Divide By Zero: Is It Ever Logical Not To?

[eNathan]

Can somebody please tell me any case where it is logical to NOT divide by zero? I know division by zero is illogical itself, but usually when you divide by zero the result should be zero anyway.

 

[cronxeh]

Try dividing 1 by 0.000000000000000000000000000001. Big number?

Now try 1/(10^-99999) - is that bigger than 0 or close to infinity?

 

[eNathan]

Try dividing 1 by 0.000000000000000000000000000001. Big number?

Now try 1/(10^-99999) - is that bigger than 0 or close to infinity?

Excellent explanation! Hmn, I actually though of this the other day. I though "hmn, when you divide by 0 shouldn't the result be infinite?"

I just though of an equation where dividing by zero violates something in 'nature' by common sense, but the result should still be infinite.

Let's say I travel 10 meters at a rate of 0 meters per second. The only way this is posible is it I have an infinite amount of time, right? Why isn't $$\frac {10 m} {0 mps} = infinite$$? (I know there is a latex symbol for infiinite I just don't know it). But thanks again for that 'infinite' explination. So my point is, what is wrong with an infinite number?

 

[HallsofIvy]

Can somebody please tell me any case where it is logical to NOT divide by zero? I know division by zero is illogical itself, but usually when you divide by zero the result should be zero anyway.

Where did you get the idea that "usually when you divide by zero the result should be zero anyway." I can't imagine any situation (except under some condition where you have 0/0 but you don't seem to be talking about that) where dividing by 0 could reasonably be interpreted at resulting in 0.
Some people say, loosely, that dividing by 0 results in infinity- but surely not 0!

 

[eNathan]

Where did you get the idea that "usually when you divide by zero the result should be zero anyway." I can't imagine any situation (except under some condition where you have 0/0 but you don't seem to be talking about that) where dividing by 0 could reasonably be interpreted at resulting in 0.
Some people say, loosely, that dividing by 0 results in infinity- but surely not 0!

hmn, well that is why I started this thread because I was not quite sure about it. Like I said, the other day I was thinking about it and I though that the result should either be infinite or 0.

 

[matt grime]

And here we go again.

At least you want to look at it logically.

So, let's do so. When we divide by x where x is any nonzero number, what we're doing is "logically" multiplying by 1/x, where 1/x is the symbol that satisfies x*(1/x)=(1/x)*x=1.

Now, it is easy to show (you should do so) that in any field, it is logical that 0*x=0 for all x.

Thus, if we were to wish for dividing by zero to be defined logically then we should have

0=0*(1/0)=1

so that'd be a problem.

If we wish to divide by zero we need to pass to a larger system that wouldnt't be a field.

 

[snoble]

Two reasons

First off you got a unique solution to your operation because the nature of the reality (how long does it take to get from here to there) precludes negative numbers. But in general who's to say that $$2/0 \ne -\infty$$ (infinity is \infty in tex by the way). So you have an ambiguity. But you also have an ambiguity with the square root and we work around that right? In fact if you construct numbers from a projective plane (borrow something by Coxeter from your local math library) you are allowed to divide by zero but in that case
infinity is the same as negative infinitity and addition is not defined on infinity. So this is not a situation you want to be dealing with.

Also if you allow x/0 other questions arrise. Like what is x/0 -y/0. What is 0/0. etc. Then you truly do get problems if you allow those to have answers as seen here http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html

(Copied directly from site)
Let a=b.
then $$a^2 =ab$$
$$a^2 +a^2 = ab+ a^2$$
$$2a^2 = ab+ a^2$$
$$2a^2 -2ab= ab+ a^2-2ab$$
and $$2a^2 -2ab= a^2 -ab$$
This can be written as $$2(a^2-ab) = 1(a^2-ab)$$
and cancelling $$a^2-ab$$ from both sides gives 1=2.

So in the end division by 0 is bad because it may lead to greater problems and there is almost no value in allowing division by 0

 

[eNathan]

Hmn I just found a time where zero messes up an equation. We can establish that
if $$z = xy$$ then $$x = \frac {y} {z}$$ and $$y = \frac {x} {z}$$
But this is NOT true when x or y = 0.

