I just cited myself. : science

[–] [email protected] 2 points 4 months ago (1 children)

The explanation I've seen is that ... is notation for something that can be otherwise represented as sums of infinite series.

In the case of 0.999..., it can be shown to converge toward 1 with the convergence rule for geometric series.

If |r| < 1, then:

ar + ar² + ar³ + ... = ar / (1 - r)

Thus:

0.999... = 9(1/10) + 9(1/10)² + 9(1/10)³ + ...

= 9(1/10) / (1 - 1/10)

= (9/10) / (9/10)

= 1

Just for fun, let's try 0.424242...

0.424242... = 42(1/100) + 42(1/100)² + 42(1/100)³

= 42(1/100) / (1 - 1/100)

= (42/100) / (99/100)

= 42/99

= 0.424242...

So there you go, nothing gained from that other than seeing that 0.999... is distinct from other known patterns of repeating numbers after the decimal point.

[–] [email protected] 1 points 4 months ago* (last edited 4 months ago) (1 children)

The explanation I've seen is that ... is notation for something that can be otherwise represented as sums of infinite series

The ellipsis notation generally refers to repetition of a pattern. Either ad infinitum, or up to some terminus. In this case we have a non-terminating decimal.

In the case of 0.999..., it can be shown to converge toward 1

0.999... is a real number, and not any object that can be said to converge. It is exactly 1.

So there you go, nothing gained from that other than seeing that 0.999... is distinct from other known patterns of repeating numbers after the decimal point

In what way is it distinct?
And what is a 'repeating number'? Did you mean 'repeating decimal'?

[–] [email protected] 1 points 4 months ago* (last edited 4 months ago) (1 children)

The ellipsis notation generally refers to repetition of a pattern.

Ok. In mathematical notation/context, it is more specific, as I outlined.

This technicality is often brushed over or over simplified by math teachers and courses until or unless you take some more advanced courses.

Context matters, here's an example:

Generally, pdf denotes the file format specific to adobe reader, while in the context of many modern online videos/discussions, it has become a colloquialism to be able to discuss (accused or confirmed) pedophiles and be able to avoid censorship or demonetization.

0.999... is a real number, and not any object that can be said to converge. It is exactly 1.

Ok. Never said 0.999... is not a real number. Yep, it is exactly 1 because solving the equation it truly represents, a geometric series, results in 1. This solution is obtained using what is called the convergence theorem or rule, as I outlined.

In what way is it distinct?

0.424242... solved via the convergence theorem simply results in itself, as represented in mathematical nomenclature.

0.999... does not again result in 0.999..., but results to 1, a notably different representation that causes the entire discussion in this thread.

And what is a 'repeating number'? Did you mean 'repeating decimal'?

I meant what I said: "know patterns of repeating numbers after the decimal point."

Perhaps I should have also clarified known finite patterns to further emphasize the difference between rational and irrational numbers.

EDIT: You used a valid and even more mathematically esoteric method to demonstrate the same thing I demonstrated elsewhere in this thread, I have no idea why you are taking issue with what I've said.

[–] [email protected] 1 points 4 months ago (1 children)

Ok. In mathematical notation/context, it is more specific, as I outlined.

It is not. You will routinely find it used in cases where your explanation does not apply, such as to denote the contents of a matrix.

Furthermore, we can define real numbers without defining series. In such contexts, your explanation also doesn't work until we do defines series of rational numbers.

Ok. Never said 0.999... is not a real number

In which case it cannot converge to anything on account of it not being a function or any other things that can be said to converge.

because solving the equation it truly represents, a geometric series, results in 1

A series is not an equation.

This solution is obtained using what is called the convergence theorem

What theorem? I have never heard of 'the convergence theorem'.

0.424242... solved via the convergence theorem simply results in itself

What do you mean by 'solving' a real number?

0.999... does not again result in 0.999..., but results to 1

In what way does it not 'result in 0.999...' when 0.999... = 1?

You seem to not understand what decimals are, because while decimals (which are representations of real numbers) '0.999...' and '1' are different, they both refer to the same real number. We can use expressions '0.999...' and '1' interchangeably in the context of base 10. In other bases, we can easily also find similar pairs of digital representations that refer to the same numbers.

I meant what I said: "know patterns of repeating numbers after the decimal point."

What we have after the decimal point are digits. OTOH, sure, we can treat them as numbers, but still, this is not a common terminology. Furthermore, 'repeating number' is not a term in any sort of commonly-used terminology in this context.

