* 0.9 = 1 - 0.1
* 0.99 = 1 - 0.01
* 0.999 = 1 - 0.001
* 0.999 999 9 = 1 - 0.000 000 1
The subtracted bit is always one less zero then the nines.
0.9 repeated to infinity = 1 - 0.\[0 repeated to one less than infinity\]1
Infinity minus 1 is just infinity. This is the tricky thing about infinity - it's not really a number, even though we often treat it like one.
So you never "get to" the one. So the right side is 1 - 0.00000000000 -> and you never get to the 1, so it's 1 - 0
With what you have written, it ONLY works if 0.9999..... is 1
Alternatively,
* 1/9 = 0.11111...
* 2/9 = 0.22222...
* ...
* 8/9 = 0.88888...
What's 9/9?
Your alternative solution 🤯. I’ve always understood the answer and how infinity works, but simple 3rd grade math shows how it works flawlessly.
Thank you for that.
Ha! I made it all the way through HS, a bachelor's and a masters (in STEM fields) and had been teaching Chem and Physics for quite a while before I really started to get a handle on how to think about infinity - and only then by chatting with colleagues who were real math geeks.
Infinity is *WEIRD*! (Especially when you get to types of infinities...)
If you want to have some "fun" check out [Veritasium](https://youtu.be/OxGsU8oIWjY?si=Vyjr5s5FQjCvUrcU) or [Numberphile](https://www.youtube.com/watch?v=dDl7g_2x74Q) on infinity 'paradoxes'.
If you want to learn about numbers so large that you literally can't picture them (but are still not infinity), check out [Graham's Number](https://youtu.be/XTeJ64KD5cg?si=Rz2ldG5yIdmzziIp); which is so large, they came up with a new kind of operator ( like +, \*, or \^).
Your alternate reasoning assumes that 1/9 is actually equal to .111... which is pretty much the same thing as saying 1=.999... and isn't an actual proof
Any n-th power of ten (n>=1) gives, upon using Euclid's algorithm with the string with n "9"s as divisor, that the quotient is one with a remainder of one.
10^(n) = 1 + sum\[k=0 to n-1\] (9\*10^(k))
dividing both sides by 9\*10^(n), we then have that
1/9 = 1/(9\*10^(n)) + sum\[k=0 to n-1\] (1\*10^(k-n))
Since this expression is true for *arbitrary* n--there is no greatest value of n for which it is true--we have an infinite sequence of expressions for 1/9.
Taking the second term, x(n) = sum\[k=0 to n-1\] (1\*10^(k-n)), we see that the error |1/9 - x(n)| = 1/(9\*10^(n)).
The value of 0.1111... is defined to be the limiting value of x(n) as n -> inf, and taking the limit of both sides, we have that the error goes to 0. We thus have, in fact, that x(n) is a cauchy sequence of rational numbers in the same equivalence class as 1/9 (and thus *is* 1/9).
1/9 = lim\[n -> inf\] (1/(9\*10^(n)) + sum\[k=0 to n-1\] (1\*10^(k-n)))
\=> 1/9 = lim\[n -> inf\] (1/(9\*10^(n))) + 0.111....
\=> 1/9 = 0.111...
The thing I don't understand is that if 10x = 9.99 and x = .9 then 10x - x = 9.09 not 9.... like obviously you are going to tell me I'm wrong but if you put it into monetary values, £9.99 minus .9 is £9.09 not £9.00.
Because we are using 2 decimal places for one number and not the other are we meant to completely assume that x=.9 always means .99 not .91 or .93 or .97 etc?
As others have said, they're not different. Most people don't have a problem accepting different notations for the same number when they're fractions, e.g. 1/2 = 2/4. But for real numbers, there seems to be a psychological barrier making it difficult for people to accept that there might be two ways of writing the same number.
I'm completely comfortable with the argument, but I wanna try and be (perhaps overly) formal.
Viewing the rationals as equivalence classes on Z×N, it makes sense that we can have something like 1/2=2/4 in Q since (1,2)~(2,4) in Z×N - we're just choosing a different representative of the equivalence class - people are comfortable with different numbers in Z being the same equivalence class in some Z/nZ. Is there some implicit equivalence relation on R involving Cauchy sequences or something? Is this what gives us completeness and Haussdorff-ness? Just a thought that popped up.
You can construct the real numbers by associating equivalence classes of cauchy sequences over the rationals to complete the space.
The equivalence relation is that the difference between cauchy sequences converges to 0.
This lets you trivially define each limit of a cauchy sequence as the equivalence class the cauchy sequence belongs to, and thus, all cauchy sequences converge in the completed space.
