# Division by Zero: 1/0 = 1

• eNathan
In summary, after considering division by zero as being "un-defined", it is clear that the expression 1/0 divided by 1/0 is undefined, as division is not defined for meaningless symbols. Additionally, the concept of undefined is different from that of a variable, as a variable represents an unknown or unspecified value, not an undefined one. Furthermore, algebraic operations cannot be applied to concepts such as infinity or divide by zero, as they do not follow the rules of arithmetic. While it may be possible to define infinity and perform operations on it, this would result in the loss of certain properties of fields. Thus, it is not recommended to redefine algebraic structures in order to accommodate infinity and divide by zero.

#### eNathan

I have came up with this conclusion after considering division by zero as being "un-defined".

$$\frac {1/0} {1/0} = 1$$
Or maybe
$$\frac {1/0} {1/0} = undefined$$
I choose the former

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It's the latter. 1/0 is undefined, so it's certainly wrong that:

(meaningless symbol)/(meaningless symbol) = 1

Division is not defined for meaningless symbols, so there's no way to compute the quotient, let alone compute that it is 1.

AKG said:
It's the latter. 1/0 is undefined, so it's certainly wrong that:

(meaningless symbol)/(meaningless symbol) = 1

Division is not defined for meaningless symbols, so there's no way to compute the quotient, let alone compute that it is 1.

Is there any reason, mathematicly, however, that undefined / undefined <> 1?

I mean 1/0 is undefined, but so is "x", but we all know x/x = 1 even though we don't know the value of x. Maybe my definition of "undefined" is wrong though :tongue:

eNathan said:
Is there any reason, mathematicly, however, that undefined / undefined <> 1?

I mean 1/0 is undefined, but so is "x", but we all know x/x = 1 even though we don't know the value of x. Maybe my definition of "undefined" is wrong though :tongue:
I'm guessing you don't know what undefined means. Tell me, is a;kljdfa;lkjasdf greater than ;kljasdf;kljsdf? Does that previous question even make sense? No, it doesn't. x is a variable, and it's value is unknown or unspecified, not undefined. x stands for a number, and normally, we want to figure out what the number is by solving for x. 1/0 does not stand for any number. Division is an operation defined on numbers, so elephant/rhinoceros doesn't make sense. Similarly, (1/0)/(1/0) doesn't make sense, since 1/0 is not a number.

I'm assuming that you think (1/0)/(1/0) is 1 because by inverse multiplication, it is equal to (1/0)*(0/1) and the 0's somehow cancel out. You're assuming, of course, that 0/0 = 1.

eNathan said:
Is there any reason, mathematicly, however, that undefined / undefined <> 1?

I mean 1/0 is undefined, but so is "x", but we all know x/x = 1 even though we don't know the value of x. Maybe my definition of "undefined" is wrong though :tongue:
If one were being carefull one would say something like
x/x=1 when x is an elements of some division algebra and x is not zero.
0/0 is undefined as is 1/0
it must be because 0 has the property
0x=0 for all x
thus if we attempt to define
1/0
0(1/0)=0
but the whole point of defining 1/0 was so that0(1/0) was not 0
so no definition could be consistient.

AKG, good point. I was thinking that perhaps 1/0 had a value that is never known or something, which I did not belive, I was just woundering.

While we are on this strange topic, what about
inf / inf = 1
Therefore inf - inf = 0
eh?

Division is also not defined for infinity, neither is subtraction (or addition, or multiplication).

eNathan said:
While we are on this strange topic, what about
inf / inf = 1
Therefore inf - inf = 0
eh?

As AKG points out, $$\infty$$ is a mathematical concept; it is not a number, thus cannot be used as one in arithmetical operations.

In other threads it has been pointed out that basic algebra cannot be applied to concepts such as infinity and divide by 0.

Is this true?

$$100\infty = 200\infty$$

$$100 = 200*\frac{\infty}{\infty}$$

$$100=200$$

I think not.

In medieval times and before, both 0 and 1 used not to be considered numbers. Fractions with numerators other than one were not accepted as numbers for a long while. And infinity could very well be defined over all the operations if we choose to define it that way. But as we all agree, we see no reason to redefine in order to fit them in; the Indians came to the same conclusion in the 1300s.

Infinitesmals have always interested me. The whole idea that the limit of (0)*infinity could approach any number based on the setup is astounding. (It also made me question the irrational number, e.)

I still agree with the logic that Jameson, AKG, and Dave use to dispel eNathan's preponderances.

GlauberTeacher said:
And infinity could very well be defined over all the operations if we choose to define it that way.

how could this be done?

Jameson said:
In other threads it has been pointed out that basic algebra cannot be applied to concepts such as infinity and divide by 0.

Is this true?

$$100\infty = 200\infty$$

$$100 = 200*\frac{\infty}{\infty}$$

$$100=200$$

I think not.

yes, but that is also not true if you used any number (except 0).

quetzalcoatl9 said:
how could this be done?
It depends what is you want to do. For many purposes we wand to work with fields. No fields can be defined with zero division and infinite opererations. You could define a an extension so h^2=0 h not 0. Then we lose the zero product propert among other things (that is x*y can be zero without x or y being 0). Another approch is to define hyperreal numbers as is done in nonstandard analysis.
Here is a link to a intro to calculus book that uses nonstandard analysis.
http://www.math.wisc.edu/~keisler/calc.html
The problem adding yucky stuff to an algabraic structure is that you distroy the nice properties that that structure originally enjoyed. In particular nonstandard analysis introduces an infinite number of positive numbers less than all positive reals, a disturbing addition.

