Divison by zero


by Jeff Ford
Tags: divison
Jeff Ford
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#1
Oct18-05, 11:17 AM
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Forgive me if I'm being ignorant, but this recently occured to me. We all know division by zero is undefinfed, but [itex] \sqrt {-1} [/itex] used to be undefined too, until [itex] i [/itex] was created.
Has anyone ever proposed an imaginary number to indicate the result after division by zero?
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arildno
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#2
Oct18-05, 11:25 AM
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Well, you might replace the axiom in the real number system that says [itex]1\neq{0}[/itex] with the axiom [itex]1=0[/itex] instead.
It won't be a terribly interesting number system, but the number [tex]1/0[/tex] would be defined there.
El Hombre Invisible
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#3
Oct18-05, 11:27 AM
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There's a huge difference. A number divided by zero is 'undefined' since it can take any possible value. For instance, if a = bc, and a is zero and c is zero, what is b? b can be anything at all!

Root -1 is not 'undefined' in this sense (in fact, I've never heard of anyone past or present refering to it as undefined) - it is just not real, that is: any real number squared is always greater than or equal to zero. No-one ever suggested, to my knowledge, that root -1 is equal to root -2, or any other number, so the two really aren't comparable.

TD
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#4
Oct18-05, 11:34 AM
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Divison by zero


When ab = c, we can find b if we divide c by a, i.e. c/a. This inverse operation of multiplication, using the multiplicative inverse of a number x being 1/x, gives a unique number b here, at least that's how we want it to be.
By allowing division by 0, we lose this uniqueness since for every x, 0*x = 0.
Jeff Ford
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#5
Oct18-05, 11:55 AM
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I figured I was wrong. Just wanted to know why.

Thanks
Hurkyl
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#6
Oct18-05, 05:08 PM
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Actually, [itex]\sqrt{-1}[/itex] is undefined... when working over the real numbers!

Compiling what the others have said, if you take the field axioms, that is the list of properties like a+(b+c)=(a+b)+c that a "good" number system should have, together with the axiom [itex]1 \neq 0[/itex], you can prove that one cannot extend the definition of division to allow division by zero.

Similarly, if you take the [u]ordered field axioms[/i], that is the field axioms together with additional properties defining "<", then you can prove that one cannot extend the definition of square root to allow square roots of negative numbers.

The moral is that if you want these extra things, such as division by zero to be allowed, or negative numbers to have square roots, you're going to have to give up one of the "good" properties.

For example, to get square roots of negative numbers, one gives up the ordering: "<" does not make sense for complex numbers.

One can give up things to get division by zero. The most obvious is the assumption that 1 is nonzero. There are other things you can do: for example, I've seen a number system where you can compute x/0 for any nonzero x, but sometimes x+y is undefined.
masudr
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#7
Oct18-05, 07:41 PM
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If we have the set of all matrices
[tex]\left(
\begin{array}{clrr}
x&-y\\
y&x
\end{array}
\right)[/tex]
where [itex]x, y \in \mathbb{R}[/itex] then under addition and multiplication of matrices we have a field.

We then can associate any matrix in this form with a special number of the form [itex]x+iy[/itex] where [itex]i[/itex] has the special property such that [itex]i^2=-1[/itex]. Of course, these are complex numbers.

As long as our field axioms don't contradict each other, we can be safe in the knowledge that we will not run into trouble with complex numbers.
Robokapp
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#8
Oct20-05, 07:41 PM
P: 218
i was created for electronics purposes, and for the e^(i*Pi)+1=0

1/0 would not have a value, but maybe a constant. Graph 1/x and look at it at x=0. you'll see that y= both positive and negative infinity. in other words at x=0, the absolute value of y= infinity.

well, it's not that easy to swallow a concept like that...but who knows.

