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
Comment on I just cited myself.
Valthorn@feddit.nu 1 week ago
x=.9999… 10x=9.9999… Subtract x from both sides 9x=9 x=1
There it is, folks.
Blum0108@lemmy.world 1 week ago
wieson@feddit.org 6 days 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.
Shampiss@sh.itjust.works 1 week ago
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
ArchAengelus@lemmy.dbzer0.com 1 week ago
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
Shampiss@sh.itjust.works 1 week ago
The context doesn’t make a difference
In base 10 --> 1/3 is 0.333…
In base 12 --> 1/3 is 3
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
ColeSloth@discuss.tchncs.de 6 days ago
They’re different numbers. Base 10 isn’t perfect and can’t do everything just right, so you end up with irrational numbers that go on forever, sometimes.
chaonaut@lemmy.world 1 week ago
This seems to be conflating
0.333…3with0.333…One is infinitesimally close to 1/3, the other is a decimal representation of 1/3. Indeed, if1-0.999…resulted in anything other than 0, that would necessarily be a number with more significant digits than0.999…which would mean that the…failed to be an infinite repetition.
ColeSloth@discuss.tchncs.de 6 days ago
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.
WldFyre@lemm.ee 6 days ago
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.
ColeSloth@discuss.tchncs.de 6 days ago
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.
pyre@lemmy.world 6 days ago
non repeating
it’s literally repeating
yetAnotherUser@discuss.tchncs.de 6 days ago
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 is equal to -1:
...999 = x ...990 = 10x Calculate x - 10x: x - 10x = ...999 - ...990 -9x = 9 x = -1
And while this is true for 10-adic numbers, it is certainly not true for the real numbers.
Valthorn@feddit.nu 6 days ago
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.
a0=b0, thus
a=b, meaning of course your …999 can equal -1.
yetAnotherUser@discuss.tchncs.de 6 days 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.
sp3tr4l@lemmy.zip 5 days ago
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.
Tomorrow_Farewell@hexbear.net 4 days ago
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 as 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’?sp3tr4l@lemmy.zip 4 days ago
The ellipsis notation generally refers to repetition of a pattern.
Ok. In mathematical notation/context, it is more specific, as I outlined.
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.
Tomorrow_Farewell@hexbear.net 3 days ago
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.
ColeSloth@discuss.tchncs.de 6 days ago
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.
blue@ttrpg.network 6 days 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.
Redex68@lemmy.world 6 days ago
? Where did you get 9x=5 -> x=1 and 5/9 is 0.555… so it checks out.
lazyViking@lemmy.world 6 days ago
Quick maffs
force@lemmy.world 6 days ago
Lol what? How did you conclude that if
9x = 5thenx = 1? Surely you didn’t pass algebra in high school, otherwise you could see that gettingxfrom9x = 5requires dividing both sides by 9, which yieldsx = 5/9, i.e.0.555… = 5/9.
barsoap@lemm.ee 1 week ago
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 0.333… vs. ⅓, then ask them what multiplying each by 3 is.
ColeSloth@discuss.tchncs.de 6 days ago
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.
rockerface@lemm.ee 6 days ago
pi isn’t even a fraction. like, it’s actually an important thing that it isn’t
I_am_10_squirrels@beehaw.org 6 days ago
pi=c/d
it’s a fraction, just not with integers, so it’s not rational, so it’s not a fraction.
Wandering_Uncertainty@lemmy.world 6 days 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.
force@lemmy.world 6 days ago
Pi isn’t a fraction – it’s an irrational number, i.e. a number with no fractional form in integer bases. Furthermore, it’s a transendental 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 integer ratios/fractions.
Since
|r|<1 => ∑[n=1, ∞] arⁿ = ar/(1-r), and0.999…is that sum witha = 9andr = 1/10(visually,0.999… = 9(0.1) + 9(0.01) + 9(0.001) + …), it’s easy to see after plugging in,0.999… = 9(1/10) / (1 - 1/10) = 0.9/0.9 = 1). This was a proof in Euler’s Elements of Algebra.sp3tr4l@lemmy.zip 5 days ago
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.
myslsl@lemmy.world 3 days ago
i has nice real world analogues in the form of rotations by pi/2 about the origin.
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 though 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.
ccunning@lemmy.world 6 days ago
I want to go to there.
DeanFogg@lemm.ee 6 days ago
Cut a banana into thirds and you lose material from cutting it hence .9999
wholookshere@lemmy.blahaj.zone 5 days ago
That’s not how fractions and math work though.
SkyezOpen@lemmy.world 6 days ago
Oh shit, don’t think I saw that before. That makes it intuitive as hell.