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## Why is infinite technically not a number.

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### #21Cornstalks  Members

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Posted 02 June 2014 - 09:44 PM

This is in fact, not true. If you evaluate the sum of all the natural numbers (1 + 2 + 3 + 4...) it is infinitely large. To evaluate this sum to have a value of -1/12th is not really correct. =/
If you want to learn the mathematical reasoning behind why the answer is infinity, but why you could evaluate a similar looking sum to have a value of -1/12th, I suggest looking up the zeta function.If you have an interest in maths, I really recommend it, there are some surprising results and really beautiful mathematics to be found there. If you don't want to learn the maths, just take it as 1 + 2 + 3 + 4... = infinity

I'm not a mathematician, but according to wikipedia and wolframalpha, ζ(−1) = -1/12

Mats1 is actually kind of correct. The thing is that the sum of the natural numbers is indeed infinite. In order to get -1/12 you have to use a different concept of numbers, called p-adic numbers.

For the curious, this question and answer give good a good introduction to the subject.

Anyway, this is further complicated by the fact that we aren't actually talking about 1 + 2 + 3 + ... = -1/12. What we're really talking about are limits and convergence, which isn't necessarily the same (or as strict) as equality. Because it's a limit we're computing, there are more ways to show 1 + 2 + 3 + ... = -1/12 than just the zeta function. So you might say 1 + 2 + 3 + ... is infinity just as much as you might say it's -1/12.

Edited by Cornstalks, 02 June 2014 - 09:45 PM.

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### #22Álvaro  Members

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Posted 02 June 2014 - 10:57 PM

I don't know what 1+2+3+4+...=-1/12 has to do with p-adic numbers. Care to explain?

### #23Mats1  Members

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Posted 03 June 2014 - 01:22 AM

I'm not a mathematician, but according to wikipedia and wolframalpha, ζ(−1) = -1/12

It might not be true under every system of mathematics, but it's certainly a correct answer under some of them. It's even used in physical calculations where the mathematical prediction matches up correctly with observations!

ζ(−1) = -1/12

This is the zeta function. Zeta of -1 DOES evaluate to - 1/12. The simple sum of writing 1 + 2 + 3 + 4 etc does NOT evaluate to - 1/12.

The zeta of -1 is the sum:

1/1^-1 + 1/2^-1 + 1/3^-1 + 1/4^-1

Edited by Mats1, 03 June 2014 - 01:23 AM.

### #24Hodgman  Moderators

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Posted 03 June 2014 - 01:29 AM

Isn't 2-1 equivalent to 1/2, so 1/2-1 would be 1/(1/2), which would be equivalent to 2? So 1/1-1 + 1/2-1 + 1/3-1 + 1/4-1 ... = 1+2+3+4...?

Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.

i.e. sum n where n=1..∞ != sum 1/(1/n) where n=1..∞

### #25Felix Ungman  Members

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Posted 03 June 2014 - 02:18 AM

Isn't it simply because they use different summation methods? I similar problem occurs when measuring an area with riemann integrals vs lebesgue integrals, some areas can be measured with one but not the other.

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### #26Paradigm Shifter  Members

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Posted 03 June 2014 - 02:31 AM

Isn't 2-1 equivalent to 1/2, so 1/2-1 would be 1/(1/2), which would be equivalent to 2? So 1/1-1 + 1/2-1 + 1/3-1 + 1/4-1 ... = 1+2+3+4...?

Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.

i.e. sum n where n=1..∞ != sum 1/(1/n) where n=1..∞

It's because you can only do arithmetic on limits of series if they are absolutely convergent (i.e. the absolute value of each term, summed up, converges to a finite limit). Otherwise you can expect nonsense in general.

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### #27Álvaro  Members

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Posted 03 June 2014 - 05:21 AM

Isn't 2-1 equivalent to 1/2, so 1/2-1 would be 1/(1/2), which would be equivalent to 2? So 1/1-1 + 1/2-1 + 1/3-1 + 1/4-1 ... = 1+2+3+4...?

Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.
i.e. sum n where n=1..∞ != sum 1/(1/n) where n=1..∞

The problem is that you can't sum infinitely many things. So when you write something like 1 + 1/2 + 1/4 + 1/8 + ... = 2, you are actually saying something like "the limit of the sequence (1, 1 + 1/2, 1 + 1/2 + 1/4, 1 + 1/2 + 1/4 + 1/8, ...) is 2". That is by far the most common way to interpret an "infinite sum", and according to that one 1 + 2 + 3 + 4 + ... = infinity.

There are other interpretations of "infinite sums". See the Wikipedia page on divergent series for details.

### #28Felix Ungman  Members

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Posted 03 June 2014 - 05:57 AM

But writing the terms as k or 1/k^-1 shouldn't matter, should it? Same values in the same order, same series, right?

