division by zero: is it really undefined?!

Grizwald · 2003-05-10T09:12:18

what about the square-root of -1? we gave that an ''i''? This has had me for awhile? What makes one even slightly more rational than another?

Math and Physics Programming

Started by Grizwald April 22, 2003 09:50 PM

74 comments, last by Grizwald 20 years, 11 months ago

EvilTachikoma

122

April 24, 2003 04:05 AM

quote:Original post by smitty1276
If you want to try to prove that it exists, the first step would be to show me an example... ANY example... of a scenario in which the division of a number by zero would ever be useful in anyway.

Since the number doesn''t exist, you will not find one.

Oh really, how about the relativistic relationship between mass and velocity ?

mr = m0 /sqrt(1 - v^2/c^2);

if v^2 = c^2, then we have a situation where m0 is devided by zero, IOW, it equals infinity, thus it would be impossible to accelerate to the speed of light. But you are right, there is no number that can describe infinity, it''s a concept. Nonetheless Division by zero can be quite usefull once in a while.

higherspeed

230

April 24, 2003 06:40 AM

Division by 0 is never useful, although the case surrounding it can.

eg in a system with 3 variables and 3 equations connecting them. Stick them in a 3x3 matrix. Now to solve the equations find the inverse matrix. This involves division by the determinant, if this is zero then that''s impossible, so you can say there is no inverse matrix and therefore no one solution, either no solution or a line of solutions.

There was no division by 0 involved in this.

Maybe some physicists find it useful for proving their model matches the real world, but in maths it is not much use.

Please prove me wrong and show me how dividing by zero can be of use.

owl

376

April 24, 2003 06:55 AM

quote:Original post by CoffeeMug
1 / 0 = x
0x = 1
0 = 1

:O, maybe our maths are too limited? Could there be somewhere/sometime in the universe when 0 = 1? Why not?

[size="2"]I like the Walrus best.

Cedric

158

April 24, 2003 12:09 PM

quote:Original post laeuchli
1/0 is undefined, however the limit of 1/x as it approches 0 from either side is defined

quote:My post
If it's defined, what is it??? lim (x->0) |1/x| is defined to be infinity, but 1/x doesn't have a limit at x=0.

quote:Original post by Nervo
Ever heard of limits AT infinity? It is terminology used.
That's not the point; nobody talked about limits AT infinity.

But now that I've reread laeuchli's original post, my comment was not really warranted in the first place; we could say that the limit of 1/x as it approaches 0 from either side is defined, but the limit as it approaches 0 (implicitly: from both sides) is not defined.

Cédric

EDIT: Didn't they allow nested quotes?

[edited by - cedricl on April 24, 2003 1:12:02 PM]

sjelkjd

171

April 24, 2003 01:47 PM

quote:Original post by Anonymous Poster
My background in mathematics is very solidly in abstract algebra, so I''m used defining numbers in sets like fields and rings among others and in either a ring or field of those defining 1 == 0 is not only invalid, but utterly rediculous.

Yeah, that''s about what I figured =P

tententrample

122

April 24, 2003 02:53 PM

I think the reasion the square root of -1 is i is more rational than defining division by zero is that the square root of negative one is always i. Making up i is like, "What number when multiplied by itself equals -1?" Well i multiplied by itself equals -1. But then division by zero is like, "How many zeros are in this number?" It has no certain answer.

Enselic

829

April 24, 2003 03:21 PM

Division by zero is undefined since it can equal many things depending on the situation:

Consider:

  f(x) = 4x²

then:

  f'(x) = lim(h->0) (f(x + h) - f(x)) / h  f'(x) = lim(h->0) (4(x + h)² - 4x² ) / h

If we set h to 0 we'll see what happends:

  f'(x) = (4(x+0)² - 4x² ) / 0

Intressting! f'(x) seems to be the quotient with a denominator of zero . Lets just replace 0 with h and simplify the expression a bit:

  f'(x) = (4(x² + 2xh + h² ) - 4x² )/h  f'(x) = (4x² + 8xh + 4h² - 4x² )/h  f'(x) = (8xh + 4h² )/h  f'(x) = 8x + 4h

and when h = 0:

  f'(x) = 8x

We have proved that in this case, a division by zero lead to the result 8x. If we change f(x) we will get another result, hence a division by zero is undefined.

EDIT: Got rid of the smileys :-)

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[edited by - Enselic on April 24, 2003 4:24:50 PM]

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alvaro

21,604

April 24, 2003 04:35 PM

It''s amazing how much confusion infinity generates. Let''s try to make a few things clear.

x/y is usually defined to mean x * inverse(y), where inverse(y) is an element such that y*inverse(y) = 1. Since in any non-trivial ring 0 doesn''t have an inverse, 1/0 doesn''t make sense.

There is one context in which 1/0 does make sense, and it will be defined to be infinity (without any signs). The context is the slope of a straight line on the plane. To be precise, the slope is not a real number, but a point in the projective line, which is pretty much either a real number or infinity.

If we are talking about dividing complex numbers, there is also a concept of a projective complex line (which is topologically a sphere), and this is again either a complex number or infinity, and in that context, it makes sense to say that 1/0 = infinity.

higherspeed

230

April 24, 2003 05:46 PM

Enselic: that argument is flawed, because h can''t be 0, as there is no triangle with one side of width 0. You are working out what f''(x) tend to as h tends to 0, by a flawed method, admittedly a totally effective method, which works. But it can not be used in the argument of the validity of 1/0.

Your approximation for f''(x) is valid for all h except 0, which is why we work it out as it tends to 0. From your argument you clearly understand that, just be careful.

Enselic

829

April 25, 2003 03:27 AM

quote:Original post by higherspeed
Enselic: that argument is flawed, because h can't be 0, as there is no triangle with one side of width 0. You are working out what f'(x) tend to as h tends to 0, by a flawed method, admittedly a totally effective method, which works. But it can not be used in the argument of the validity of 1/0.

Your approximation for f'(x) is valid for all h except 0, which is why we work it out as it tends to 0. From your argument you clearly understand that, just be careful.

I know this method isn't a real proof, it was just my way of complicate this discussion :-)

Anyways, what triangle are you talking about? The defination of the derivate is delta y divided by delta x ,the slope, between two points on the f(x) graph with the delta x as h. And h can indeed be zero. Since we are able to remove the h from the denomiator, we can set h to 0 and hence get the best result. if f(x) = 4x² then f'(x) is exactly 8x.

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[edited by - Enselic on April 25, 2003 4:28:32 AM]

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