I've always pondered this question when it came to Fourier series and I think I've just proved it.

Let any even number be equal to 2k for all k in Z, then if k=0, 2k=0, an even number. Just what was to be shown.

 

That is an interesting question, I had never considered it before. I'll sit back and watch the other folk discuss it.

 

I agree, an even number is one that does not result in a fractional part when divided by 2.

 

I've scanned the web high and low and it always comes down to this divide by 2 business. Zero divided by anything is zero. But if I contradict my first post then,

Let any odd number be equal to 2k+1 for all k in Z, then if k=0, 2k+1=1, an odd number. Just not what was to be shown.

 

Another way to look at it is to ask if zero is odd? If not, then zero must be even.

An integer n is odd, if n+1 is even. 0+1=1 which is odd. Hence, zero must be even.

Look at http://mathforum.org/library/drmath/view/57104.html

 

Another way to look at it is to ask if zero is odd? If not, then zero must be even.

An integer n is odd, if n+1 is even. 0+1=1 which is odd. Hence, zero must be even.

Look at http://mathforum.org/library/drmath/view/57104.html

That's what I just did, I contradicted it. Look, Let any odd number be equal to 2k+1 for all k in Z, then if k=0, 2k+1=1, an odd number. Just not what was to be shown.

 

That's what I just did, I contradicted it. Look, Let any odd number be equal to 2k+1 for all k in Z, then if k=0, 2k+1=1, an odd number. Just not what was to be shown.

But that doesn't say anything about whether k is even or odd. In fact, for all integers 2k+1 is an odd number.
Reread what I was saying.

 

Say what? I'm choosing the integer, LET K BE EQUAL TO ZERO.

 

Say what? I'm choosing the integer, LET K BE EQUAL TO ZERO.

And the result is odd. So what? Your question is whether zero is even. That test will always be odd. Let me ask another question. Is 3 even? 2*3+1=7, which is odd. What does that mean about 3? Is 2 even? 2*2+1=5, which is odd. What does that say about 2? In both cases, it doesn't tell us anything about the original number.

The test I referred to is different. Is 3 even? 3+1=4, which is even. So, 3 is odd. Is 2 even? 2+1=3, which is odd. So, 2 is even.

 

I've re-read your post and I'll try and explain. I'm using the set of integers Z (positive whole numbers, negative whole numbers and zero), you can't make 2k+1 = 0 for any integer, only if it's rational. So I choose the integer k=0 because it's in the set I'm using and then in your post you put "0+1=1 which is odd. Hence, zero must be even. " But that's what I just did by letting k=0 i.e. 2k +1 = 1 if k =0. That's why I contradicted what I was saying and now you're now telling me it's wrong when you've posted what I've posted.

 

If you think of integers in binary representation, the least significant bit (LSB) = 1 indicates that the number is odd.
Hence zero is an even number.

 

If you think of integers in binary representation, the least significant bit (LSB) = 1 indicates that the number is odd.
Hence zero is an even number.

Wouldn't this be a corollary to the divide by two test?

 

Folks often confuse whether zero is even or odd (it's even) with whether its positive or negative (it's neither).

 

I am not going argue with Wikipedia:

http://en.wikipedia.org/wiki/Parity_of_zero

 

Folks often confuse whether zero is...positive or negative (it's neither).

There's utility in allowing zero to have a sign:

http://people.freebsd.org/~das/kahan86branch.pdf

 

As one would expect, zero is a special case that breaks all the rules.

However, in computer programming logic, zero is even and positive.

 

As one would expect, zero is a special case that breaks all the rules.

However, in computer programming logic, zero is even and positive.

Its not even, not odd, not positive, not negative.

It is zero, not a number.

Any number multiplied with zero again is zero.
A division by zero isnt possible.
When you add or subtract zero, nothing changes.
The real world representation of zero is nothing.

As for geometry, it is the origin by some definition.

 

TheElectrician
There's utility in allowing zero to have a sign:

http://people.freebsd.org/~das/kahan86branch.pdf

There are contradictions that can be stated if you allow one or the other, the simplest being that if you allow it to be say positive there are then more positive numbers than negative ones.

Some would have it as both, but you then require rules as to which to choose and mathematicians do not like ambiguous cases.

In any case it is not necessary since it can always be multiplied by plus or minus 1, just as we form all imaginary numbers by multiplying real ones by i.

As a matter of interest your link contains some 'forbidden' mathematical operations.

Takao
It is zero, not a number.

It is indeed a valid number that solves the equation
a+x = a

 

if zero was not a nuumber, then how would we do the place setting for decimals? 10, 100, and such? which are all even numbers by the way.

 

If you add 1 to (or subtract 1 from) an odd number you get an even number.

If you add 2 to (or subtract 2 from) an even number you get an even number.

By my arithmetic, zero is an even number, unless somebody changed arithmetic on me.