- Valthorn ( @Valthorn@feddit.nu ) English64•6 months ago
x=.9999…
10x=9.9999…
Subtract x from both sides
9x=9
x=1
There it is, folks.
- barsoap ( @barsoap@lemm.ee ) English45•6 months 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 the difference, or rather non-difference, between 0.333… and ⅓, then ask them what multiplying each by 3 is.
- ColeSloth ( @ColeSloth@discuss.tchncs.de ) English2•6 months 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 🇺🇦 ( @rockerface@lemm.ee ) English6•6 months ago
pi isn’t even a fraction. like, it’s actually an important thing that it isn’t
- I_am_10_squirrels ( @I_am_10_squirrels@beehaw.org ) English2•6 months ago
pi=c/d
it’s a fraction, just not with integers, so it’s not rational, so it’s not a fraction.
- sp3ctr4l ( @sp3tr4l@lemmy.zip ) English1•6 months 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.
- DeanFogg ( @DeanFogg@lemm.ee ) English1•6 months ago
Cut a banana into thirds and you lose material from cutting it hence .9999
- wholookshere ( @wholookshere@lemmy.blahaj.zone ) English3•6 months ago
That’s not how fractions and math work though.
- yetAnotherUser ( @yetAnotherUser@discuss.tchncs.de ) English6•6 months 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 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.
- Valthorn ( @Valthorn@feddit.nu ) English2•6 months 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.
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.
- yetAnotherUser ( @yetAnotherUser@discuss.tchncs.de ) English3•6 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.
- sp3ctr4l ( @sp3tr4l@lemmy.zip ) English2•6 months 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.
- ColeSloth ( @ColeSloth@discuss.tchncs.de ) English2•6 months 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 ( @blue@ttrpg.network ) English9•6 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.
- Jerkface (any/all) ( @jerkface@lemmy.ca ) English55•6 months ago
Okay, but it equals one.
- aname ( @lauha@lemmy.one ) English8•6 months ago
No, it equals 0.999…
- Karcinogen ( @Karcinogen@discuss.tchncs.de ) English46•6 months ago
2/9 = 0.222… 7/9 = 0.777…
0.222… + 0.777… = 0.999… 2/9 + 7/9 = 1
0.999… = 1
No, it equals 1.
- Crozekiel ( @Crozekiel@lemmy.zip ) English14•6 months ago
That’s the best explanation of this I’ve ever seen, thank you!
- YTG123 ( @ytg@sopuli.xyz ) English14•6 months ago
Similarly, 1/3 = 0.3333…
So 3 times 1/3 = 0.9999… but also 3/3 = 1Another nice one:
Let x = 0.9999… (multiply both sides by 10)
10x = 9.99999… (substitute 0.9999… = x)
10x = 9 + x (subtract x from both sides)
9x = 9 (divide both sides by 9)
x = 1 - ColeSloth ( @ColeSloth@discuss.tchncs.de ) English2•6 months ago
If you can’t do it without fractions or a … then it can’t be done.
- WldFyre ( @WldFyre@lemm.ee ) English5•6 months ago
1/3=0.333…
2/3=0.666…
3/3=0.999…=1
- ColeSloth ( @ColeSloth@discuss.tchncs.de ) English2•6 months ago
Fractions and base 10 are two different systems. You’re only approximating what 1/3 is when you write out 0.3333…
The … is because you can’t actually make it correct in base 10.
- WldFyre ( @WldFyre@lemm.ee ) English7•6 months ago
The fractions are still in base 10 lmfao literally what the fuck are you talking about and where are getting this from?
You keep getting basic shit wrong, and it makes you look dumb. Stop talking and go read a wiki.
- Jerkface (any/all) ( @jerkface@lemmy.ca ) English9•6 months ago
THAT’S EXACTLY WHAT I SAID.
- pruwyben ( @pruwybn@discuss.tchncs.de ) English49•6 months ago
I thought the muscular guys were supposed to be right in these memes.
- Kairos ( @LodeMike@lemmy.today ) English27•6 months ago
0.9 is literally equal to 1
- jonsnothere ( @jonsnothere@beehaw.org ) English1•6 months ago
0.9 is most definitely not equal to 1
- mathemachristian[he] ( @mathemachristian@lemm.ee ) English9•6 months ago
Hence the overbar. Lemmy should support LaTeX for real though
- jonsnothere ( @jonsnothere@beehaw.org ) English7•6 months ago
Oh, that’s not even showing as a missing character, to me it just looks like 0.9
At least we agree 0.99… = 1
- Kairos ( @LodeMike@lemmy.today ) English3•6 months ago
Oh lol its rendering as HTML for you.
- magic_lobster_party ( @magic_lobster_party@kbin.run ) 25•6 months ago
If 0.999… < 1, then that must mean there’s an infinite amount of real numbers between 0.999… and 1. Can you name a single one of these?
- ns1 ( @ns1@feddit.uk ) English16•6 months ago
Sure 0.999…95
Just kidding, the guy on the left is correct.
- magic_lobster_party ( @magic_lobster_party@kbin.run ) 6•6 months ago
You got me
- bluewing ( @bluewing@lemm.ee ) English9•6 months ago
Meh, close enough.
- uis ( @uis@lemm.ee ) English7•6 months ago
0.(9)=0.(3)*3=1/3*3=1
- BachenBenno ( @BachenBenno@feddit.de ) English6•6 months ago
Mathematics is built on axioms that have nothing to do with numbers yet. That means that things like decimal numbers need definitions. And in the definition of decimals is literally included that if you have only nines at a certain point behind the dot, it is the same as increasing the decimal in front of the first nine by one.
- Sop ( @Sop@lemmy.blahaj.zone ) English14•6 months ago
That’s not an axiom or definition, it’s a consequence of the axioms that define arithmetic and can therefore be proven.
- Th4tGuyII ( @Th4tGuyII@fedia.io ) 5•6 months ago
0.999… / 3 = 0.333… 1 / 3 = 0.333… Ergo 1 = 0.999…
(Or see algebraic proof by @Valthorn@feddit.nu)
If the difference between two numbers is so infinitesimally small they are in essence mathematically equal, then I see no reason to not address then as such.
If you tried to make a plank of wood 0.999…m long (and had the tools to do so), you’d soon find out the universe won’t let you arbitrarily go on to infinity. You’d find that when you got to the planck length, you’d have to either round up the previous digit, resolving to 1, or stop at the last 9.
- cumskin_genocide ( @cumskin_genocide@lemm.ee ) English2•6 months ago
The only sources I trust are the ones that come from my dreams
- Unbeelievable ( @Unbeelievable@beehaw.org ) English1•6 months ago
The way I see it is the difference between equal numbers is zero.
The difference between 0.999… and 1 is 0.000…, and since the nines don’t end, the zeros don’t end, so the difference is just zero.
Meaning 0.999… = 1