The actual term that you were looking for is 'repeating decimal'.

Perhaps I should have also clarified known finite patterns to further emphasize the difference between rational and irrational numbers

No irrational number can be represented by a repeating decimal.

[–] [email protected] 1 points 4 months ago (1 children)

https://www2.kenyon.edu/Depts/Math/Paquin/GeomSeriesCalcB.pdf

Here's a standard introduction to the concept of the Convergence/Divergence Theorem of Geometric Series, starts on page 2.

Its quite common for this to be referred to as the convergence test or rule or theorem by teachers and TA's.

[–] [email protected] 1 points 4 months ago* (last edited 4 months ago)

Now, ask yourself this question, 'is 0.999..., or any real number for that matter, a series?'. The answer to that question is 'no'.

You seem to be extremely confused, and think that the terms 'series' and 'the sum of a series' mean the same thing. They do not. 0.999... is the sum of the series 9/10+9/100+9/1000+..., and not a series itself.

EDIT: Also, the author does abuse the notations somewhat when she says '1+1/2+1/4 = 2' is a geometric series, as the geometric series 1+1/2+1/4+... does not equal 2, because a series is either just a formal sum, a sequence of its terms, or, in German math traditions, a sequence of its partial sums. It is the sum of the series 1+1/2+1/4+... that is equal to 2. The confusion is made worse by the fact that sums of series and the series themselves are often denoted in the same way. However, again, those are different things.
Would you mind providing a snippet with the definition of the term 'series' that she provides?

EDIT 2: Notably, that document has no theorem that is called 'convergence theorem' or 'the convergence theorem'. The only theorem that is present there is the one on convergence and divergence of geometric series.

[–] [email protected] 5 points 4 months ago* (last edited 4 months ago) (1 children)

Unfortunately not an ideal proof.

It makes certain assumptions:

That a number 0.999... exists and is well-defined
That multiplication and subtraction for this number work as expected

Similarly, I could prove that the number which consists of infinite 9's to the left of the decimal separator is equal to -1:

...999.0 = x
...990.0 = 10x

Calculate x - 10x:

x - 10x = ...999.0 - ...990.0
-9x = 9
x = -1

And while this is true for 10-adic numbers, it is certainly not true for the real numbers.

[–] [email protected] 1 points 4 months ago* (last edited 4 months ago) (1 children)

While I agree that my proof is blunt, yours doesn't prove that .999... is equal to -1. With your assumption, the infinite 9's behave like they're finite, adding the 0 to the end, and you forgot to move the decimal point in the beginning of the number when you multiplied by 10.

x=0.999...999

10x=9.999...990 assuming infinite decimals behave like finite ones.

Now x - 10x = 0.999...999 - 9.999...990

-9x = -9.000...009

x = 1.000...001

Thus, adding or subtracting the infinitesimal makes no difference, meaning it behaves like 0.

Edit: Having written all this I realised that you probably meant the infinitely large number consisting of only 9's, but with infinity you can't really prove anything like this. You can't have one infinite number being 10 times larger than another. It's like assuming division by 0 is well defined.

0a=0b, thus

a=b, meaning of course your ...999 can equal -1.

Edit again: what my proof shows is that even if you assume that .000...001≠0, doing regular algebra makes it behave like 0 anyway. Your proof shows that you can't to regular maths with infinite numbers, which wasn't in question. Infinity exists, the infinitesimal does not.

[–] [email protected] 2 points 4 months ago

Yes, but similar flaws exist for your proof.

The algebraic proof that 0.999... = 1 must first prove why you can assign 0.999... to x.

My "proof" abuses algebraic notation like this - you cannot assign infinity to a variable. After that, regular algebraic rules become meaningless.

The proper proof would use the definition that the value of a limit approaching another value is exactly that value. For any epsilon > 0, 0.999.. will be within the epsilon environment of 1 (= the interval 1 ± epsilon), therefore 0.999... is 1.

[+] [email protected] 1 points 4 months ago

[deleted]

[+] [email protected] -14 points 4 months ago (4 children)

X=.5555...

10x=5.5555...

Subtract x from both sides.

9x=5

X=1 .5555 must equal 1.

There it isn't. Because that math is bullshit.

[–] [email protected] 3 points 4 months ago* (last edited 4 months ago)

Lol what? How did you conclude that if 9x = 5 then x = 1? Surely you didn't pass algebra in high school, otherwise you could see that getting x from 9x = 5 requires dividing both sides by 9, which yields x = 5/9, i.e. 0.555... = 5/9 since x = 0.555....