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### #29Paradigm Shifter  Members

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Posted 03 June 2014 - 06:46 AM

The zeta function is only equal to the summation if the argument (corresponding to the power in the sum) is greater than 1. (Well, it is also true if the argument is complex and the real part is > 1).

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### #30apatriarca  Members

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Posted 03 June 2014 - 06:57 AM

The series in the definition of the zeta function is convergent (in the usual sense) for complex numbers with real parts greater than 1. We may however give different definitions of this function with a greater domain and this analytic continuation is what we use to compute the zeta function of -1.

### #31Cornstalks  Members

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Posted 03 June 2014 - 09:25 AM

I don't know what 1+2+3+4+...=-1/12 has to do with p-adic numbers. Care to explain?

Because using the "normal"/simple definitions of summation and integers, that summation does diverge. The natural numbers are closed over addition, and yet -1/12 is not a natural number, which breaks that closure of natural numbers. So in order to make sense of this contradiction, alternative/fancier definitions of summations and numbers must be used. Specifically, p-adic numbers, which converge for large values rather than diverge. Once you're using p-adic numbers, you're not using the natural numbers and aren't restricted to the closure of natural numbers, and so can achieve 1 + 2 + 3 + ... = -1/12.

Isn't 2-1 equivalent to 1/2, so 1/2-1 would be 1/(1/2), which would be equivalent to 2? So 1/1-1 + 1/2-1 + 1/3-1 + 1/4-1 ... = 1+2+3+4...?

Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.
i.e. sum n where n=1..∞ != sum 1/(1/n) where n=1..∞

As others have said, you can't treat infinity like a variable and do algebra with it. You can do some things (which actually involve evaluating a limit), but there are several things one might be tempted to do with infinity that would seem valid, but in reality aren't.

But maybe I'm not understanding, as I can't see the contradiction/inequality in sum n where n=1..∞ != sum 1/(1/n) where n=1..∞.

Edited by Cornstalks, 03 June 2014 - 09:29 AM.

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### #32Álvaro  Members

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Posted 03 June 2014 - 11:44 AM

I am going to do a point-by-point refutation of Conrstalks's post, and I know that can come through as rude. I just want to say I don't mean to be rude or sound upset in any way.

I don't know what 1+2+3+4+...=-1/12 has to do with p-adic numbers. Care to explain?

Because using the "normal"/simple definitions of summation and integers, that summation does diverge. The natural numbers are closed over addition, and yet -1/12 is not a natural number, which breaks that closure of natural numbers. So in order to make sense of this contradiction, alternative/fancier definitions of summations and numbers must be used. Specifically, p-adic numbers, which converge for large values rather than diverge. Once you're using p-adic numbers, you're not using the natural numbers and aren't restricted to the closure of natural numbers, and so can achieve 1 + 2 + 3 + ... = -1/12.

I understand what you are saying, but it doesn't apply in this case. Even when working with p-adic numbers, an infinite series can only converge if the terms have limit 0 (it turns out in p-adic numbers this condition is also sufficient). That means that you can compute things like 1*1 + 2*2 + 3*4 + 4*8 + 5*16 + 6*32 + ... in the 2-adic numbers (I think the sum is 1). But the general term in 1 + 2 + 3 + 4 + ... does not converge to zero even using the p-adic norm.

Isn't 2-1 equivalent to 1/2, so 1/2-1 would be 1/(1/2), which would be equivalent to 2? So 1/1-1 + 1/2-1 + 1/3-1 + 1/4-1 ... = 1+2+3+4...?

Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.
i.e. sum n where n=1..∞ != sum 1/(1/n) where n=1..∞

As others have said, you can't treat infinity like a variable and do algebra with it. You can do some things (which actually involve evaluating a limit), but there are several things one might be tempted to do with infinity that would seem valid, but in reality aren't.

But maybe I'm not understanding, as I can't see the contradiction/inequality in sum n where n=1..∞ != sum 1/(1/n) where n=1..∞.

1 + 2 + 3 + 4 + ... and 1/(1/1) + 1/(1/2) + 1/(1/3) + 1/(1/4) + ... are the exact same series, because their terms are the same. Of course their sums can't be different, no matter how you define them.

### #33Hodgman  Moderators

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Posted 03 June 2014 - 06:07 PM

Isn't 2-1 equivalent to 1/2, so 1/2-1 would be 1/(1/2), which would be equivalent to 2? So 1/1-1 + 1/2-1 + 1/3-1 + 1/4-1 ... = 1+2+3+4...? Like I said, I'm not good at the math, so it's interesting why two seemingly equivalent statements aren't equivalent.i.e. sum n where n=1..∞ != sum 1/(1/n) where n=1..∞

As others have said, you can't treat infinity like a variable and do algebra with it. You can do some things (which actually involve evaluating a limit), but there are several things one might be tempted to do with infinity that would seem valid, but in reality aren't.But maybe I'm not understanding, as I can't see the contradiction/inequality in sum n where n=1..∞ != sum 1/(1/n) where n=1..∞.
1 + 2 + 3 + 4 + ... and 1/(1/1) + 1/(1/2) + 1/(1/3) + 1/(1/4) + ... are the exact same series, because their terms are the same. Of course their sums can't be different, no matter how you define them.
That question of mine was in response to Mats1's post above it, where he told me that one of those sums did equal -1/12, but the other did not. Seeing they're equivalent, I was asking why this is so.