 

[gregmead]

thats where you've got an asymptote in the graph.

eg, the graph y = 1/x is undefined when x=0

 

[eNathan]

Oh ye, if $$\frac {x} {0} = \infty$$ then $$x \ne \frac {x} {\frac{1}{0}}$$ But that depends on how you do the math.

 

[gregmead]

Oh ye, if $$\frac {x} {0} = \infity$$ then $$x != \frac {x} {\frac{1}{0}}$$

Unfortunately you can't just state things like if x/0= becuase that makes no sence. :-)

 

[gregmead]

your statement x/0 = Infinity is not technically correct, sorry I'm not good at latex so you'll have to make do with shoddey normal writing.

Its better to say that x/0 is underfined or things start getting a bit messed up, becasue otherwise it leads onto saying things like

0/0 = 1
and
Inf/Inf = 1

which is not allways right.

 

[eNathan]

your statement x/0 = Infinity is not technically correct, sorry I'm not good at latex so you'll have to make do with shoddey normal writing.

Its better to say that x/0 is underfined or things start getting a bit messed up

Well I was not stating it as fact. But I do understand this concept of $$\frac {x} {0}$$ being underfined.

thanks a lot! :tongue2:

 

[gregmead]

I edited my last post ;-) a bit

 

[eNathan]

One more question. $$111\infty = 999\infty$$

Is this correct?

 

[gregmead]

hmmm, I'd say no, becuase it implies that

$$\frac {111} {999}$$ = $$\frac {\infty} {\infty}$$

I'm learning latex ;-)

$$\frac {\infty} {\infty}$$ can equal all sorts of things, but you really need an equation defined before you can calculate it.

Edit: Like moo said below

 

[Moo Of Doom]

You don't really want to do arithmetic operations with $$\infty$$, but yeah generally it is true that $$a*\infty = b*\infty$$ so long as a and b aren't zero.

 

[cronxeh]

I think we all agreed that division by zero is immoral..

 

[gregmead]

doing algebra with infinities generally turns pretty ugly.

you can say that $$\frac {1} {\infty} = 0$$ but then things get messed up when you start rearranging that formula to stuff like:

$$\infty = \frac {1} {0}$$ then start to try doing divisions with this defanition if infinity like: $$\frac {\infty} {\infty} = \frac {\frac {1} {0}} {\frac {1} {0}}$$

becasue you end up with crazy stuff like $$\frac {\infty} {\infty} = \frac {0} {0}$$ which implies that this is the only answer for $$\frac {\infty} {\infty}$$ which is not technically correct, and also seems to show if you take $$\infty$$ to be some sort of algebraic argument then it gives $$\infty = 0$$ which is clearly wrong.

Think of infinities in this case more as some useful device for working out limits of equations rather than some kind of 'number' you can use in algeba.

 

[gregmead]

You don't really want to do arithmetic operations with $$\infty$$, but yeah generally it is true that $$a*\infty = b*\infty$$ so long as a and b aren't zero.

If a and b = 0 then you end up with 0=0 which is correct. I would argue that this is the only time that $$a*\infty = b*\infty$$ is correct, becuase here your suggesting that the two infinities are actual numbers, and are somehow different from each other.

 

[Icebreaker]

On a similar note, do you consider $$lim f(x) = \infty$$ (when it is the case) a misleading notation?

 

[gregmead]

I don't see anything wrong with that notation, is this a trick question ? ;-)

 

[Moo Of Doom]

You can never assume $$0*\infty=0$$. That's indeterminate form.

What I was getting at was more set theory cardinality, like $$\aleph_0=2*\aleph_0$$.

You see how infinity should not be so manipulated? :P

 

[HallsofIvy]

On a similar note, do you consider $$lim f(x) = \infty$$ (when it is the case) a misleading notation?

I have no problem with the notation,as long as you remember that it means "this has no limit" (but in a partcular way).

 

[Icebreaker]

I don't see anything wrong with that notation, is this a trick question ? ;-)

Sort of. Although you read the equation as "the limit of f equals infinity", its meaning, as pointed out of HallsOfIvy, is different. Which is kind of misleading.