Also, you shouldn't just use uppercase X in place of lowercase x or vice versa. Case is usually significant for variable names.

[–] [email protected] 10 points 4 months ago* (last edited 4 months ago)

?
Where did you get 9x=5 -> x=1
and 5/9 is 0.555... so it checks out.

[–] [email protected] 16 points 4 months ago

x = 5/9 is not 9/9. 5/9 = .55555...

You're proving that 0.555... equals 5/9 (which it does), not that it equals 1 (which it doesn't).

It's absolutely not the same result as x = 0.999... as you claim.

[–] [email protected] 7 points 4 months ago

Quick maffs

[–] [email protected] 48 points 4 months ago (1 children)

I was taught that if 0.9999... didn't equal 1 there would have to be a number that exists between the two. Since there isn't, then 0.9999...=1

[–] [email protected] 3 points 4 months ago

Not even a number between, but there is no distance between the two. There is no value X for 1-x = 0.9~

We can't notate 0.0~ ....01 in any way.

[–] [email protected] 23 points 4 months ago (2 children)

Divide 1 by 3: 1÷3=0.3333...

Multiply the result by 3 reverting the operation: 0.3333... x 3 = 0.9999.... or just 1

0.9999... = 1

[+] [email protected] -9 points 4 months ago (2 children)

You're just rounding up an irrational number. You have a non terminating, non repeating number, that will go on forever, because it can never actually get up to its whole value.

[–] [email protected] 5 points 4 months ago

non repeating

it's literally repeating

[–] [email protected] 11 points 4 months ago (1 children)

1/3 is a rational number, because it can be depicted by a ratio of two integers. You clearly don't know what you're talking about, you're getting basic algebra level facts wrong. Maybe take a hint and read some real math instead of relying on your bad intuition.

[+] [email protected] -11 points 4 months ago (2 children)

1/3 is rational.

.3333....... is not. You can't treat fractions the same as our base 10 number system. They don't all have direct conversions. Hence, why you can have a perfect fraction of a third, but not a perfect 1/3 written out in base 10.

[–] [email protected] 5 points 4 months ago (1 children)

.333... is rational.

at least we finally found your problem: you don't know what rational and irrational mean. the clue is in the name.

[–] [email protected] 1 points 4 months ago (1 children)

TBH the name is a bit misleading. Same for "real" numbers. And oh so much more so for "normal numbers".

[–] [email protected] 3 points 4 months ago (1 children)

not really. i get it because we use rational to mean logical, but that's not what it means here. yeah, real and normal are stupid names but rational numbers are numbers that can be represented as a ratio of two numbers. i think it's pretty good.

[–] [email protected] 1 points 4 months ago (1 children)

I know all of that, but it's still misleading. It's not a dumb name by any means, but it still causes confusion often (as evidenced by many comments here)

[–] [email protected] 3 points 4 months ago

fair enough, but i think the confusion for that commenter comes from a misunderstanding of the definition of the mathematical concept rather than the meaning of the English word. they just think irrational numbers are those that have infinite decimal digits, which is not the definition.

[–] [email protected] 14 points 4 months ago

0.333... exactly equals 1/3 in base 10. What you are saying is factually incorrect and literally nonsense. You learn this in high school level math classes. Link literally any source that supports your position.

[+] [email protected] -20 points 4 months ago (2 children)

In this context, yes, because of the cancellation on the fractions when you recover.

1/3 x 3 = 1

I would say without the context, there is an infinitesimal difference. The approximation solution above essentially ignores the problem which is more of a functional flaw in base 10 than a real number theory issue

[–] [email protected] 9 points 4 months ago

This seems to be conflating 0.333...3 with 0.333... One is infinitesimally close to 1/3, the other is a decimal representation of 1/3. Indeed, if 1-0.999... resulted in anything other than 0, that would necessarily be a number with more significant digits than 0.999... which would mean that the ... failed to be an infinite repetition.

[–] [email protected] 24 points 4 months ago* (last edited 4 months ago) (1 children)

The context doesn't make a difference

In base 10 --> 1/3 is 0.333...

In base 12 --> 1/3 is 0.4

But they're both the same number.

Base 10 simply is not capable of displaying it in a concise format. We could say that this is a notation issue. No notation is perfect. Base 10 has some confusing implications

[–] [email protected] 70 points 4 months ago* (last edited 4 months ago) (4 children)

Somehow I have the feeling that this is not going to convince people who think that 0.9999... /= 1, but only make them madder.

