- #1

danielassayag

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Therefore if 1/ ∞ = 0,

∞ * 0 should be equal to 1

and

1/0 = ∞

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In summary, division by zero or by infinity is undefined because it leads to mathematical inconsistencies and confusion. While division by infinity can be thought of as tending towards zero, it is not a defined operation in mathematics and can lead to ambiguous answers. Similarly, division by zero is also undefined and can result in conflicting answers. Therefore, it is best to avoid dividing by zero or infinity in mathematical equations and calculations.

- #1

danielassayag

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Therefore if 1/ ∞ = 0,

∞ * 0 should be equal to 1

and

1/0 = ∞

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- #2

jedishrfu

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https://math.stackexchange.com/questions/44746/one-divided-by-infinity

- #3

danielassayag

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Why multiplying 0 by infinity should be zero and not one, when dividing 1 by infinity is 0?

the a/b = c equivalent to a = b*c relation doesn't apply here?

the a/b = c equivalent to a = b*c relation doesn't apply here?

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- #4

danielassayag

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Division by infinity is not undefined. Division by infinity tends to zero, The plot is stretched oppositely whether the number is smaller or greater than one.jedishrfu said:Division by zero or by infinity is undefined because they lead to mathematical inconsistencies as you have discovered.

plot | 1/x | x = -0.1 to 0.1

plot 1/x from x=-100 to 100

- #5

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You can say that over and over as many times as you like, but saying it is not going to make it true.danielassayag said:Division by infinity is not undefined.

- #6

danielassayag

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You are right. I just realized that 1/x when x gets larger tends to 0.phinds said:You can say that over and over as many times as you like, but saying it is not going to make it true.

But if x is infinitely smaller, example 1/0.1, 1/0.01.., then it tends to infinity again.

So 1/infinity tends both to zero and infinity. So i guess you can call it undefined.

- #7

danielassayag

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I didn't mean to outsmart wolframalpha btw..phinds said:You can say that over and over as many times as you like, but saying it is not going to make it true.

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It's nice that you are now in sync with the rest of the world.danielassayag said:So i guess you can call it undefined.

- #9

jedishrfu

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As you noted, zero times any number is 0 and a number times its reciprocal is 1 but when using infinity as a number you get confusing ambiguous answers and so mathematicians declare it to be undefined.

- #10

danielassayag

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Wolfram alpha and microsoft don't seem to implement your understanding of maths, as they don't comprehend that 1/infinity should be undefined. Though they understand the concept of undefined itself, when it comes to dividing 1/0.phinds said:It's nice that you are now in sync with the rest of the world.

- #11

Mark44

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And in mathematics symbols we write this: $$\lim_{x \to \infty}\frac 1 x = 0$$danielassayag said:You are right. I just realized that 1/x when x gets larger tends to 0.

But what if x approaches zero from the negative side? Then the result gets unboundedly negative.danielassayag said:But if x is infinitely smaller, example 1/0.1, 1/0.01.., then it tends to infinity again.

In terms of limits,

$$\lim_{x \to 0^-}\frac 1 x = -\infty$$

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

For this reason, ##\frac 1 0## is

Absolutely not, and this makes no sense. The symbol ##\infty## is not a number in the real number system, and cannot be used in expression involving arithmetic or algebraic operations. Note that the Extended Reals do include ##\infty##, however.danielassayag said:So 1/infinity tends both to zero and infinity. So i guess you can call it undefined.

- #12

Mark44

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Because ##\infty## is not a number in the real number system, so multiplication of it is not defined.danielassayag said:

the a/b = c equivalent to a = b*c relation doesn't apply here?

##\frac a b = c \Leftrightarrow a = b \cdot c## provided that ##b \ne 0##. You need to read the fine print.

- #13

Infrared

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I think some of the above comments are unnecessarily dogmatic. There are definitely contexts where it makes sense to write ##1/\infty=0.## My issue with the logic in the OP is that you end up with the quotient ##\infty/\infty##, which is definitely undefined- you can't just say that it is ##1.##

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- #14

danielassayag

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- #15

Infrared

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This isn't true. A function may tend towards a limit, but a simple quotient doesn't. Even if you want to interpret ##1/\infty## as a limit, limits still equal their value. For example, ##\lim_{x\to\infty}1/x## equals zero; it doesn't "tend to zero" (though the function ##x\mapsto 1/x## does in the limit of large x.)danielassayag said:1/infinity tends to zero, but only tends, which mean it's not completely 0, there is still some infinitely small value left to it.