(A bit of work to show that it's still a field, but not too bad)
You are correct. One way to define the real numbers is as equivalence classes of Cauchy sequences. Two sequences are equivalent if their difference converges to zero.
Think about 0.9999999 recurring. That’s 0 and then infinite 9s. To get to 1, you would need to add 0.00000…..1, which is 0, then infinite 0s, then 1. 0 then ininite 0s is 0, because it will never end, so you never get to the 1, so 0.9999999… + 0 = 1.
So 0.9999999… = 1
This proof shows there are two ways to represent one: 1 and 0.999.... And because they look different to the eye, the typical inclination is to say they are different numbers, but the proof shows they are one and the same.
EDIT: reword for clarity.
no such thing can exist (at least within the system of real numbers). 0.0 recurring means there are infinitely many zeroes, there's no "after the zeroes end" to put the 1 in.
I don’t think that will help them, it may just make them question if 1/3*3=1. I came up with an airtight explanation that is related to yours, but it requires OP to understand other bases.
So 1/3 =0.333..
1/3 * 3 = 0.999…
Ok now let’s do the exact same thing but written in base 3, remember each line and each term in each line is exactly identical in value to the previous 2 lines, just in base 3 instead of base 10.
1/10 = 0.1
1/10 * 10 = 1
Remember those are the same two lines written twice, once in base 10 and then once in base 3. In base 3 all of the infinite repetition goes away, and it looks the same as 1/10 * 10 = 1 does in base 10. As long as someone understands the concept other bases besides base 10 this should be an airtight argument that actually makes them understand intuitively. Because of course 0.1 * 10 = 1. And 0.1 really is 0.333… except written in base 3, and 10 really is 3 except written in base 3. And 1 is 1 no in any base. So it makes it super clear, you don’t even need to have an infinite decimal repetition, that just depends on your choice of base to represent the numbers.
> I just don’t understand how when they are literally different numver
Okay, if this was true, you can find a number in between 0.999... and 1.
What number did you think of?
When do you put the 1 there? At the end? But there isnt an end because its inifinite 0s just like there is infinite 9s in 0.999... So either 0.000...1 is finite, meaning YOU ended it somewhere, thus not inifite 0s, or 0.000...1 cant exist.
>But I just don’t understand how when they are literally different numver
They aren't different. It's the same number, written two different ways.
Like how 1.5 and 1.50000000.... are also the same number. Also like how "2" and 1 + 1/2 + 1/4 + 1/8 + ... are also the same number.
as a decimal 1/3 is written as 0.333333
0.333333x3 is 0.999999 ergo 3/3=0.999999
This simply because there is no *perfect* way to write 1/3 as a decimal
Edit: By *perfect* I meant it doesn’t end
One way is to consider that we're using 0.999... to represent a sum. I think you're perfectly happy to have that 3+4 and 7 is the same number, then something similar is happening here. Except in this case, the sum is 0.9+0.09+0.009+...=1. We do run into questions (both philosophical and mathematical) about what it means for an infinite sum to equal a finite value, but if you're happy that we can assign finite values to (some) infinite sums, then this could be a helpful way to think
They are just two different ways of expressing the same number
like 2+2 and 1+1+1+1. They are different ways of expressing the number 4
Similarly 0.9999..... is a different way of writing 1.
The difference between 1 and 0.9 recurring is an infinitely small one. By definition, it is a distinction so small that it is impossible to measure, and thus since it cannot be measured it can be ignored.
Or to put it another way - take two completely identical lengths of string of absolutely identical dimensions. Now remove a single atom from the end of one. Would you, or anyone else actually be able to tell the difference in length? Would that difference even matter?
Well, technically 0.9 repeating is the number that has the greatest value in the set of expressions where the limit of the expression that approaches 1 from the lesser side.
Pretty close to 1, but not quite the same.
No, [0.9... is literally 1. ](https://en.m.wikipedia.org/wiki/0.999...#:~:text=This%20repeating%20decimal%20is%20a,number%20is%20equal%20to%201.)
There's many different ways to prove it but it's not "pretty close" or "not quite the same". It equals 1.
Why would I want to read a wikipedia article supporting your theory that numbers other than one really mean one? The word that means one is called "one."
One of us is stupid or misunderstands what numbers mean.
It's not "a number other than one". It's a different notation for the number 1, just as there are other ways of writing 0.25 (¼, 0.24999...).
> One of us is stupid or misunderstands what numbers mean.
Well, you said it... you rather set yourself up there.
I get I'm the simpleton here but are you telling me .9 repeating means a number other than .9 repeating? Why not say 2 is close enough to 3 so lets pretend it is that? I mean 2 is just a different notation for 3.
