I thought “zero has no identity” meant that it wasn’t odd or even, among other things.
But I think I am misintepreting it:
Today in my trig class the question was raised as to whether 0 was odd or even. I of course said neither. This was asked due to how we could tell if the functions were odd even or neither and how degrees of a function determined its outcome. I say that it just buy chance makes a function even if there is no variable in the equation…
Well we got into this debate and I figured that n, n+2,…,n+2x would make the set of all odd start with 1 and that the set of all even would start with 2, despite the fact that n=0.
So please tell me, what are all the identities of zero?
Also, are there any other ways to prove that zero is even?
(Flame me all you like… I am in no way good with mathematics)
To be honest, I’ve never heard the phrase “zero has no identity”. Perhaps you’re confusing it with “zero has no (multiplicative) inverse”? The multiplicative inverse, of course, is the number b by which we multiply another number a to get the multiplicative identity (i.e., 1). Of course, we denote b by a[sup]-1[/sup]. Clearly, since 0 times anything is still 0 (i.e., never 1), zero has no multiplicative inverse.
As for even/odd – the even integers are defined as those that can be written as n=2k (where k is an integer). The odd ones are those that have the form n=2k+1. Since 0=2*0, it is even.
I’ve never heard that phrase either. Perhaps you are thinking of the multiplicative and additive identities:
Zero is the additive identity because x + 0 = x.
One is the multiplicative identity because x × 1 = x.
I have a vague recollection that zero is defined as even (to preserve odd/even continuity when crossing from negative to positive integers), but I am willing to be corrected by people more current in math.
I think I made that up about the identities… but perhaps properties sounds better… sorry for that…
Yea I figured… but I just feel like the lowest number of a sequence of multiples (I just made that term up I think… but like 5,10,15,20… multiples of 5 obviously) should still be a multiple of the sequence. By that definition, I would figure that you shouldn’t be able to say that the sequence of even numbers, the same as a sequence of multiples, should begin with 0. I also think that you should be able to divide by the first term of the sequence, and thus you would get undefined…
Then again a)I invented new laws of sequences and b)that is the same as saying that the sequence of odd doesn’t begin with 1.
No such luck eh?
Mathematics is a universal langauge but unfortunatly I speak English.
It seems like what was being discussed was whether a polynomial function is odd or even. The definition is, if f(x)=f(-x), then f(x) is even, whereas if f(x)=-f(-x), then f(x) is odd. Some functions are neither even nor odd. For polynomials, f(x)=x[sup]n[/sup], f(x) is even when n is even and odd when n is odd.
As has been discussed in other posts, the number 0 is even. A polynomial where n=0 is f(x)=x[sup]0[/sup]=0. So a zero-order polynomial (the zero function) is an even function. Another way to look at it is to consider that for f(x)=0, f(x)=f(-x), so f(x)=0 is even.
True, but f(x) = 0 is also odd. It’s the only function that’s both even and odd. I think what you meant to say is f(x) = x[sup]0[/sup] = 1, which is indeed an even function.
The problem with this is that by talking about the “lowest” term in the sequence, you’ve pretty much already left out 0. After all, 0=50, so 0 is a multiple of 5. Similarly, -5=5-1, -10=5*-2, -15=5*-3, and so on. So, what is the “lowest” term, then? Even if you ignore the negative numbers, 0 is still a multiple of 5, and so would be included in the sequence. Of course, you could just allow positive numbers, but then your analogy to the evens/odds would only apply to the positive numbers in those sequences, which ignores 0 altogether, and does nothing to resolve the issue.
Good point. My brain’s not with it. a multiple of 2n+1 can indeed be odd or even, depending on whether the mulitplicative factor is odd or even. Surely this argument then implicitly assumes that zero is even, so that 0*(2n+1) is even?
Yeah, I am embarrased. However, I feel that the argument that 2*0=0 => 0 is even is a bit weak; there are better arguments certainly, but I do accept that 0 is even.
We did this a bit in IM, so sorry to bring it up here. But I have to ask. I maybe agree there’s something unsatisfactory about “2*0=0 -> 0 even”. But what is a better argument?
BTW Let me check.
Do we mean “Better argument, starting from the definition that n is even if n=2i for some integer i.”? I assume not as then the 2*0=0 is pretty much concisest and decisivest.
Do we mean “Best argument for making the definition so 0 is even.”? There’s a variety of ways of describing 2, 4, 6, 8… that can be extended. “every other one counting from 2”, etc. But I think “multiples of 2” is as good if not better than any other.
The problem with arguments like “every other one counting [up] from 2” is that they ignore the negatives altogether. That is, with an agrument like this, would I say -4 is even or odd? Well, since we’re starting at 2, and going up, we’ll never get to -4. So, it isn’t even, by that rule. Even worse, if you define the odds similarly (i.e., every other one counting up from 1) you’d get that -4 isn’t odd, either. So, is it neither, or what? Well, perhaps changing the starting point for counting will help. But where do you set it? There is no “lowest” integer, so there doesn’t really seem to be any good starting point for a counting method.
On the other hand, using a definition like n is even if n=2k, for some integer k completely eliminates this problem. We get the even positives, and even negatives, and it of course gives 0 as being even (the whole point of the discussion. ). It’s a lot less messy on the whole.
Also, if you think about it, since we’ve defined even numbers to be n=2k, (and odd ones to be n=(2k)+1), there’s no real ambiguity about whether or not 0 (or any other integer, for that matter), is even or odd, since 0 most definitely equals 2*0.
My point was that if you know which natural numbers are even, and are trying to decide which integers should be even, there’s a few ways to do it. Normally, take the definition, and see if it can be extended. So “every other number from 2 up” becomes “every other number from 2 up or down” and “all 2n where n is a natural number” becomes “all 2n where n is an integer.” Then check all the rules “odd plus odd = even” etc work how you’d like.
(Of course, I do agree that the actual definition is “2n, where n is an integer” but I’m discussing why that is.)
