division by zero (1 Viewer)

[cwag]

can it be defined? could a new number system be created for it? any thoughts?

 

[Aplus]

No .

 

[cwag]

aww..my hopes were so high.....but seriously..havn't you ever thought about it...i am intriged.

 

[Aplus]

Unless you formulate your own personal numerical system which only you would know.

 

[Sarah182]

Put pretty simply no.

 

[kony]

by taking a limit, we can see that it approaches +- infinity.

 

[minijumbuk]

Don't rush your way to the Nobel Prize by luck, mate =]

 

[cwag]

Don't rush your way to the Nobel Prize by luck, mate =]

haha good call

 

[Slidey]

can it be defined? could a new number system be created for it? any thoughts?

Yep, certainly. Basically any algebra that forms a commutative ring (e.g. real numbers, integers, and some matrix subsets like the complex numbers, any integral domain) can be extended to include division by zero using wheel theory: http://en.wikipedia.org/wiki/Wheel_theory

Don't rush your way to the Nobel Prize by luck, mate =]

There is no Nobel Prize for maths. The Fields Medals and Abel Prize were invented because of this.

 

[flicka08]

no

 

[tommykins]

If division by 0 were to be defined, then the number system used would imply that all intergers a can be equal to any othe integer b.

a/0 = x
a = 0*x = 0

3a/0 = x
3a = 0*x = 0

.:. 3a = a

Something along those lines.

 

[Slidey]

no

Even though I just posted proof that the answer is yes?

Astounding.

 

[YannY]

Would you agree that any number times 0 is zero? But with a division of zero, i,e a/0 . 0 then the zero cancels out.

Another way to prove this is via polynomials. p(x)=x^2+a
P(0)=a
but p(x)=x (x+a/x)
this p(0)=0 . a/0
supposedly if a/0 is defined then 0 . a/0 = a??? this then defies the law where all numbers x 0 is zero.

 

[Slidey]

Would you agree that any number times 0 is zero? But with a division of zero, i,e a/0 . 0 then the zero cancels out.

Another way to prove this is via polynomials. p(x)=x^2+a
P(0)=a
but p(x)=x (x+a/x)
this p(0)=0 . a/0
supposedly if a/0 is defined then 0 . a/0 = a??? this then defies the law where all numbers x 0 is zero.

Actually, you don't need anything so complex. One of the axioms for the real numbers specifically states 0 has no multiplicative inverse.

Or are you trying to give some misguided proof against Wheel algebras?

The thread starter specifically asked if there exists any number system or modification that includes a definition of division by zero.

The answer is an unequivocal yes.

Like-wise there exists an alteration of the real number system that allows a definition of calculus without recourse to limits. The hyperreals. It extends the number system to include infinite and infinitesimal numbers.

 

[Foxodi]

It's pretty much infinite - I dont think you can do anything more then accept that it's 'untouchable' maths