Poll: 0.999... = 1

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Delta342

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Aaah the old question me and my housemates have argued about since our first year in doing mathematics, (we're now all 4th year masters students). Technically the proof is sound in that the limit of 0.999... is in fact equal to one.. However a lot of Mathematicians will disagree and say that it is in fact infinitesimally close to 1.. This makes use of non-standard analysis (feel free to google it). So yes, in a way 0.9999999 etc is equal to one.. But in fact it is only infinitesimally close to 1. Although to tell the truth equality isn't exactly a particularly strong relation in Mathematics.
 

Piflik

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Coldie said:
Piflik said:
I didn't add anything, I shifted brackets (that is no mathematical operation, but additions are commutative, so the order is irrelevant). And I already said that this 'proof' is nonsensical (twice), but if you take infinity into consideration, this 'proof' is just as valid as 0.9999... = 1. I know there is a -1 missing at the end that I omitted, but you (among others) argue that there is no end to infinity, so this -1 doesn't exist.

The difference between 0.99999... and one might be infinitesimal (or infinitely) small, but it is there.
As you say, there's no operation "shift brackets". What you did is redefine the sequence. You, essentially, created a new, independent sum. With an incorrect result, but oh well - there's no -1 missing after the infinity, it's missing from the right part.

inf.sum(1 - 1) = 0
1 + inf.sum(-1 + 1) = 1
The result of Sum[n=0..inf](-1)[sup]n[/sup] is, of course, {1 for even n, 0 for odd n} and is undetermined when n = infinity, because infinity is both odd and even.

And the difference between 0.(9) and 1 is 0.(0), which is to say, exactly 0.
The difference is 0.000000000....000001 which is exactly 0.000000000....000001 and it is not 0.
 

SextusMaximus

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BlacklightVirus said:
Piflik said:
Athinira said:
You can't get any "closer" than "equivalent", at least not in the system of Real Numbers, because that number system doesn't allow infinitesimal values [http://en.wikipedia.org/wiki/Infinitesimal].
If you want to allow infinity, you also have to allow infinitesimal values, since that is nothing other than 1/infinity.
SextusMaximus said:
waxwingslain said:
How about this one:

The difference between 0.9999.... and 1.0 is 0.0000... and since the difference between them is 0, the two numbers must be equal.
*EPIC FACEPALM*

Unless you were being sarcastic... the difference is 0.0000000...(infinity)...001
Epic facepalm to you too. To write infinity in between two decimal places is illogical because you will never reach the decimals after "infinity".

Also I have another proof which I don't think has been posted yet:



A and B are two real numbers. The midpoint (M) of A and B (i.e. the value between them such that AM = MB) can be found with the formula M = (A + B)/2.

Let A = 0.999... and let B = 1.

M = (1 + 0.999...)/2
∴M = 1.999.../2
∴ M = 0.999... (check that on a calculator if you must)

Hence A = 0.999... = M
∴A = M

Therefore in terms of the original equation:

(A + B)/2 = M = A
∴(A + B)/2 = A
∴A + B = 2A
∴B = 2A - A
∴B = A

And substituting the original values:
0.999... = 1


*Note, at no point in that proof before the last line did I use 1 = 0.999... so the proof is valid.
Nevertheless, that is the difference.
 

Piflik

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BlacklightVirus said:
Piflik said:
BlacklightVirus said:
Piflik said:
Athinira said:
You can't get any "closer" than "equivalent", at least not in the system of Real Numbers, because that number system doesn't allow infinitesimal values [http://en.wikipedia.org/wiki/Infinitesimal].
If you want to allow infinity, you also have to allow infinitesimal values, since that is nothing other than 1/infinity.
SextusMaximus said:
waxwingslain said:
How about this one:

The difference between 0.9999.... and 1.0 is 0.0000... and since the difference between them is 0, the two numbers must be equal.
*EPIC FACEPALM*

Unless you were being sarcastic... the difference is 0.0000000...(infinity)...001
Epic facepalm to you too. To write infinity in between two decimal places is illogical because you will never reach the decimals after "infinity".

Also I have another proof which I don't think has been posted yet:



A and B are two real numbers. The midpoint (M) of A and B (i.e. the value between them such that M-A = B-M) can be found with the formula M = (A + B)/2.

Let A = 0.999... and let B = 1.

M = (1 + 0.999...)/2
∴M = 1.999.../2
∴ M = 0.999... (check that on a calculator if you must)
A calculator rounds...and 0.99999....*2 = 1.99999....98...