Natural numbers 5
Whole numbers -5
Rational numbers -5/3
Real numbers i
Complex numbers i-1
-------------------
"Jeff" numbers 2/0+(7)^(1/2)*i-5/3


????? Would that do?
Divisionbyzer0
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#9
Oct20-05, 10:19 PM
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While I would agree that electronics and electrical engineering, in general, greatly benefits from the conventions of complex numbers I'm not so sure that they were created for "electronics purposes"? I think rather that complex numbers were at first seen as a way to naturally expand the solutions of polynomial equations.
hypermorphism
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#10
Oct20-05, 11:25 PM
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The complex numbers were first seriously considered when they arose in the solution of cubic polynomials with real roots.
matt grime
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#11
Oct21-05, 08:01 AM
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Quote Quote by Robokapp
i was created for electronics purposes, and for the e^(i*Pi)+1=0
Only if Cardano et al knew about electronics 3 centuries before the electron was discovered.
Tzar
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#12
Oct21-05, 09:13 AM
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Quote Quote by matt grime
Only if Cardano et al knew about electronics 3 centuries before the electron was discovered.
I found out today that Cardano didn't actually come up with the solution for the cubic. Infact it was shown to him by another mathematician (whos name I forgot) and he just published it. Did anyone also hear this?

Also, it was mentioned earlier about the axiom that 0 doesn't equal 1. I've done a bit of field theory and set theory, and never encountered such an axiom. Where does it appear? What is its precise formulation?
matt grime
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#13
Oct21-05, 09:27 AM
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The set of nonzero elements of a field form a multiplicative group and that set is non-empty. That is one way to formulate the fact that 0=/=1.

of course there is another intrinsic reason why this is true:

if 0 were also the mutl identity then a.0=a for any a and one can also prove that 0.a=0 for any a (without appeal to what 1 might be) and thus in a putative field where 0=1 it follows that every number is zero. Thus there is a degenerate case that implies {0,+,.} is a field, though with the troubling notion that its multiplicative subgroup is no longer the non-zero elements but all elements including zero, which is not how one usually phrases it.
hypermorphism
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#14
Oct21-05, 09:33 AM
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Quote Quote by Tzar
I found out today that Cardano didn't actually come up with the solution for the cubic. Infact it was shown to him by another mathematician (whos name I forgot) and he just published it. Did anyone also hear this?
This is referred to in "An Imaginary Tale" by Nahin and "Mathematics and its History" by Stillwell. Cardano attributed the credit to del Ferro (who passed it to Tartaglia) in his papers, even though he had to rederive the solution himself.
Quote Quote by Tzar
Also, it was mentioned earlier about the axiom that 0 doesn't equal 1. I've done a bit of field theory and set theory, and never encountered such an axiom. Where does it appear? What is its precise formulation?
One of the field axioms is that [itex]a\in F[/itex] implies aa-1 = a-1a = 1 for all nonzero a. If you remove the qualifier "nonzero", you can rewrite the axiom a + 0 = a as 0-1*a + 1 = 0-1*a which leads to 1=0.
Robokapp
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#15
Oct21-05, 01:58 PM
P: 218
I know for a fact that it has something to do with electronic aplications, especially with electronics like cell phones.

I also know it was "figured out" although it's poorly siad by an indian in the 16th century or around that time anyway. someone called Omar something? i don't know. Anyway...substituting (-1)^(1/2) by i is no real accomplishment in my eyes...

i can substitute 0^0 by z and have a constant that is undefined...and i'd be cool like that, but until someone uses my 0^0 for something...it's pointless.
hypermorphism
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#16
Oct21-05, 02:46 PM
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The constant 'i' was integrated for purely mathematical purposes, without reference to physics. Physics may use any part of established mathematics at any time to build quantitative models, but the inclusion of said mathematics thereof isn't a justification for the mathematics itself. The mathematics must be self-consistent and yield general meaningful theorems/tie into existing mathematical structures to be useful.
Your renaming of 0^0 is not analogous to the inclusion of the square root of -1 into a system of numbers that enhanced our understanding of real numbers/geometry. It had nothing to do with the name of the element, whether it was called i or j. Renaming an undefined object doesn't serve a mathematical purpose; it is only done as a shorthand reference.
Quote Quote by Robokapp
I also know it was "figured out" although it's poorly siad by an indian in the 16th century or around that time anyway.
i was algebraically used to solve equations in Bombelli's 1572 Algebra, but I am unaware of any other recorded formal use of i before this, or independent discovery at this time.
Robokapp
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#17
Oct21-05, 03:02 PM
P: 218
1572 is 16th century...right? yea. 1950 were in 20th century.
hypermorphism
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#18
Oct21-05, 03:31 PM
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Quote Quote by Robokapp
1572 is 16th century...right? yea. 1950 were in 20th century.
Bombelli is an Italian.


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