 

[TsunamiJoe]

well...i think to solve this you ought not try to use such math above the 4th grade level of schooling...take it by its basic terms

$$5/0=x$$ which states 5 objects put into 0 groups is equal to a number, which is obviously false, because say x=1, 5 objects put into 0 groups is 1 group...which is obviously false...

 

[Alkatran]

Remember that infinity is just a point at the end of the reals (the extended reals, I suppose). This means that the way things behave at infinity is just an inference of how they should behave. IE: as x becomes arbitrarily large, 1/x becomes arbitrarily close to 0. So if x is infinity, 1/x is 0.

This is better stated by the equation:

$$\lim_{x \rightarrow \infty} \frac{1}{x} = 0$$

Now note that this limit is approaching from the left (since you can't approach infinity from the right) so x/0 doesn't equal infinity BUT:

$$\lim_{x \rightarrow 0+} \frac{1}{x} = \infty$$


The importance is remembering that infinity is not a number, just a behavior as numbers become really large (in so far as limits are concerned, in any case!).

 

[Hurkyl]

Even in the extended reals, 1/0 is undefined. The limit as you approach 0 from the positive side is different than if you approach from the negative side, so division cannot be extended continuously to (1,0).

 

[Alkatran]

Even in the extended reals, 1/0 is undefined. The limit as you approach 0 from the positive side is different than if you approach from the negative side, so division cannot be extended continuously to (1,0).

That was my point.

I was just saying that when you write infinity, you're really using a limit, and that this limit is approaching from the left (in the case of positive infinity)

That's why 1/infinity = 0 but 1/0 = undefined != infinity, because 1/0 isn't a one sided limit.

 

[eNathan]

I just found a case where x/0 'should' result in 0. Ok you know the graph equation $$Y = mx + b$$ where m is the slope represented by $$\frac {y_1} {x_1} - \frac {y_2} {x_2}$$ So we have established that I assume? Now look at this

Using the second equation, when $$x = 0$$ you have to divide by zero. And the result should be zero to make the graph look right. Correct me if I am wrong.

 

[gregmead]

then the line would have no slope because change in y would also be zero

Edit: read next post

 

[eNathan]

then the line would have no slope because change in y would also be zero

That is exatly my point. $$\frac {y_1} {x_1}$$ Or $$\frac {y_2} {x_2}$$ have to result in 0, and then after you add the interception point of y you can get your straight line.

 

[gregmead]

Edit: sorry misunderstood

if dx=0 then the slope WILL be infinate, eg undefined, not equal to 0. It will be vertical, which when you think about what slope means and what the graph is telling you, makes sence.

 

[Moo Of Doom]

Since when is slope $$\frac{y_1}{x_1}-\frac{y_2}{x_2}$$? Isn't it $$\frac{y_1-y_2}{x_1-x_2}$$?

In that case, if the two x values are the same, then the slope is vertical. Vertical slope is orthogonal to zero slope, so you can hardly say that division by zero results in zero.

 

[C4rlne55]

Since when is slope $$\frac{y_1}{x_1}-\frac{y_2}{x_2}$$? Isn't it $$\frac{y_1-y_2}{x_1-x_2}$$?

In that case, if the two x values are the same, then the slope is vertical. Vertical slope is orthogonal to zero slope, so you can hardly say that division by zero results in zero.

you have a point there, but if the two x values are the same then what exactly are we trying to prove? that the graph will tend to infinity in the y direction as the function approaches the limit where the gradient is $$\frac{1}{0}$$, ie undefined, ie a number much greater than we can work with, ie the gradient is mahoosive, ie vertical?

which has been shown on oh so many graphs over the years, for example the classical y=$$\frac{1}{x}$$, where at the point x=0, the function is undefined and has an asymptote

it is worth pointing out gregmead's point on the fact that $$\frac{1}{0}$$ is undefined as opposed to "infinity". "infinity" is actually a bungled job on the part of physicists, because they like to say (for example) bring a charge in from infinity... well, that raises the question "what does that mean?"