1.9999999.../2 = 0.9999...95 (note that the 0.9999....95 is 0.0000...05 more than 0.99999...)

so the following is invalid

BlacklightVirus said:
Hence A = 0.999... = M
∴A = M

Therefore in terms of the original equation:

(A + B)/2 = M = A
∴(A + B)/2 = A
∴A + B = 2A
∴B = 2A - A
∴B = A

And substituting the original values:
0.999... = 1


*Note, at no point in that proof before the last line did I use 1 = 0.999... so the proof is valid.
It is not.
Except that I did not use a calculator and I did not at any point multiply 0.999... by 2. Because I was establishing 0.999... = 1 I could not say 0.999... + 1 = 2*0.999... in my proof.
You just invalidated your own proof...

You said (1 + 0.99999...)/2 = 0.999999 (this is a line from your proof)
Times 2 on both sides: 1 + 0.9999999.... = 2 * 0.99999...
Minus 0.99999... on both sides: 1 = 0.9999999...
...circular reasoning...

Your proof is invalid.
 

Piflik

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Redingold said:
Piflik said:
Redingold said:
Piflik said:
I didn't say it never makes a difference. But in this case it doesn't (as long as you keep infinity in mind, or else there will be a -1 at the end, but as multiple people here said, there is no end to infinity, so there is no -1 and 1 = 0).
But it does make a difference, because 1 + (-1 + 1) + (-1 + 1) + ... = 1

Infinite addition is not necessarily commutative.
So you agree that traditional maths is not necessarily valid when dealing with infinity, because in traditional maths additions are always commutative...so for all intends and purposes you agree with me ;)
Of course traditional maths isn't always relevant when we're dealing with infinity. After all, 2 * infinity is infinity, which implies infinity = 0, or that 1 = 2. However, 0.999... is still equal to 1.

Redingold said:
M'kay. The number 0.999... is equal to an infinite series 0.9 + 0.09 + 0.009 + 0.0009 and so on. If you know anything about slightly advanced maths, you'll know that the sum of an infinite geometric series is equal to a/(1-r) when |r| < 1 (explained below for those who aren't so good at maths)

In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).

So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.

Explanation of maths involved:

A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].

The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]

Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]

Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]

This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])

And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)

Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero.

Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1

Satisfied now?
See?

If you can find a flaw in that, well done.
It has the same flaw like any other such 'proof'...the usage of infinity. You said yourself that you cannot use traditional maths when dealing with infinity at least not always...

For example the geometric sequence...a geometric sequence is converging to a certain number, but will never ever (ever EVER) reach this number. If you find an expression where an infinite sum equals a number, then this is just an abbreviation for 'the limit of that sequence for n approaching infinity equals to said number'. Still it is not the same as this number, it just converges to it.
 

Lyx

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Piflik said:
Coldie said:
Piflik said:
Actually, there really is no such thing as infinity. It is a theoretical concept, but it doesn't exist. Every number has an end. Always.
Infinity is very prominent a mathematical concept. It's quite real, I assure you. And, as redundant as it is redundant, infinitely long numbers are, in fact, infinitely long. They never have an end. Here, let me show you, an infinitely long number that has an infinite number of digits after the decimal point and no end ever:



Quite a famous transcendent number, especially prominent in trigonometry and geometry. Also it's infinitely long and has a finite value.

Some infinitely long numbers have a repeating pattern, for instance 1/3 is just infinitely repeating threes. The notation for such numbers is 0.(periodic pattern), so 1/3 = 0.(3) or 1/7 = 0.(142857).

So if you take 0.(9): 1 - 0.(9) = 1 x 10[sup]-infinity[/sup] = 0.
You can find the detailed calculations in the first couple pages of the thread.

Don't try to understand it, just accept it as the universal and absolute truth. Because that's what it is. Elementary Arithmetics.
So you agree with my prof that 0 = 1? Because if you want to do traditional maths (or Elementary Arithmetics as you call it...) with infinity, you would have to...
A very good argument :) Let me debunk it *g*

At first, i will backtrack from some earlier statements and partially agree with what jaime wolf said earlier - though, probably in a different way than would be expected of me.

I'll abolish numbers. Or rather, i'll define that "number" is just the dividend of the base, in a base-unitsystem. In more simple words: 1/3 -> 1 is the dividend/number, and 3 the base.

With VALUE, i will mean the following: 1/3 and 1/4 - the "number/dividend" in both cases is identical, but the value is not.

-----

Okay, lets rock:

1. Let's start easy. We'll enter 1/3 into our calculator. Notice that this is actually an operation in the style number/base. This is important. Our "value" is actually represented as a formula/process/function/howeverYouWannaCallIt.