This doesn't follow.danielassayag said:If you multiply that infinitely small value of 0 decimal, to an infinitely large integer number, you end multiplying the depth of decimal against the depth of integer, which is equal ; therefore you get back you an equality, 1.

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That's their problem, not mine.danielassayag said:Wolfram alpha and microsoft don't seem to implement your understanding of maths, as they don't comprehend that 1/infinity should be undefined.

- #17

jedishrfu

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There are many examples of software doing the wrong thing mathematically or arithmetically:

https://rootly.com/blog/you-do-the-math-reliability-issues-triggered-by-math-errors

https://listverse.com/2019/08/30/10-simple-but-costly-math-errors-in-history/

- #18

Infrared

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I don't think it's a "problem" of theirs. There are lots of contexts in which ##1/\infty## is defined to be zero and it makes sense to me for it to be implemented that way.phinds said:That's their problem, not mine.

- #19

danielassayag

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danielassayag said:

1/infinity tends to zero, but only tends, which mean it's not completely 0, there is still some infinitely small value left to it.

to then respond:

but here is what I wrote: (as you can see it hasnt been edited on the post)Infrared said:This isn't true. A function may tend towards a limit, but a simple quotient doesn't.

I'm interpreting 1/x with x getting infinitely larger and larger as a function, that when x gets larger, it gets closer and closer to 0.danielassayag said:in some case, (when x get larger and larger) 1/x tends to zero, but only tends, which mean it's not completely 0, there is still some infinitely small value left to it.

I realize that 1/x (with x getting infinitely smaller and smaller than 1) tends to a greater and greater number.

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Pretty simple to demonstrate and Wolfram agrees.danielassayag said:Wolfram alpha and microsoft don't seem to implement your understanding of maths, as they don't comprehend that 1/infinity should be undefined. Though they understand the concept of undefined itself, when it comes to dividing 1/0.

View attachment 294454 View attachment 294455

View attachment 294456

First, what does Wolfram says about ##0 \times \infty##: undefined

But Wolfram also says that ##\frac{1}{\infty} = 0##. It is not undefined. So why isn't ##0 \times \infty = 1##?

Because Wolfram also says that:

This is why ##0 \times \infty## is undefined: it can lead to an infinite number of answers. ##1## is just one of them.

The same problem arises with ##\frac{\infty}{\infty}## which is also undefined even though ##1 \times\infty = \infty##. That's because ##a \times \infty = \infty## no matter the value of ##a##.

Another way to look at it is that if we multiply anything by ##0##, it gives ##0##, and anything multiplied by ##\infty## equals ##\infty##. What if we multiplied ##0 \times \infty##? It is undefined because it could at least be either ##0## or ##\infty## (or anything else as demonstrated previously).

- #21

ergospherical

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- #22

Mark44

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But mathematics (excluding the Extended Reals) isn't one of those contexts.Infrared said:There are definitely contexts where it makes sense to write 1/∞=0.

More to the point, this is an indeterminate form, often written in mathematics textbooks as ##\left[\frac \infty \infty \right]##. The usual context for indeterminate forms is in limits such as the following.Infrared said:My issue with the logic in the OP is that you end up with the quotient ∞/∞, which is definitely undefined- you can't just say that it is

$$\lim_{x \to \infty} \frac {3x} x = 3$$

$$\lim_{x \to \infty} \frac {x} {3x} = \frac 1 3$$

$$\lim_{x \to \infty} \frac {x} {x^2} = 0$$

In all three examples, both the numerator and denominator are increasing without bound, but the resulting limits are all different.

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I remember a student who was asked to give an example of an unbounded sequence and his answer was:$$1,2,3, \infty, 4, 5 \dots $$Infrared said:I don't think it's a "problem" of theirs. There are lots of contexts in which ##1/\infty## is defined to be zero and it makes sense to me for it to be implemented that way.

- #24

Mark44

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Because wolframalpha is inconsistent.jack action said:Pretty simple to demonstrate and Wolfram agrees.

