Is zero even or odd?

[Virgil]

Yes!

 

[George Cox]

That's not true. A prime can be divided by anything, but an integer
greater than one is prime if its only positive divisors are itself and
one, but zero isn't prime because it's even.

And two isn't prime because it's even?

 

[Franz Heymann]

I read in sci.electronics.design that Nicholas O. Lindan

wrote (in <[email protected]>) about
'Is zero even or odd?', on Mon, 20 Dec 2004:


One possible solution, given the enormous lack of rigour in

'infinity'.

There is no lack of rigour in the definition of infinity. Read anbout
the work of Cantor, Dedekind and others.

[snip]

Franz

 

[Franz Heymann]

[snip]

(-0)^2 = -0

Not on my Casio calculator.

Franz

 

[David Kastrup]

0/0 can take ANY value.

Well, and in the above it takes on 1, so that would be quite legal,
right?

 

[Hue-Bond]

Shawn Corey, lun20041220@21:43:59(CET):

Zero is even. You cannot divide by zero. Limits are not division.
Infinity is not a number. Computers bugger up the system.

That could be said louder, but not clearer .

 

[David Kastrup]

[snip]

(-0)^2 = -0

Not on my Casio calculator.

What else is it there?

 

[Alfred Z. Newmane]

The divisor would have to be something smaller than 0 like -2.
Therefore zero is both even and negative.

I'm sorry, sir, but 0 is universally defined in math to be neither - nor
+, it's jsut 0. Many graphing and soem scientific calculators have a
sign function. On any calc I've tried this on, it gives -1 for any
negative number, it gives a 1 for ant positive value, and 0 (zero) for,
well, 0 (zero.) As is defined in basic math.

 

[Alfred Z. Newmane]

Sure it can: 0 / 0 = 0 * (1 / 0) = 0 * infinity = 1

It works if the only three numbers in the universe are
0, 1, and infinity -- A number system that seems very
suited to usenet.

Except for the fact that: 0 / 0 = undefined

Or actually more correct: n / 0 = undefined

 

[Alfred Z. Newmane]

I read in sci.electronics.design that David Kastrup <[email protected]>


0/0 can take ANY value.

n / 0 = undefined

 

[Shawn Corey]

1 + 1 = 0/0 + 0/0 = (0 + 0)/0 = 2 * 0/0 = 2

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a+b = b
but a = b
a+a = a
2a = a
2 = 1

What could be clearer?

--- Shawn

 

[Jim Thompson]

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a+b = b
but a = b
a+a = a
2a = a
2 = 1

What could be clearer?

--- Shawn

I remember that one from when I was in high school ;-)

...Jim Thompson

 

[Fred Bloggs]

That's a deduction, not a law. But this type of deduction fits right
in with summations and series where zero is considered even.

It's a number theoretic definition- we are not talking physics and there
are no laws to invoke here. Even/odd is partition of the set of integers
in accordance with modulo-2 equivalence classes: EVEN={m: m MOD 2=0} and
ODD=INTEGERS-EVEN. The set of even numbers is all integers of the form
2n, everything else is odd.

 

[Nicholas O. Lindan]

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a+b = b
but a = b
a+a = a
2a = a
2 = 1

Now if we can just get the IRS to agree.

 

[Fred Bloggs]

Except for the fact that: 0 / 0 = undefined

Or actually more correct: n / 0 = undefined

0/0={ SET OF ALL INTEGERS }

n/0= NULL SET for n<>0

It is very well-defined.

 

[Alfred Z. Newmane]

(-0)^2 = -0

No, it's just 9. You dont have +0 or -0, it's just 0 (zero.) Writing it
as +0 or -0 still is jsut 0 sicne 0 has no sign. You have +, -, and 0.
Or for what you get from any proper sign function, either 1, -1, or 0.

 

[Alfred Z. Newmane]

[snip]

(-0)^2 = -0

Not on my Casio calculator.

Thats becuase, when translated to reality, that statement becomes (0)^2
= 0, because 0 has no sign. I really wish people would stop trying to
spread the false hood that0 actually has a sign.

 

[Jim Ward]

I know 0 is neither negative or positive but what about odd/even? I think
it's even.

0 has no flat edges, so it can't be even.

 

[Richard Tobin]

You dont have +0 or -0, it's just 0 (zero.)

No, you *do* have +0 and -0, and they are both equal to 0.

-- Richard

 

[Ed Murphy]

a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b(a-b)
a+b = b
but a = b
a+a = a
2a = a
2 = 1

What could be clearer?

I used to know a proof along these lines:

-----

x = +1
A = [integral] f(x) dx
x = -1

By examining a graph, it is obvious that A > 0.

Let y = 1/x, therefore x = 1/y, and substitute this into the equation:

x = +1
A = [integral] f(1/y) f'(y) dy
x = -1

The right-hand side is clearly equivalent to -A

A = -A

A = 0