2. Next, we want to store this in decimal. So, a num/1 system with digits ranging from 0-9. Problem is that this system cannot accurately represent 1/3. No matter how many fractions we add, we'll never reach the accuracy of 1/3. Therefore, the value changes. If we have to discard something, we lose something. A difference is a difference. 0.333... therefore cannot be equal to 1/3, for the plain simple reason that this isn't a matter of digits... decimal simply cannot represent this value accurately, no matter what you do to the digits.

3. For the same reason, 0.999... can never become 1. Why? Quite simply because it is the wrong operation to get what we want. Before i explain this, remember that what we're defining is a formula. The "0.999"-part actually means "0.999 / 1". Okay?

Good. If we add infinity, the formula is this: "zero point infiniteTimes9 / 1" (this isnt absolutely correct, but it doesnt affect what i'm gonna show). So, we just added a command to append "9" digits forever. But just as with 1/3, no matter how many digits you add, the value is not identical to 1, for the simply reason that there again is a loss in accuracy, and therefore there is a loss, which in turn means, there is a difference. The accuracy-loss may become ever smaller, but it never vanishes, as long as all you're doing is appending 9's.

So, it is simply the wrong operation to achieve a result of 1. What we'd need to do, is more like - as so many "naive" posters proposed - ROUND UP.... but then we'd have to acknowledge, that we have to change the value to get it to 1, and that therefore it cannot have been 1 before the rounding.

---

4. Ready for pi? Okay. So far, the formula-style we used to represent values has been of the scheme num/base. But since we've thrown the traditional idea of "numbers" overboard, we can say that it really doesn't need to be that way. We can represent a value with any formula we like (which is why relations work at all). Some values can be represented with multiple formulas ( 2/4 = 1/2 ). There are however also formulas for which currently there is no representation in the num/base format possible. Which is why they run "forever" in those. The supposed "infinity" in pi, is nothing else, then the infinity in 1/3 when converted to decimal. In other words: The only reason why pi is running forever in decimal, is that the decimal system cannot accurately represent it.

Notice a pattern? In the above example, values always could "forever not resolved" in a num/base system, precisely BECAUSE of a difference! Only because the system cannot accurately represent the desired value, we never reach total accuracy. Or to put it super bluntly: 1/3 converted to decimal results in endless 3's BECAUSE there is forever a difference!

---

To finish this wall of text, a little addition about infinities. The above "infinite inability to be equal" has very little to do with what i'd call singularities. You know, stuff like "infinitely small" or "infinitely high" - totally different thing. So, that unit-conversions results in such weird things, proves in no way at all that singularities exist in reality - that is a different kind of "infinity". Bodies in space may follow curves according to some formula called pi, that by coincidence doesn't fit into our dumb num/base system.... but that does not mean that "there is some kind of infinity in those bodies flying around" - it's just an anomaly in our number system, nothing else - call it a maths-bug if you want to.

Infinity is very prominent a mathematical concept. It's quite real, I assure you.
Yes, indeed - very "real".
 

TheTejs

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Sturmdolch said:
Naheal said:
Actually, it didn't :-/ That proof works.
Third last step, at 2a(a-b) = a(a-b). a = b, so dividing by (a-b) is impossible. Yet they still do. So no.

The third last step is still valid, because a-b = 0, so both sides equal 0.

You could just as easily say

532531521215a(a-b) = a(a-b)
532531521215 = 1

which is equally as wrong.


havass said:
If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...

In conclusion, I have just proven 1 = 0.9999...
You're doing it wrong, too. At step 4, 9x = 9, that is not true.

9x = 8.99999999



Edit: I'm not even a math major, but the amount of ignorant false intellectualism in this thread is about to make me cry.
9x=8,000...1
and no 0,999...=1 is false
see:
100/(0,9999-1)
100/(1-1)
guess which one cannot be done
 

Rainforce

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Redingold said:
Really? I suppose you have some strong, alien mathematics where 0.999... is not equal to 1, then. Could I see it?
No, but our interpretation has severe problems with some things, because we pretty much always operate/think in modeled enviroments that are far from reality. So I guess we still have much to learn.
 

Coldie

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Piflik said:
The difference is 0.000000000....000001 which is exactly 0.000000000....000001 and it is not 0.
Now you're just trolling. Stop making up nonsense and start studying math from the very beginning.
 

Rainforce

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Coldie said:
Piflik said:
The difference is 0.000000000....000001 which is exactly 0.000000000....000001 and it is not 0.
Now you're just trolling. Stop making up nonsense and start studying math from the very beginning.
ok, a much cheaper perspective: why do we differentiate between 0.999~ and 1 then?
 

