- #1
danielassayag
- 8
- 0
As we know those relations are true: if a/b = c, then a = b*c and b = a/c
Therefore if 1/ ∞ = 0,
∞ * 0 should be equal to 1
and
1/0 = ∞
Therefore if 1/ ∞ = 0,
∞ * 0 should be equal to 1
and
1/0 = ∞
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.
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.
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.
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.
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.
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.
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.
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.
Because ##\infty## is not a number in the real number system, so multiplication of it is not defined.danielassayag said: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?
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.
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.
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.
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.
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
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
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.
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.
Mark44 said:But mathematics (excluding the Extended Reals) isn't one of those contexts.
It's not inconsistent: ##0\cdot\infty## is indeterminate, but ##1/\infty## is not.Mark44 said:Because wolframalpha is inconsistent.
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.
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.
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.
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##.
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?
Mark44 said:Given that this thread is marked 'B', discussions about the extended reals and the Riemann sphere are way off topic.
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##.