- #1

eNathan

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[tex]\frac {1/0} {1/0} = 1[/tex]

Or maybe

[tex]\frac {1/0} {1/0} = undefined[/tex]

I choose the former

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- Thread starter eNathan
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- #1

eNathan

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[tex]\frac {1/0} {1/0} = 1[/tex]

Or maybe

[tex]\frac {1/0} {1/0} = undefined[/tex]

I choose the former

- #2

AKG

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(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

eNathan

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AKG said:

(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.

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

AKG

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

- #5

- #6

lurflurf

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If one were being carefull one would say something likeeNathan 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 dont know the value of x. Maybe my definition of "undefined" is wrong though :tongue:

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

eNathan

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While we are on this strange topic, what about

inf / inf = 1

Therefore inf - inf = 0

eh?

- #8

AKG

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

- #9

DaveC426913

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eNathan said:While we are on this strange topic, what about

inf / inf = 1

Therefore inf - inf = 0

eh?

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

- #10

Jameson

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Is this true?

[tex]100\infty = 200\infty[/tex]

[tex]100 = 200*\frac{\infty}{\infty}[/tex]

[tex]100=200[/tex]

I think not.

- #11

GlauberTeacher

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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

quetzalcoatl9

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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?

- #13

quetzalcoatl9

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Jameson said:

Is this true?

[tex]100\infty = 200\infty[/tex]

[tex]100 = 200*\frac{\infty}{\infty}[/tex]

[tex]100=200[/tex]

I think not.

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

- #14

lurflurf

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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.quetzalcoatl9 said:how could this be done?

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

Hurkyl

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

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)

- #16

lurflurf

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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.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)

- #17

mathmike

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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

Jameson

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- #19

mathmike

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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

HallsofIvy

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One example is not a proof!

- #21

Jameson

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mathmike said:

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

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

[tex]\lim_{x\rightarrow\infty}\frac{x}{e^x}[/tex]

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

Hessam

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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 dont 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

- #23

mathmike

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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

Jameson

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

[tex]\frac{\frac{x}{0}}{\frac{x}{0}}=0[/tex]

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

[tex]\frac{\frac{x}{0}}{\frac{x}{0}}=0[/tex]

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

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- #25

mathmike

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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

- #26

Hurkyl

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it shows that it is 0.

No, it doesn't show that either.

- #27

mathmike

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- #28

Jameson

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

- #29

Jameson

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mathmike said:

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.

[tex]\lim_{x\rightarrow{0}}\frac{\sin{x}}{x}=1[/tex]

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

- #30

ron damon

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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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- #31

Hurkyl

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

- #32

ron damon

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- #33

mathmike

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- #34

ron damon

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

- #35

Hurkyl

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Then, how can an undefined entity differ from itself?

It is just as invalid to say "a;kljdfa;lkjasdf [itex]\neq[/itex] a;kljdfa;lkjasdf" as it is to say "a;kljdfa;lkjasdf = a;kljdfa;lkjasdf".

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