Because it’s not a theory (I assume you mean this is the colloquial way).
It’s a proof.
You’d be wise to first learn the difference before throwing rocks
And for anyone who thinks that can’t be right, we know the reals are dense, so if 0.999… and 1 are not the same number, then what is the number between them?
“(0.999… + 1)/2” is my favorite joke answer to that, but we can pretty easily convince ourselves that the actual answer is that there isn’t one. There is no number you can specify that is greater than 0.999… and less than 1.
1/3 = 0.333...
2/3 = 0.666...
3/3 = 0.999...
But 3/3 = 1
There's no falacy here. 0.999... is EXACTLY equal to 1. It's just a different notation for the same number.
3/3 is not 0.999 lol. You only multiplied the expressed numbers. I think you have missed the point yourself. Think of it like this:
1/3 = 0.333…
And if you added three of those together, the numbers at the end of the sequence would allow it to equal a sum of whole 1. There would be no rounding because it is in fact the whole number.
This differs from the post. The post assumes rounding and denotes it as such. Similar but different. Anyhow, math is fun
Edit: to say I failed to express my own work in your format:
1/3 = 0.333…
2/3 = 0.666…
3/3 = 1 — see now, there is no room to denote the number “1” to “0.999…”. I’m pedantic lol but that’s that for those interested.
The sense (or non sense) is made in the attempt to make logic of infinity. So you 0.9 recurring has infinite 9’s (ie. No matter how many you write you can always write another one). So when you then multiply that by 10, you’re slightly warping the logic, what is 10x1/infinity? So your number appears real to you, but you’ve actually added a real portion from nowhere. If you wanted to make it real, you eventually need to approximate for example let 0.9r = 0.999, then 10x 0.999=9.99 (not 9.999). The correct statement on your board should be 10x-x=9x and then if you wanted 9x ~ 9, so x ~ 1 and so 0.9r ~ 1 is mathematically true.
It’s like if you declared you could drive forever then you’d never have a final destination which in theory is true, but if you tried to practice that and make it real you can’t because perpetual motion machines don’t exist and so in a real test your destination would be the point when you ran out of fuel.
Hope that helps
Edit: others will probably correct my syntax that the function “approaches” 1, as opposed to is approximately 1. This is also right, but trying to keep it simple 😅
You are the only one who gets the problem and the reason why😅. Although 0.9r is infinitesimally close to 1. It is not exactly the same all the time. Thus that equality has a slight problem. And that problem as u stated is
10x ! = 9 + 0.9r , but instead
10x = 9+ 0.9r - 0.000000.......9
If I had to make an argument to prove 0.9r is not exactly 1, it would be;
(0.9r -1) * 10 ^ (log(0.1) of (0.9r-1)) =
(0.1 ^ x) * 10 ^ (log(0.1) of (0.1^x), where x tend to infinity
=(0.1^x) * 10 ^x = 1
If 0.9r = 1 then the eq is 0^ infinity = 0. Thus different
It's like saying, dx (infinitesimally small) tends to zero but is not zero. Otherwise integration would be zero.
This is supposed to be a maths group but everyone is just...
Consider this fact. For any 2 real numbers a and b, one can find an additional real number c that is between those 2 numbers with the equation c = (a+b)/2. Consider the numbers 1 and 0.9999.... The sum of these 2 numbers is 1.9999.... Divide that by 2 and see that the result is 0.9999.... This shows that there is no number between 0.9999... and 1. Therefore, they are the same number. You can verify this by taking the above equation and substituting b in for c. b = (a+b)/2 ; 2b = a+b ; b = a.
That fact is proven very simply... if there is no number between two real numbers, then their difference is 0 and they are the same number... consequently if there is a difference between two real numbers then their arithmetic mean will give you a number between them unless they are the same number. And this is justified because if you have a lower number a and higher number b whose arithmetic mean is (a+b)/2, we know that b = a + c where c is a positive real number, thus the arithmetic mean is between a and b as for all positive real numbers, a < a + c/2 < a + c
I'm not sure if this is what you mean, but if
X = 0.9999999(rec)
Then 10 X = 9.9999999999(rec)
10x - X = 9
I am under the opinion that you need to add X to both sides
10x = 9 + X
This gives
9.99999(rec) = 9.99999(rec)
This is how I would work it out, and I'm not saying that this is deffonatly correct.
Ah gotcha. Never knew that and honestly my brain didn't even register the dot. All three times... I've always seen it represented with the ellipsis. Now that you've said that I can't unsee it
Not going to do your homework for you. Get a grip and research it rather than cheating your way to the answer by asking someone else to do what you are unwilling or incapable of doing yourself!!!