Redingold

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Piflik said:
Redingold said:
Piflik said:
Redingold said:
Piflik said:
I didn't say it never makes a difference. But in this case it doesn't (as long as you keep infinity in mind, or else there will be a -1 at the end, but as multiple people here said, there is no end to infinity, so there is no -1 and 1 = 0).
But it does make a difference, because 1 + (-1 + 1) + (-1 + 1) + ... = 1

Infinite addition is not necessarily commutative.
So you agree that traditional maths is not necessarily valid when dealing with infinity, because in traditional maths additions are always commutative...so for all intends and purposes you agree with me ;)
Of course traditional maths isn't always relevant when we're dealing with infinity. After all, 2 * infinity is infinity, which implies infinity = 0, or that 1 = 2. However, 0.999... is still equal to 1.

Redingold said:
M'kay. The number 0.999... is equal to an infinite series 0.9 + 0.09 + 0.009 + 0.0009 and so on. If you know anything about slightly advanced maths, you'll know that the sum of an infinite geometric series is equal to a/(1-r) when |r| < 1 (explained below for those who aren't so good at maths)

In our example here, a, the first term, is 0.9, and r, the common ratio, is 0.1 (because each term is the previous term multiplied by 0.1).

So we have 0.9/(1-0.1) which equals 0.9/0.9 which equals 1.

Explanation of maths involved:

A geometric sequence is one where each term is the previous term multiplied by some number r. The first term is a, the second term is ar, the third term is ar[sup]2[/sup] and so on. The nth term is ar[sup]n-1[/sup].

The sum of a geometric series to n terms, which we shall call S[sub]n[/sub], is therefore equal to a + ar + ar[sup]2[/sup]...+ ar[sup]n-2[/sup] + ar[sup]n-1[/sup]

Multiplying by r, we get rS[sub]n[/sub] = ar + ar[sup]2[/sup] + ar[sup]3[/sup]...+ ar[sup]n-1[/sup] + ar[sup]n[/sup]

Subracting rS[sub]n[/sub] from S[sub]n[/sub] leads to S[sub]n[/sub] - rS[sub]n[/sub] = a - ar[sup]n[/sup]

This means S[sub]n[/sub](1-r) = a(1 - r[sup]n[/sup])

And S[sub]n[/sub] = a(1 - r[sup]n[/sup])/(1-r)

Now, to find the sum to infinity, n must be equal to infinity. If |r| > 1, r[sup]infinity[/sup] is infinite. If |r| < 1, r[sup]infinity[/sup] is equal to zero.

Thus, S[sub]infinity[/sub] = a(1 - r[sup]infinity[/sup])/(1-r) = a(1-0)/(1-r) = a/(1-r) when |r| < 1

Satisfied now?
See?

If you can find a flaw in that, well done.
It has the same flaw like any other such 'proof'...the usage of infinity. You said yourself that you cannot use traditional maths when dealing with infinity at least not always...

For example the geometric sequence...a geometric sequence is converging to a certain number, but will never ever (ever EVER) reach this number. If you find an expression where an infinite sum equals a number, then this is just an abbreviation for 'the limit of that sequence for n approaching infinity equals to said number'. Still it is not the same as this number, it just converges to it.
Sure, when you're writing it down, you'll never get to the sum, but the sum still exists, and can be proven mathematically. If you were to actually take an infinite number of terms (which is exactly what we've done with 0.999...), it would add up to this amount. The fact that you can't take an infinite amount of terms in reality doesn't matter. It will converge on a point after an infinite number of terms, and because it's a recurring decimal, WE HAVE an infinite number of terms.
 

FolkLikePanda

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havass said:
If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...

In conclusion, I have just proven 1 = 0.9999...
...you have answered a question that has haunted me ever since Year 6
 

Piflik

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Coldie said:
Piflik said:
The difference is 0.000000000....000001 which is exactly 0.000000000....000001 and it is not 0.
Now you're just trolling. Stop making up nonsense and start studying math from the very beginning.
Read Lyx' post on this page...I am not trolling...there is a difference between 0.99999...and one. As Lyx said, these irrational numbers are infinite for just the reason that there is always a difference between any representation possible with limited amount of decimals and that numbers real value for example pi that you injected in this discussion. So there is a difference between the 0.99999... and any possible representation of that value, including 1.
 

grammarye

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Lyx said:
2. Next, we want to store this in decimal. So, a num/1 system with digits ranging from 0-9. Problem is that this system cannot accurately represent 1/3. No matter how many fractions we add, we'll never reach the accuracy of 1/3. Therefore, the value changes. If we have to discard something, we lose something. A difference is a difference. 0.333... therefore cannot be equal to 1/3, for the plain simple reason that this isn't a matter of digits... decimal simply cannot represent this value accurately, no matter what you do to the digits.[/b]
You would seem to have missed the point where it's infinite. The entire point of the recurring nature of it is 'turtles all the way down' - it is an acceptance that a decimal alone cannot precisely represent some fractions, but if we state 'and this just keeps going, tending towards 1/3' then we have precisely expressed this.

