I just cited myself.
I just cited myself.
231 comments
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x=.9999...
10x=9.9999...
Subtract x from both sides
9x=9
x=1
There it is, folks.
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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.
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Oh shit, don't think I saw that before. That makes it intuitive as hell.
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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)
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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.
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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
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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.
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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
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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
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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:
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...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.
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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.
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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.
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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'?
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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.
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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.
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?
Where did you get 9x=5 -> x=1
and 5/9 is 0.555... so it checks out. -
Quick maffs
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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... = 5/9sincex = 0.555....Also, you shouldn't just use uppercase
Xin place of lowercasexor vice versa. Case is usually significant for variable names.
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I thought the muscular guys were supposed to be right in these memes.
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Nah. They are supposed to not care about stuff and just roll with it without any regrets.
It's just like the wojak crying with the mask on, but not crying behind it.
There's plenty of cases of memes where the giga chad is just plainly wrong, but they just don't care. But it's not supposed to be in a troll way. The giga chad applies what it believes in. If you want a troll, there's troll face, who speak with the confidence of a giga chad, but know he is bullshiting
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0.9<overbar.> is literally equal to 1
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There's a Real Analysis proof for it and everything.
Basically boils down to
- If 0.(9) != 1 then there must be some value between 0.(9) and 1.
- We know such a number cannot exist, because for any given discrete value (say 0.999...9) there is a number (0.999...99) that is between that discrete value and 0.(9)
- Therefore, no value exists between 0.(9) and 1.
- So 0.(9) = 1
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That actually makes sense, thank you.
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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?
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Sure 0.999...95
Just kidding, the guy on the left is correct.
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You got me
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(0.999... + 1) / 2
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That number happens to be exactly 1
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This is why we can't have nice things like dependable protection from fall damage while riding a boat in Minecraft.
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Reals are just point cores of dressed Cauchy sequences of naturals (think of it as a continually constructed set of narrowing intervals "homing in" on the real being constructed). The intervals shrink at the same rate generally.
1!=0.999 iff we can find an n, such that the intervals no longer overlap at that n. This would imply a layer of absolute infinite thinness has to exist, and so we have reached a contradiction as it would have to have a width smaller than every positive real (there is no smallest real >0).
Therefore 0.999...=1.
However, we can argue that 1 is not identity to 0.999... quite easily as they are not the same thing.
This does argue that this only works in an extensional setting (which is the norm for most mathematics).
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Easiest way to prove it:
1 = 3/3 = 1/3 3 = 0.333... 3 = 0.999...
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Ehh, completed infinities give me wind...
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Are we still doing this 0.999.. thing? Why, is it that attractive?
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People generally find it odd and unintuitive that it's possible to use decimal notation to represent 1 as .9~ and so this particular thing will never go away. When I was in HS I wowed some of my teachers by doing proofs on the subject, and every so often I see it online. This will continue to be an interesting fact for as long as decimal is used as a canonical notation.
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Welp, I see. Still, this is way too much recurting of a pattern.
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The rules of decimal notation don't sipport infinite decimals properly. In order for a 9 to roll over into a 10, the next smallest decimal needs to roll over first, therefore an infinite string of anything will never resolve the needed discrete increment.
Thus, all arguments that 0.999... = 1 must use algebra, limits, or some other logic beyond decimal notation. I consider this a bug with decimals, and 0.999... = 1 to be a workaround.
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don't sipport infinite decimals properly
Please explain this in a way that makes sense to me (I'm an algebraist). I don't know what it would mean for infinite decimals to be supported "properly" or "improperly". Furthermore, I'm not aware of any arguments worth taking seriously that don't use logic, so I'm wondering why that's a criticism of the notation.
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i don't think any number system can be safe from infinite digits. there's bound to be some number for each one that has to be represented with them. it's not intuitive, but that's because infinity isn't intuitive. that doesn't mean there's a problem there though. also the arguments are so simple i don't understand why anyone would insist that there has to be a difference.
for me the simplest is:
1/3 = 0.333...
so
3×0.333... = 3×1/3
0.999... = 3/3
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Meh, close enough.
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I wish computers could calculate infinity
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Computers can calculate infinite series as well as anyone else
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As long as you have it forget the previous digit, you can bring up a new digit infinitely
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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.
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That's not how it's defined. 0.99.. is the limit of a sequence and it is precisely 1. 0.99.. is the summation of infinite number of numbers and we don't know how to do that if it isn't defined. (0.9 + 0.09 + 0.009...) It is defined by the limit of the partial sums, 0.9, 0.99, 0.999... The limit of this sequence is 1. Sorry if this came out rude. It is more of a general comment.
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I study mathematics at university and I remember it being in the definition, but since it follows from the sum's limit anyways it probably was just there for claritie's sake. So I guess we're both right...
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The decimals '0.999...' and '1' refer to the real numbers that are equivalence classes of Cauchy sequences of rational numbers (0.9, 0.99, 0.999,...) and (1, 1, 1,...) with respect to the relation R: (aRb) <=> (lim(a_n-b_n) as n->inf, where a_n and b_n are the nth elements of sequences a and b, respectively).
For a = (1, 1, 1,...) and b = (0.9, 0.99, 0.999,...) we have lim(a_n-b_n) as n->inf = lim(1-sum(9/10k) for k from 1 to n) as n->inf = lim(1/10n) as n->inf = 0. That means that (1, 1, 1,...)R(0.9, 0.99, 0.999,...), i.e. that these sequences belong to the same equivalence class of Cauchy sequences of rational numbers with respect to R. In other words, the decimals '0.999...' and '1' refer to the same real number. QED.
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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.
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Math doesn't care about physical limitations like the planck length.
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Any real world implementation of maths (such as the length of an object) would definitely be constricted to real world parameters, and the lowest length you can go to is the Planck length.
But that point wasn't just to talk about a plank of wood, it was to show how little difference the infinite 9s in 0.999... make.
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Except it isn't infinitesimally smaller at all. 0.999... is exactly 1, not at all less than 1. That's the power of infinity. If you wanted to make a wooden board exactly 0.999... m long, you would need to make a board exactly 1 m long (which presents its own challenges).
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It is mathematically equal to one, but it isn't physically one. If you wrote out 0.999... out to infinity, it'd never just suddenly round up to 1.
But the point I was trying to make is that I agree with the interpretation of the meme in that the above distinction literally doesn't matter - you could use either in a calculation and the answer wouldn't (or at least shouldn't) change.
That's pretty much the point I was trying to make in proving how little the difference makes in reality - that the universe wouldn't let you explore the infinity between the two, so at some point you would have to round to 1m, or go to a number 1x planck length below 1m.
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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
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I can honestly say I learned something from the comment section. I was always taught the .9 repeating was not equal to 1 but separated by imaginary i ... Or infinitely close to 1 without becoming 1.
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The only sources I trust are the ones that come from my dreams
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Even the hyperreal numbers *R, which include infinitesimals, define 1 == .999...
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Sort of how 0.0000001 = 0
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No, not like that.
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231 comments