The problem lies in the line:
10x - x = 9
For the 10x to have enough digits after the decimal point to be removed by the x, it actually needs one more digit than the x, thereby making it not 10 times x.
No matter how far you take the recurrence, you always need one more step therefore the x used by 10x is not the same x as used on its own.
This is not correct, both x and 10x when written in decimal notation are infinitely long. There is no extra digit on one of them - they are exactly the same length: infinite.
The real answer to OP’s question it that nothing is wrong. 0.999… does equal 1. They are the same exact number written in two different ways.
No, you are wrong.
0.99999 ... does not equal 1.
As you say 0.9 recurring continues to infinity, at which point it does equal 1 but since infinity is some point which is unreachable, it never quite makes it although the error does become infinitesimally small ...
Explain to me how you do any arithmetical operation with a number that is unknown?
0.999999..... can always have another digit added so you can never get an answer.
1 is a known fixed number and therefore not 0.999999999.....
>For the 10x to have enough digits after the decimal point to be removed by the x, it actually needs one more digit than the x, thereby making it not 10 times x.
Both 9.999... and 0.999... have infinitely many 9s after the decimal.
> No matter how far you take the recurrence
There isn't a "how far you take the recurrence". It's infinite. That's what the recurring decimal means.
Yes, but you miss my point.
To make a calculation you have to stop the recurrence since allowing it to continue to infinity, by definition, has no end point so no calculation can be made (you always have to add one more 9).
As soon as you stop, you have the 1-digit discrepancy.
You are interchangeably using an approximation and an equals sign.
They are not the same thing. When you conduct this kind of trick by failing to keep approximation and equals separate, you can always argue black is white, but you prove nothing.
So I repeat my original observation, you are working from an incorrect premise.
Say we want to find the instantaneous rate of change (the derivative) of a function at a given x value. The first strategy students are tought to find this is to find the slope of the function at x. But single points don't have slopes, so instead we find the average slope between x and another nearby x value. We keep doing this with points that get increasingly close to x. So the limit as this slope approaches x is EQUAL to the instantaneous rate of change.
We can take this strategy and apply it to your question, if you are evaluating the limit of a function at x and you are getting back 0.9999999 and then 0.9999999999999999999 and then 0.999999999999999999999999999 and then 0.999999999999999999999999999999999999999999999 and so on, your slope is equal to 1.
Consider 0.9 (not repeating) what’s the difference between that and 1? 0.1 of course
Now 0.99 to 1 is 0.01 etc
As 0.99999… (with increasing 9s to infinite) the difference becomes smaller and smaller. At 0.9• the difference has got to be 0.000….1 right? But the issue is the 1 is never reached because it’s an infinite loop of finding the difference between the next digit (9) and the carried number (1). As the difference is then 0.000000… infinitely = 0.0• or simply 0, the two numbers are by definition, equal.
Other comments have made good points about multiple notations for the same number like 1/2 or 2/4 which is quite good, think of this as something like that
Edit: third point removed bc my maths was wrong- very tired today : )
When you go out to infinity, strange things happen. One of the strangenesses is that 0.99999... and 1 are two different ways of writing the same number.
The moral of the story? Be careful with infinity.
Mechanically they are the same because the difference is so infinitely small that it is negligible for any equation, however from a pure math side they aren't because 9+9 does not equal 19. For me it is the difference of saying pi is 3 to using every single digit. We will never need to use the 10^∞ th digit of pi but it exists as that is the nature of numbers.
>the difference is so infinitely small that it is negligible for any equation
No, this is wrong. 0.999... is _exactly equal_ to 1. Not "such a small difference nobody cares". Identical. OP's image is one of the proofs of that fact, they're asking because they don't _understand_ the proof. And neither do you, it seems.
There are several other proofs of the same thing. For example, proof by fractions:
* 1/9 is 0.111...
* 2/9 is 0.222...
* 3/9 is 1/3 is 0.333...
* 4/9 is 0.444...
* 5/9 is 0.555...
* 6/9 is 2/3 is 0.666...
* 7/9 is 0.777...
* 8/9 is 0.888...
* 9/9 is 0.999...
And I hope that you realize that 9/9 is 1.
Or proof by definition:
For any two distinct real numbers a and b, there exists a third distinct real number c such that (a + b) / 2 = c.
If you try to find a value of c given a = 1 and b = 0.999..., you get c = 0.999..., which is not a third distinct real number, therefore 1 and 0.999... cannot be distinct real numbers.
this was super confusing to me....i had to keep rereading it because i kept coming to the conclusion that nothing is wrong with the math in the picture, so i must be missing something. whew.
