/2, 3/2, ...}, then n would be an integer but not specifically an even integer.
Again, zero, an integer, is necessarily even because it has a zero remainder when divided by two.
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https://en.wikipedia.org/wiki/Parity_of_zero
Plus 0 lies exactly halfway between -1 and 1 in the x-axis; so even
Plus when lookagain pays me back the $5 he owes me,
then he'll owe me $0 and he'll say: "we're EVEN stevens"
Plus 0 lies exactly halfway between -1 and 1 in the x-axis; so even
Plus when lookagain pays me back the $5 he owes me,
then he'll owe me $0 and he'll say: "we're EVEN stevens"
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