First, what does Wolfram says about 0×∞: undefined

But Wolfram also says that ##\frac{1}{\infty} = 0##. It is not undefined. So why isn't 0×∞=1?

An expression that can lead to multiple answers is called an indeterminate form. Other such forms include ##[\infty \pm \infty], [1^\infty], [\frac 0 0]## and others.jack action said:This is why 0×∞ is undefined: it can lead to an infinite number of answers.

- #25

Infrared

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Mark44 said:But mathematics (excluding the Extended Reals) isn't one of those contexts.

Of course there are mathematical contexts where it makes sense! The one that first comes to my mind is that the Riemann sphere ##\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}## has the natural structure of a complex curve, and its automorphism group is the group of Mobius transformations ##z\mapsto\frac{az+b}{cz+d},## but in order for this to make sense on the whole sphere, you need to set ##1/\infty=0## and ##1/0=\infty.##

It's not inconsistent: ##0\cdot\infty## is indeterminate, but ##1/\infty## is not.Mark44 said:Because wolframalpha is inconsistent.

As an aside, I can think of at least case where ##0\cdot\infty## is sometimes defined: in measure/integration theory it is in fact usually set to 0 because integrating a function which is (almost) always zero on an infinitely large set, or integrating an infinitely large function on a set of measure zero give zero. But it's still indeterminate...

- #26

Mark44

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I explicitly excluded the extended reals.Infrared said:Of course there are mathematical contexts where it makes sense! The one that first comes to my mind is that the Riemann sphere ##\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}## has the natural structure of a complex curve, and its automorphism group is the group of Mobius transformations ##z\mapsto\frac{az+b}{cz+d},## but in order for this to make sense on the whole sphere, you need to set ##1/\infty=0## and ##1/0=\infty.##

Mark44 said:Because wolframalpha is inconsistent.

But WA reports that ##\frac 1 \infty## is a number, and outside of the extended reals, it's not legitimate to use the symbol ##\infty## in arithmetic expressions.Infrared said:It's not inconsistent: ##0\cdot\infty## is indeterminate, but ##1/\infty## is not.

- #27

Infrared

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The Riemann sphere is not the extended reals.Mark44 said:I explicitly excluded the extended reals.

It depends on what arithmetic you're doing. I agree it's not an equality of real numbers, but I think in any context where you can do arithmetic with ##\infty##, then ##1/\infty## will be zero.Mark44 said:But WA reports that ##\frac 1 \infty## is a number, and outside of the extended reals, it's not legitimate to use the symbol ##\infty## in arithmetic expressions.

- #28

Mark44

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Infrared said:Riemann sphere ##\hat{\mathbb{C}}=\mathbb{C}\cup\{\infty\}##

Tomato, tomahto -- you've extended the complex numbers to include ##\infty##.Infrared said:The Riemann sphere is not the extended reals.

- #29

Infrared

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They're very different. The Riemann sphere only has one infinity; the extended reals has two. The Riemann sphere has the natural structure of a complex curve, which is why the Mobius transformations are applicable. This isn't some obscure object: it's the most basic example of a Riemann surface.Mark44 said:Tomato, tomahto -- you've extended the complex numbers to include ##\infty##.

And why are we excluding the extended reals in the first place?

- #30

Mark44

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Because in the ordinary reals, the symbol ##\infty## does not represent a number, so expressions such as ##\frac 1 \infty## are not defined.Infrared said:And why are we excluding the extended reals in the first place?

Given that this thread is marked 'B', discussions about the extended reals and the Riemann sphere are way off topic.

- #31

Infrared

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Mark44 said:Given that this thread is marked 'B', discussions about the extended reals and the Riemann sphere are way off topic.

This could well be the case, but then I don't think it's fair to say that wolfram alpha is wrong or inconsistent. It doesn't know about the limited usage which may or may not be relevant here.

I don't plan to message more on this issue, for fear of further derailing the thread.

- #32

Office_Shredder

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- #33

martinbn

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Just to add to the fun, there are contexts in which ##\infty \times 0 = 0##.

- #34

Infrared

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I think I mentioned this already in post 25 :)martinbn said:Just to add to the fun, there are contexts in which ##\infty \times 0 = 0##.

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