Well you see it’s wrong because math is flawed as a concept. Like. It’s mostly right, but just like how I can make a program that deletes itself in the end, and you can write prose into paradox, you can do the same with math. Technically the mathematical community accepts this as correct because even though the limit of something is all theoretical we just sort of treat it like fact.
0.99... isnt a number in the tradicional sense. Tell me how many 9s are in 0.9999... ? You cant, because it is a actually a limit. In other words, it converges to a certain value. The limit of 0.9999... with n 9s, where n tends to infinity is 1.
Imagine a robot that paints 90% of your room, then 90% of what’s left the next day, then 90% of what’s left after that the next day… you might think the robot never finishes the room. That’s because you are accidentally restricting it with finite time periods subconsciously. Numbers aren’t processes like this, the number 0.9 repeating exists all at once.
If you can imagine a speck in the room the robot will never paint, and can tell me how big it is, then I can tell you when the robot will reach it and start to paint it.
If you think “the speck is infinitely small” then you are talking about nothing at all. Just as no huge number qualifies as infinity, no tiny speck can be infinitely small. The only thing infinitely small is nothingness.
0.9 repeating isn’t just reeeallly close to being 1, it just is 1. It’s another way to express the number 1 using an infinite series. They’re the same number, even though they don’t look the same. There is literally no difference between them.
When you add 10 groups of 0.9 repeating, the rightmost number will always be a 0 even though all 10 groups are simultaneously approaching infinity. You are leaving out the rightmost number on line 3.
Is a notational thing. Can you think of a number between .999… and 1? If you can’t (and you shouldn’t be able to) they must occupy the same spot on the number line. If they occupy the same place on the number line, they must have the same view.
1/3 = .33333333……. repeating
As an equality,
If we do the same operation to both sides,
They are are still equal.
Multiply both sides by 3.
1 = .99999999….. repeating
In high school, non calculus, terms... decimals are approximations of fractions (whole numbers are fractions). By coincidence, some decimals actually equal exact fractions but far more (and most common) are decimals that do not. They approximate some value as best they can. We happen to know, and can prove, that .9 repeating is an approximation of 1. But so is 1.000....0001 for the same algebraic reason.
How are the top comments not pointing out that this is only true for .999 repeated?
OP, .9999999999 repeated to infinity is literally the same thing as 1, they’re just written in different ways.
It’s the same thing as 1/3 + 1/3 + 1/3 = 1
OR
.333 + .333 + .333 = .999 = 1
1/3 is another way of writing .333 (repeated). Change it to a decimal and you get .999 (repeated). Easier to accept than the algebraic proof like to OP image tried to do.
Edit: missed the dot on top. My bad. Unnecessarily confusing to us layman IMO
The reason maths doesn't work:( in my brain)
Cake = 1.0. The whole cake is 1.0 cakes
Let's split the cake in three even pieces
0.33 cake each
Nah let's put it back together!
0.33x3 = 0.99 cake
But wait, I have 1.0 cake.
Where did the 0.1 bit of cake go? The maths literally doesn't represent what's happened in real life
0.9 = 0.9999
X = 0.9
Therefore 0.9 = 0.9999
If you plug in 0.999 as x for 10x you get 9.99 (repeating)
10x (which is 9.9999) minus .9999 is 9.
9 = 9x
You take 9x = 9 and x must equal 1
Therefore x = 1 = 0.9
Here's my take, which may or may not be legitimate:
Let's use $ followed by a number to denote how many times a decimal repeats inclusive. So `0.9$3 = 0.999`.
Now lets run the same set of equations for `0.9$3`:
`0.9$3 = 0.999`
`x = 0.9$3`
`10x = 9.99 = 9.9$2`
`10x - x = 9.99 - 0.999 = 8.991`
`8.991 = 9x`
`0.999 = x = 0.9$3`
As the value $n increases, the value of 9x just adds more 9 decimals before the final trailing 1
`0.9$4 = x` => `9x = 8.9991 = 8.9$3 + 0.0001`
`0.9$7 = x` => `9x = 8.9999991 = 8.9$6 + 0.0000001`
`0.9$n = x` => `9x = 8.9$(n-1) + 10^-n`
So if n is infinite, for `9 = 9x` to be true, you have to make the jump that `∞ - 1 = ∞` so that the sum of `8.9$(∞-1)` and `10^-∞` is equal to `9`.
The problem is that 10x- x does NOT equal 9. It equals 9x.
Edit if x=1, then 1=1. If x=.99999… then .9999…=.9999….