Personally I like to point to the difference, or rather non-difference, between 0.333... and ⅓, then ask them what multiplying each by 3 is.

[–] [email protected] 1 points 1 month ago

The thing is 0.333... And 1/3 represent the same thing. Base 10 struggles to represent the thirds in decimal form. You get other decimal issues like this in other base formats too

(I think, if I remember correctly. Lol)

[–] [email protected] 0 points 4 months ago (1 children)

Cut a banana into thirds and you lose material from cutting it hence .9999

[–] [email protected] 3 points 4 months ago

That’s not how fractions and math work though.

[–] [email protected] 1 points 4 months ago

Oh shit, don't think I saw that before. That makes it intuitive as hell.

[+] [email protected] -8 points 4 months ago (5 children)

I'd just say that not all fractions can be broken down into a proper decimal for a whole number, just like pie never actually ends. We just stop and say it's close enough to not be important. Need to know about a circle on your whiteboard? 3.14 is accurate enough. Need the entire observable universe measured to within a single atoms worth of accuracy? It only takes 39 digits after the 3.

[–] [email protected] 1 points 4 months ago (1 children)

There are a lot of concepts in mathematics which do not have good real world analogues.

i, the _imaginary number_for figuring out roots, as one example.

I am fairly certain you cannot actually do the mathematics to predict or approximate the size of an atom or subatomic particle without using complex algebra involving i.

It's been a while since I watched the entire series Leonard Susskind has up on youtube explaining the basics of the actual math for quantum mechanics, but yeah I am fairly sure it involves complex numbers.

[–] [email protected] 0 points 4 months ago* (last edited 4 months ago)

i has nice real world analogues in the form of rotations by pi/2 about the origin (though this depends a little bit on what you mean by "real world analogue").

Since i=exp(ipi/2), if you take any complex number z and write it in polar form z=rexp(it), then multiplication by i yields a rotation of z by pi/2 about the origin because zi=rexp(it)exp(ipi/2)=rexp(i(t+pi/2)) by using rules of exponents for complex numbers.

More generally since any pair of complex numbers z, w can be written in polar form z=rexp(it), w=uexp(iv) we have wz=(ru)exp(i(t+v)). This shows multiplication of a complex number z by any other complex number w can be thought of in terms of rotating z by the angle that w makes with the x axis (i.e. the angle v) and then scaling the resulting number by the magnitude of w (i.e. the number u)

Alternatively you can get similar conclusions by Demoivre's theorem if you do not like complex exponentials.

[–] [email protected] 1 points 4 months ago

pie never actually ends

I want to go to there.

[–] [email protected] 3 points 4 months ago* (last edited 4 months ago)

Pi isn't a fraction (in the sense of a rational fraction, an algebraic fraction where the numerator and denominator are both polynomials, like a ratio of 2 integers) – it's an irrational number, i.e. a number with no fractional form; as opposed to rational numbers, which are defined as being able to be expressed as a fraction. Furthermore, π is a transcendental number, meaning it's never a solution to f(x) = 0, where f(x) is a non-zero finite-degree polynomial expression with rational coefficients. That's like, literally part of the definition. They cannot be compared to rational numbers like fractions.

Every rational number (and therefore every fraction) can be expressed using either repeating decimals or terminating decimals. Contrastly, irrational numbers only have decimal expansions which are both non-repeating and non-terminating.

Since |r|<1 → ∑[n=1, ∞] arⁿ = ar/(1-r), and 0.999... is equivalent to that sum with a = 9 and r = 1/10 (visually, 0.999... = 9(0.1) + 9(0.01) + 9(0.001) + ...), it's easy to see after plugging in, 0.999... = ∑[n=1, ∞] 9(1/10)ⁿ = 9(1/10) / (1 - 1/10) = 0.9/0.9 = 1). This was a proof present in Euler's Elements of Algebra.

[–] [email protected] 4 points 4 months ago

The problem is, that's exactly what the ... is for. It is a little weird to our heads, granted, but it does allow the conversion. 0.33 is not the same thing as 0.333... The first is close to one third. The second one is one third. It's how we express things as a decimal that don't cleanly map to base ten. It may look funky, but it works.

[–] [email protected] 10 points 4 months ago (1 children)

pi isn't even a fraction. like, it's actually an important thing that it isn't

[–] [email protected] 2 points 4 months ago

pi=c/d

it's a fraction, just not with integers, so it's not rational, so it's not a fraction.

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