No fields can be defined with zero division and infinite opererations.

I presume things you mean things that behave like 1 &infin; = 2 &infin;... because, for example, the hyperreals have lots of infinite numbers. (Meaning that they're larger in magnitude than any integer)

Hurkyl said:
I presume things you mean things that behave like 1 ∞ = 2 ∞... because, for example, the hyperreals have lots of infinite numbers. (Meaning that they're larger in magnitude than any integer)
Yes that sentence of mine was a bit unclear. I did mean infinity in the sense of a number where x=2x=3x=x^2. Hyper reals avoid those type of difficulties by introducing and infinite number of infinite and infinitessimal numbers instead of one infinity. And the fact that even with hyperreals one still cannot divide by zero, but can divide by an infinite number of numbers that are like zero.

have you heard of a guy named l'hopital?

say that limf(x) = 0 and lim g(x)= 0. if lim[f'(x)/g'(x)] has a finite value or if this limit is + or _ inf, then lim f(x)/g(x) = lim f'(x)/g'(x)

for example: lim [(x^2 - 4) / (x-2) as x approches 2.

lim [(x^2 - 4) / (x-2) lim [ d/dx(x^2 -4) / d/dx(x - 2) ] - lim 2x / 1 =4

Yes, but L'Hopital's Rule applies with limits that are indeterminate. This is more about algebra with indeterminance.

well here is proff that (1 / 0 ) / (1 / 0) = 0

take lim (x / e^x) as e approaches + inf.

lim (x / e^x) = lim [d/dx (x) ] / [d/dx (e^x)] = lim 1 / e^x = 0

One example is not a proof!

mathmike said:
well here is proff that (1 / 0 ) / (1 / 0) = 0

take lim (x / e^x) as e approaches + inf.

lim (x / e^x) = lim [d/dx (x) ] / [d/dx (e^x)] = lim 1 / e^x = 0

I think you meant as "x" approaches infinity, not e.

$$\lim_{x\rightarrow\infty}\frac{x}{e^x}$$

And I don't even see how this is an example... or a proof. I don't mean that negatively, I just don't see it.

eNathan said:
Is there any reason, mathematicly, however, that undefined / undefined <> 1?

I mean 1/0 is undefined, but so is "x", but we all know x/x = 1 even though we don't know the value of x. Maybe my definition of "undefined" is wrong though :tongue:

you alsot have to consider that "x/x" is a peice-wise defined function

where

f(x) = 1 if x =/= 0, and undefined when x = 0

yes your right i did mean as x approaches infinity.

say that lim f(x) = g(x) = inf.

if lim [f'(x) / g'(x)] is + or - inf.
then lim f(x) / g(x) = lim f'(x) / g'(x)

ok there is the theorem;

now here is another example

as x approaches 0 from the right lim ln(x) / csc(x);

we have lim ln(x) - - inf and lim csc(x) = + inf

which is of type inf / inf

so the lim ln(x_ / csc(x) = lim [(1/x) / (-csc(x) cot(x)]

which can be written as lim [- (sin(x) / x) * tan(x)] = -lim sin(x) / x * lim tan(x) = - (1) * (0) = 0

thus lim ln(x) / csc(x) = 0

I'm well aware of L'Hopital's Rule, but your examples don't show that

$$\frac{\frac{x}{0}}{\frac{x}{0}}=0$$

Your examples show that limits in indeterminate form can be modified to get a numerical result.

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your right it don't show that (x/0) / (x/0) = 1

it shows that it is 0. there is no way to show that it is 1 because it is not 1.

and isn't inf /inf indeterminite anyway. there is no way to show it isnt

it shows that it is 0.

No, it doesn't show that either.

your right it doesn't show that x/0 / x/0 =0 it shows that the limit is zero, therefore it can be said that it is zero

We all agree that L'Hopital's Rule works, but it just doesn't apply to the original topic.

mathmike said:
your right it doesn't show that x/0 / x/0 =0 it shows that the limit is zero, therefore it can be said that it is zero

That is not true. Just because a limit exists, it does not mean that the point that the limit approaches exists. We see this all the time.

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

But the point (0,1) does not exist on that graph.

AKG said:
I'm guessing you don't know what undefined means. Tell me, is a;kljdfa;lkjasdf greater than ;kljasdf;kljsdf? Does that previous question even make sense?.

but can't you say that "a;kljdfa;lkjasdf" / "a;kljdfa;lkjasdf" equals one, since "a;kljdfa;lkjasdf" = "a;kljdfa;lkjasdf" ?

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No, you cannot. "a;kljdfa;lkjasdf" is undefined, and thus so is any expression involving it.

You also cannot say "a;kljdfa;lkjasdf = a;kljdfa;lkjasdf"

Then, how can an undefined entity differ from itself? If they are the same, comparing sames doesn't equal one?

but the implication of [x / 0] / [x / 0] = 1 is perposterous. but you are right in saying that it can be manipulated to get a numerical result

also, can't you apply geomtric reasoning, since we DO know what happens to C/X when X approaches 0?

Then, how can an undefined entity differ from itself?

It is just as invalid to say "a;kljdfa;lkjasdf $\neq$ a;kljdfa;lkjasdf" as it is to say "a;kljdfa;lkjasdf = a;kljdfa;lkjasdf".