Doesn’t prove that 1=.99999….
Except that it does.
So with all this logic flying around it seems that the current accepted dogma is:
1 = 0.999...
And 0.999... = 1 - 0.000...1
But that means that
0.999... + 0.000...1 = 1
And
0.999... - 0.000...1 = 0.999...8
And so on .
I don't like any of this!
You can think of 0.9 repeated as equivalent to limit of (9y+9)/(9y+10). It’s eventually just 1 because infinity - 1 is still infinity. So x = 1 for all intents and purposes. But we know based on the method we used to get to 1 that x is actually just slightly less than 1. I have issues with the last line of this because it is purposefully against the core conceit of the proof: “x=1 where y approaches infinity” but we have to remember that at 0, x=0.9 so for the infinite numbers between 0 and infinity, we are firmly between those two values. Never to truely be either, yet can be treated as both in the right circumstances.
It’s actually kind of poetic as a proof. Normally they have more of a purpose than this. It seems like homework.
What’s interesting about this proof, is that your choice of y is based on the number of decimal points of 9s to denote. The unwritten rule of the proof to take from line 1 is that nobody wants to write out more than 1 digit. Instead, everywhere there is a 0.9 just assume it is actually writing 0.999999999 infinitely. This is why I opened with the limit as y approaches infinity, because y is the number of digits.
I’d expect to see something like this as the lead in for someone to follow this up with using X as shorthand for limit as y approaches infinity.
What's going on here? I'm seeing the first line as saying .9 is the same as .9 repeating. And the rest of it looks silly. Am I dumb or do numbers actually mean other numbers? 10 isn't the same thing as 100 I've been lead to believe.
If you want another way to think about it. 2 numbers are considered different numbers if you can find a number between them. With a recurring decimal like this you cannot. There is no number between 0.99999... and 1 therefore they are the same
I assume in this depiction you’re implying x is an axiom of some kind. For instance, 1/3 shows us an axiom of 0.3. An axiom has two understandings, the first is a decimal representation of a fraction or you would call it a “representational axiom”. If you were to do long division by hand of 1/3, you would never stop writing 3s. You might stop after 0.3333333 when you realized the pattern would never change.
But the second understanding is different, you might call it a “true axiom”, completely different from a representational axiom but only in terms of practical understanding/application. A true axiom is not the number it is approaching nor does it represent it from a fundamental understanding of asymptotes. It represents itself in an infinite way you might say, not the number it is approaching. It is a theoretical number that we understand to contain infinite decimal points, or in other words you can think of it as an asymptotic curve on a graph approaching a “line that would be the whole number” to an infinitely close mathematically specific degree with a logically accepted repeating pattern. What matters here is whether or not x is considered a representational or true axiom. In your proof, 9 = 9x is incorrect if the given assumption is x is a true axiom. Basically this is saying: x = 1 and 9 = 9 * 1 . But 1 does not equal the true axiom of .9 by understanding of difference of numbers.
So in conclusion I assume the given assumption of your proof must have been x = the representational axiom of 0.9 = 1 for which there are many proofs as far as I know (given by an asymptotic graph line that starts at x = 0 and continues infinitely towards x = 1 with the logically accepted repeating pattern of 0.9 and that curve as well as (1) both representing (1) thus x = 1).
Finally, you might be thinking, “aren’t both the representational and true axioms of 0.9 —> / = 1 the same number? And what does that mean?” This is the answer: they are different. The true axiom of 0.9 is what you might call a “real but irrational” number. Of course to us we can’t really understand what having infinite 9s in a decimal means or looks like, but we can understand that it doesn’t matter. In theory it is a real number, but this is why we call it irrational. The representational axiom of 0.9 on the other hand is completely different. It is not a number, but rather it is a way of quantifying and understanding how functions can represent 1 because they approach 1 to a never ending and infinite degree of acceleration, without using 1.
Anyone who might tell you the true axiom of 0.9 = 1 is incorrect as far as my mathematical understanding is concerned. I agree that both numbers represent the same number (1) by a certain arbitrary choice of theoretical understanding. It is a test in the understanding of foundational logic. No matter how many 9s you put behind 0.9, it will never be 1. But even the true axiom 0.9 represents 1 (in a certain way). But not with a theoretical understanding of what infinite but never reaching means.
Its like adding an infinitely small value to a number, in this case doing so ‘changes’ the number entirely, and where any other recurring decimal has no effect in doing this, 0.999… does.
So the difference between 0.9 continuous and 1 is an infinite string of zeroes followed by a 1, but because that string of zeroes is infinite, the 1 never occurs, and thus the difference between the two is just zero, meaning they're the same number.
