can 0 be assigned pos or neg (1 Viewer)

[..:''ooo]

0 divided by any number is zero

but, specifically wot about 0 divided by a negative number? , is it -0 ?

 

[turtle_2468]

-0 is 0... (btw, 0/0 is undefined)
proof:
0 is the additive identity ie 0 + x = x + 0 = x.
But the definition of minus in the real numbers is that (-x)+x=0
So -0=(-0)+0=0

 

[Slidey]

Hmm. Thinking in terms of vectors, wouldn't the real numbers be vectors of sorts? a=1 would be the additive inverse of the vector b=-1?

So if the real numbers are vectors, then zero is the zero vector and has undefined direction, right?

 

[turtle_2468]

I guess that way of it makes sense too

 

[Trebla]

the art of zero

Zero divided zero. So if you had something like a/a, which you simplify to 1 by simple algebra, what about a=0? So does that mean 1 is one possible answer and zero divided zero simply equals every real number or infinity?
There is a difference in having no solutions and having all real numbers as solutions.
Consider the following:
0^0 = ?
1 ÷ (0/1)^-1 = ?
0^-1 /1 = ?
If 0 can be positive and/or negative. Consider all positive and negative cases when graphing y= |0|/|0|


You may have heard of this as well:
Let x = y
[multiply both sides by x]
x² = xy
[Subtract y² from both sides]
x² - y² = xy - y²
[Factorise]
(x - y)(x + y) = y(x - y)
[(x-y) cancel out so divide by (x - y)]
.: x + y = y
[Sub x = y into LHS]
2y = y
[Divide by y in both sides]
.: 2 = 1
!!!Explain!!!
I think I already know how, but I'd like to see how you mathematicians explain it

 

[Slidey]

Zero divided zero. So if you had something like a/a, which you simplify to 1 by simple algebra, what about a=0? So does that mean 1 is one possible answer and zero divided zero simply equals every real number or infinity?
There is a difference in having no solutions and having all real numbers as solutions.
Consider the following:
0^0 = ?
1 ÷ (0/1)^-1 = ?
0^-1 /1 = ?
If 0 can be positive and/or negative. Consider all positive and negative cases when graphing y= |0|/|0|


You may have heard of this as well:
Let x = y
[multiply both sides by x]
x² = xy
[Subtract y² from both sides]
x² - y² = xy - y²
[Factorise]
(x - y)(x + y) = y(x - y)
[(x-y) cancel out so divide by (x - y)]
.: x + y = y
[Sub x = y into LHS]
2y = y
[Divide by y in both sides]
.: 2 = 1
!!!Explain!!!
I think I already know how, but I'd like to see how you mathematicians explain it

a/a is not 1. But a/a = 1, excluding where a=0. This is perhaps pedantic. But graphing y=1, the line is continuous, but graphing y=a/a, the line is DIScontinuous at a=0.

a/0 is commonly said to be 'infinity' (imagine the smallest number you can think of on, say, 1: you get a big number. The smaller you go, the close to zero you get, the bigger the number is). Personally, I'll just stick with undefined.

(x - y)(x + y) = y(x - y)
[(x-y) cancel out so divide by (x - y)]

Frist step there is fine. Take to one side:
(x-y)(x+y)-y(x-y)=0
(x-y)(x+y-y)=0
x(x-y)=0. Therefore you can't divide by (x-y), since (x-y) is zero.

 

[turtle_2468]

The "simple algebra" that shows that a/a = 1 is actually not that simple... it relies on the fact that R (the real numbers) is a group under multiplication ie that every member has an inverse. Because what you mean when you say a/a is actually a*a^(-1)... now R is actually not a group with respect to multiplication, but rather the set formed by R with 0 taken away. Hence you can't divide by zero.. because zero has no multiplicative inverse (there is no number such that a*0=1). Actually, there is... but only in the zero ring, and that has one element, where 0=1...

 

[Li0n]

x/y has nothing to do with infinity, its just that it approaches an extremely large number as y --> 0 which is pretty obvious. nothing is equal to infinity.
Undefined is the correct term as slide said.


y= 1, or x =1, aren't defined as functions

 

[Slidey]

What about when you had, say, a group which allowed zero divisors?

Such as if you had two matrices X and Y such that XY=O.
Could you not divide by XY, which would be multiplying by Y^-1X^-1?

Example:

Y=A
Y/O=A/O
YY^-1X^-1=AY^-1X^-1?
X=AY^-1X^-1?

Would manipulations like this be legal?

 

[Li0n]

ouch that hit me pretty hard :|

we are just starting on the set theory now in discrete maths
aparently, mathematicians cannot properly define what a 'set' is
now i can't explain what my lecturer said but i understand what she meant, and it scared me for abit..
infact i was in for a heart attack for a second.

 

[Slidey]

This is a set:

{shoe, flower, whale}

 

[Templar]

This is a set:

{shoe, flower, whale}

Let y={x|x is not a member of itself}

Is y a member of itself?

 

[Xayma]

ouch that hit me pretty hard :|

we are just starting on the set theory now in discrete maths
aparently, mathematicians cannot properly define what a 'set' is
now i can't explain what my lecturer said but i understand what she meant, and it scared me for abit..
infact i was in for a heart attack for a second.

Well that is obvious considering how many different sets of axioms there are. I think each of them have some problems which others dont etc.

But to answer the first question. 0 is not assigned to either positive or negative.

Ie the positive integers are 1, 2, 3, 4,....
negative intergers -1, -2, -3, -4,....

If you wanted 0, 1, 2, 3, 4, ....
you would say the non negative integers, similarly for 0, -1, -2, -3....

 

[Slidey]

Is that the null set? Doesn't it have no elements, yet every set is a member of itself, so...

 

[turtle_2468]

If XY=0, det(X)det(Y)=0. Hence WLOG det(X)=0 so X is not invertible..

 

[Li0n]

never mind slide rule

 

[Slidey]

If XY=0, det(X)det(Y)=0. Hence WLOG det(X)=0 so X is not invertible..

Thanks. I haven't done determinants, yet. What's WLOG, though?

 

[turtle_2468]

without loss of generality... basically a nice way to save on case bash. If you know that out of A,B,C,D,E that one of them say is equal to zero, and they are basically symmetric in the question, you can say WLOG A=0 (ie you don't have to then write the case B=0, C=0... which are the same anyway)