# (x/0) / (x/0) = 1

1. Jul 6, 2005

### 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

2. Jul 6, 2005

### AKG

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.

3. Jul 6, 2005

### eNathan

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 dont know the value of x. Maybe my definition of "undefined" is wrong though :tongue:

4. Jul 6, 2005

### AKG

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.

5. Jul 6, 2005

### Icebreaker

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.

6. Jul 6, 2005

### lurflurf

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.

7. Jul 6, 2005

### eNathan

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?

8. Jul 6, 2005

### AKG

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

9. Jul 6, 2005

### DaveC426913

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

10. Jul 6, 2005

### Jameson

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.

11. Jul 6, 2005

### GlauberTeacher

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.

12. Jul 7, 2005

### quetzalcoatl9

how could this be done?

13. Jul 7, 2005

### quetzalcoatl9

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

14. Jul 7, 2005

### lurflurf

It depends what is you want to do. For many purposes we wand to work with feilds. No feilds 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.

15. Jul 7, 2005

### Hurkyl

Staff Emeritus
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)

16. Jul 7, 2005

### lurflurf

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.

17. Jul 8, 2005

### mathmike

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

18. Jul 8, 2005

### Jameson

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

19. Jul 8, 2005

### mathmike

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

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

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

20. Jul 8, 2005

### HallsofIvy

One example is not a proof!

21. Jul 8, 2005

### Jameson

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.

22. Jul 9, 2005

### Hessam

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

23. Jul 9, 2005

### mathmike

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

24. Jul 9, 2005

### Jameson

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.

Last edited: Jul 9, 2005
25. Jul 9, 2005

### mathmike

your right it dont 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 isnt inf /inf indeterminite anyway. there is no way to show it isnt