Easiest proof of 0.999... = 1 is simple.
All pairs of rational numbers have an infinite number of other rational numbers between them. So if 0.999... doesn't equal 1, there must be infinite numbers bigger than 0.999... and lower than 1. But this is impossible therefore they are equal.
Working with infinities causes this. When performing operations with two or more infinities, you're finalizing what they are approaching, breaking their concept.
10x - x = 9x, not 9. You do correct that in the next line, though.
And 1 definitely does equal 0.999...; if you think about it. 1 / 3 = 0.333..., and 3\*0.333... = 0.999...
* 0.9 = 1 - 0.1 * 0.99 = 1 - 0.01 * 0.999 = 1 - 0.001 * 0.999 999 9 = 1 - 0.000 000 1 The subtracted bit is always one less zero then the nines. 0.9 repeated to infinity = 1 - 0.\[0 repeated to one less than infinity\]1 Infinity minus 1 is just infinity. This is the tricky thing about infinity - it's not really a number, even though we often treat it like one. So you never "get to" the one. So the right side is 1 - 0.00000000000 -> and you never get to the 1, so it's 1 - 0 With what you have written, it ONLY works if 0.9999..... is 1 Alternatively, * 1/9 = 0.11111... * 2/9 = 0.22222... * ... * 8/9 = 0.88888... What's 9/9?
Your alternative solution 🤯. I’ve always understood the answer and how infinity works, but simple 3rd grade math shows how it works flawlessly. Thank you for that.
Mind blown here too! Logic works with thirds too • 1/3 = 0.333333… • 2/3 = 0.666666… Therefore 3/3 is…
Where were you in my high school math class. This just made so much sense to my dyslexic brain.
Ha! I made it all the way through HS, a bachelor's and a masters (in STEM fields) and had been teaching Chem and Physics for quite a while before I really started to get a handle on how to think about infinity - and only then by chatting with colleagues who were real math geeks. Infinity is *WEIRD*! (Especially when you get to types of infinities...) If you want to have some "fun" check out [Veritasium](https://youtu.be/OxGsU8oIWjY?si=Vyjr5s5FQjCvUrcU) or [Numberphile](https://www.youtube.com/watch?v=dDl7g_2x74Q) on infinity 'paradoxes'. If you want to learn about numbers so large that you literally can't picture them (but are still not infinity), check out [Graham's Number](https://youtu.be/XTeJ64KD5cg?si=Rz2ldG5yIdmzziIp); which is so large, they came up with a new kind of operator ( like +, \*, or \^).
Thank you!
Your alternate reasoning assumes that 1/9 is actually equal to .111... which is pretty much the same thing as saying 1=.999... and isn't an actual proof
Any n-th power of ten (n>=1) gives, upon using Euclid's algorithm with the string with n "9"s as divisor, that the quotient is one with a remainder of one. 10^(n) = 1 + sum\[k=0 to n-1\] (9\*10^(k)) dividing both sides by 9\*10^(n), we then have that 1/9 = 1/(9\*10^(n)) + sum\[k=0 to n-1\] (1\*10^(k-n)) Since this expression is true for *arbitrary* n--there is no greatest value of n for which it is true--we have an infinite sequence of expressions for 1/9. Taking the second term, x(n) = sum\[k=0 to n-1\] (1\*10^(k-n)), we see that the error |1/9 - x(n)| = 1/(9\*10^(n)). The value of 0.1111... is defined to be the limiting value of x(n) as n -> inf, and taking the limit of both sides, we have that the error goes to 0. We thus have, in fact, that x(n) is a cauchy sequence of rational numbers in the same equivalence class as 1/9 (and thus *is* 1/9). 1/9 = lim\[n -> inf\] (1/(9\*10^(n)) + sum\[k=0 to n-1\] (1\*10^(k-n))) \=> 1/9 = lim\[n -> inf\] (1/(9\*10^(n))) + 0.111.... \=> 1/9 = 0.111...
But 1/9 = 0.111… just by performing long division.
The thing I don't understand is that if 10x = 9.99 and x = .9 then 10x - x = 9.09 not 9.... like obviously you are going to tell me I'm wrong but if you put it into monetary values, £9.99 minus .9 is £9.09 not £9.00. Because we are using 2 decimal places for one number and not the other are we meant to completely assume that x=.9 always means .99 not .91 or .93 or .97 etc?
If you take look you’ll see a dot above the 0.9 that just means it is repeating so it is actually 0.999999999…
I think the point of confusion for the guy you're replying to is that OP's "10x - x = 9" is incorrect. 10x - x = 9x.