Unless you're feeling like challenging the basic cores of mathematical expression, in which case, well 1=2 because I says so. After all, it's just a sequence of concepts that people have agreed upon.
Piflik said:
As Lyx said, these irrational numbers are infinite for just the reason that there is always a difference between any representation possible with limited amount of decimals and that numbers real value for example pi that you injected in this discussion. So there is a difference between the 0.99999... and any possible representation of that value, including 1.
They're not limited. That is the entire point of the notion of recurring decimal places. This is an abstract concept, not what you can fit into a computer's buffer or something.
 

Redingold

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Rainforce said:
Redingold said:
Really? I suppose you have some strong, alien mathematics where 0.999... is not equal to 1, then. Could I see it?
No, but our interpretation has severe problems with some things, because we pretty much always operate/think in modeled enviroments that are far from reality. So I guess we still have much to learn.
Are you saying maths should take after reality more?

That's the most stupid thing I've ever heard.

There are so many concepts with no physical reality, such as complex numbers, n-dimensional matrices and a lot of topology.

If we have to model maths on the real world, we lose all of these, including the useful applications they have.

So please tell me I've misinterpreted you.
 

Orekoya

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Another [http://www.escapistmagazine.com/forums/read/18.85789-Interesting-fact-0-999-1?page=1] .999999 [http://www.escapistmagazine.com/forums/read/18.104684-0-99-Repeating-1] of these threads [http://www.escapistmagazine.com/forums/read/18.174837-Infinity-1?page=1].
 

Piflik

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grammarye said:
Lyx said:
2. Next, we want to store this in decimal. So, a num/1 system with digits ranging from 0-9. Problem is that this system cannot accurately represent 1/3. No matter how many fractions we add, we'll never reach the accuracy of 1/3. Therefore, the value changes. If we have to discard something, we lose something. A difference is a difference. 0.333... therefore cannot be equal to 1/3, for the plain simple reason that this isn't a matter of digits... decimal simply cannot represent this value accurately, no matter what you do to the digits.[/b]
You would seem to have missed the point where it's infinite. The entire point of the recurring nature of it is 'turtles all the way down' - it is an acceptance that a decimal alone cannot precisely represent some fractions, but if we state 'and this just keeps going, tending towards 1/3' then we have precisely expressed this.

Unless you're feeling like challenging the basic cores of mathematical expression, in which case, well 1=2 because I says so. After all, it's just a sequence of concepts that people have agreed upon.
Piflik said:
As Lyx said, these irrational numbers are infinite for just the reason that there is always a difference between any representation possible with limited amount of decimals and that numbers real value for example pi that you injected in this discussion. So there is a difference between the 0.99999... and any possible representation of that value, including 1.
They're not limited. That is the entire point of the notion of recurring decimal places. This is an abstract concept, not what you can fit into a computer's buffer or something.
You missed the point...regardless of how many 3s you write (be it infinite, aleph one or more) there will always be a difference to 1/3 and that is precisely the reason why these numbers need infinity to be represented.
 

Rainforce

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Redingold said:
Rainforce said:
Redingold said:
Really? I suppose you have some strong, alien mathematics where 0.999... is not equal to 1, then. Could I see it?
No, but our interpretation has severe problems with some things, because we pretty much always operate/think in modeled enviroments that are far from reality. So I guess we still have much to learn.
Are you saying maths should take after reality more?

That's the most stupid thing I've ever heard.

There are so many concepts with no physical reality, such as complex numbers, n-dimensional matrices and a lot of topology.

If we have to model maths on the real world, we lose all of these, including the useful applications they have.

So please tell me I've misinterpreted you.
I guess you got a lot of things right.
So in other words: mathematical arguments can as well be invalid, because we also define the logik on which our arguments are based.
EDIT: that makes it "true" in itself, but not from outside.
EDIT2: and now I feel regret because I applied philosophy on mathematics again. XD
 

Athinira

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Piflik said:
If you want to allow infinity, you also have to allow infinitesimal values, since that is nothing other than 1/infinity.
There is a difference between using numbers with infinite lengths (eg. infinite decimals) and infinite values.

The Real Number system allows the former but not the latter.