As for treating infinity as a regular algebraic variable - yeah I've seen that used to "prove" that 0=1, etc...
However, wasn't the calulus-type method before calculus to introduce, say x/∞ for a large part of your working (allowing you to work with infinitely small 'differentials'), but then at the end make the assumption that these terms are 0 and remove it. IIRC, this is no where near as formally robust as modern calculus, but in many situations can give you the same results. I heard there were two camps at the time, those who rejected calculus out of hand as a stupid trick that doesn't make sense, and those who were willing to treat infinity as a workable value...

Don't physicists also end up with a lot of ∞'a in their algebra, and have to 'normalize' them away using different methods?

• n.b. I really don't know what I'm on about here, just a curious layman!

### #34Álvaro  Members

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Posted 03 June 2014 - 07:01 PM

That question of mine was in response to Mats1's post above it, where he told me that one of those sums did equal -1/12, but the other did not. Seeing they're equivalent, I was asking why this is so.

The answer is that you are correct: Those two things are the same, and however you define one of them, you should get the same answer as for the other.

As for treating infinity as a regular algebraic variable - yeah I've seen that used to "prove" that 0=1, etc...

Those "proofs" usually include invalid manipulation. Notions of number that include infinity often will break some common features that you might expect. For instance, in ordinal numbers, 1 + omega = omega, but omega + 1 is greater than omega. In cardinal numbers, 1 + aleph_0 = aleph_0 = aleph_0 + 1, and even aleph_0 * 2 = aleph_0. However, 2^aleph_0 > aleph_0.

The use of those funky names for infinities is not capricious: One has to be very precise when dealing with infinities, or you'll end up with paradoxes all over the place.

### #35Mats1  Members

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Posted 05 June 2014 - 12:20 AM

The answer is that you are correct: Those two things are the same, and however you define one of them, you should get the same answer as for the other.

I thought however, that when presented with the naive sum 1 + 2 + 3 + 4... This will diverge. However, a value can be assigned to this series, after some manipulation, that is - 1/12, but it does not reflect the simple summation of the terms of the sequence.  I'm no expert of the zeta function, but I was pretty sure this is how it goes down.

### #36apatriarca  Members

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Posted 05 June 2014 - 01:33 AM

@Mats1: As I said ealier, the result has been obtained using more complicated methods based on complex analysis and not using some kind of series manipulation. When you sum an infinite number of terms you have to be very careful to what you do. Most manipulations gives in fact wrong results.

### #37Ashaman73  Members

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Posted 05 June 2014 - 06:03 AM

A number must fullfill some rules to be considered a number. Eg zero is the unit value of the additive operation, like 1 is the unit value of multiplication operation. Therefor

Addition:
(1) x + 0 = x = 0 +x
and
(2) x + (-x) = 0 = (-x) + x



Now lets check infinit

Addition:
(1) inf + 0 = inf = 0 + inf

(2) inf + (-inf) = 0

If for a infinit number inf = inf + X (eg inf = 1+inf) would hold true and inf would be a valid number, then following should work
inf + (-inf)
= 1 + inf + (-inf)
= 1 +     0
= 1
!= 0
= inf + (-inf)



Even simpe add operations would not work in a mathematically way with such a definition of infinit.

Edited by Ashaman73, 05 June 2014 - 06:06 AM.

Ashaman

### #38Álvaro  Members

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Posted 05 June 2014 - 07:51 AM

A number must fullfill some rules to be considered a number. Eg zero is the unit value of the additive operation, like 1 is the unit value of multiplication operation. Therefor

Addition:
(1) x + 0 = x = 0 +x
and
(2) x + (-x) = 0 = (-x) + x



Now lets check infinit

Addition:
(1) inf + 0 = inf = 0 + inf

(2) inf + (-inf) = 0

If for a infinit number inf = inf + X (eg inf = 1+inf) would hold true and inf would be a valid number, then following should work
inf + (-inf)
= 1 + inf + (-inf)
= 1 +     0
= 1
!= 0
= inf + (-inf)



Even simpe add operations would not work in a mathematically way with such a definition of infinit.

Check out surreal numbers, which have been mentioned in this thread before. They contain many infinite numbers, but they satisfy all the properties you mention.

### #39aregee  Members

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Posted 05 June 2014 - 03:14 PM

It is weird no one has linked to this interesting video yet:

### #40Álvaro  Members

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Posted 19 June 2014 - 05:44 AM

Vi Hart just released a video about different types of infinity. If you were interested in this thread, you'll probably enjoy it. I'm sure you'll enjoy some of her other videos too.

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