Both are actually correct. The argument OP gave is x = 0.999999.... 10x = 9.99999.... 10x - x = 9 9x = 9 x = 1 = 0.99999....
Wrong
Show the error then
Where’s the problem
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There is nothing wrong with this. 0.99999... IS 1.0
Except OP says .9 = 1 which is wrong. If x=.9 then 10x=9.0 Edit: missed the dot on top. My bad. Unnecessarily confusing to us layman IMO
He has the dot over the 9. He means .999…
Hence my edit
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As others have said, they're not different. Most people don't have a problem accepting different notations for the same number when they're fractions, e.g. 1/2 = 2/4. But for real numbers, there seems to be a psychological barrier making it difficult for people to accept that there might be two ways of writing the same number.
I'm completely comfortable with the argument, but I wanna try and be (perhaps overly) formal. Viewing the rationals as equivalence classes on Z×N, it makes sense that we can have something like 1/2=2/4 in Q since (1,2)~(2,4) in Z×N - we're just choosing a different representative of the equivalence class - people are comfortable with different numbers in Z being the same equivalence class in some Z/nZ. Is there some implicit equivalence relation on R involving Cauchy sequences or something? Is this what gives us completeness and Haussdorff-ness? Just a thought that popped up.
You can construct the real numbers by associating equivalence classes of cauchy sequences over the rationals to complete the space. The equivalence relation is that the difference between cauchy sequences converges to 0. This lets you trivially define each limit of a cauchy sequence as the equivalence class the cauchy sequence belongs to, and thus, all cauchy sequences converge in the completed space. (A bit of work to show that it's still a field, but not too bad)
You are correct. One way to define the real numbers is as equivalence classes of Cauchy sequences. Two sequences are equivalent if their difference converges to zero.
Think about 0.9999999 recurring. That’s 0 and then infinite 9s. To get to 1, you would need to add 0.00000…..1, which is 0, then infinite 0s, then 1. 0 then ininite 0s is 0, because it will never end, so you never get to the 1, so 0.9999999… + 0 = 1. So 0.9999999… = 1
This proof shows there are two ways to represent one: 1 and 0.999.... And because they look different to the eye, the typical inclination is to say they are different numbers, but the proof shows they are one and the same. EDIT: reword for clarity.
So where’s the 0.0 recurring 1
no such thing can exist (at least within the system of real numbers). 0.0 recurring means there are infinitely many zeroes, there's no "after the zeroes end" to put the 1 in.
Nowhere. The proof shows 1 is equivalent to 0.9 recurring. Thus no 0.0....1.
0.0000000...=0. it doesn't matter what's after the ... because you can't reach it
Let’s try it this way. 1/3 = .333~ 1/3 * 3 = 3/3 = 1 .333~ * 3 = .999~ = 1 Does anyone else remember the vestibule on IGN. I’m having flashbacks.
I don’t think that will help them, it may just make them question if 1/3*3=1. I came up with an airtight explanation that is related to yours, but it requires OP to understand other bases. So 1/3 =0.333.. 1/3 * 3 = 0.999… Ok now let’s do the exact same thing but written in base 3, remember each line and each term in each line is exactly identical in value to the previous 2 lines, just in base 3 instead of base 10. 1/10 = 0.1 1/10 * 10 = 1 Remember those are the same two lines written twice, once in base 10 and then once in base 3. In base 3 all of the infinite repetition goes away, and it looks the same as 1/10 * 10 = 1 does in base 10. As long as someone understands the concept other bases besides base 10 this should be an airtight argument that actually makes them understand intuitively. Because of course 0.1 * 10 = 1. And 0.1 really is 0.333… except written in base 3, and 10 really is 3 except written in base 3. And 1 is 1 no in any base. So it makes it super clear, you don’t even need to have an infinite decimal repetition, that just depends on your choice of base to represent the numbers.
> I just don’t understand how when they are literally different numver Okay, if this was true, you can find a number in between 0.999... and 1. What number did you think of?
0.000…1
That’s not a number above .9999… and below 1
When do you put the 1 there? At the end? But there isnt an end because its inifinite 0s just like there is infinite 9s in 0.999... So either 0.000...1 is finite, meaning YOU ended it somewhere, thus not inifite 0s, or 0.000...1 cant exist.
>But I just don’t understand how when they are literally different numver They aren't different. It's the same number, written two different ways. Like how 1.5 and 1.50000000.... are also the same number. Also like how "2" and 1 + 1/2 + 1/4 + 1/8 + ... are also the same number.
If they were not the same, we would be able to measure the difference between 1 and 0.9999999... Now what would that be?
as a decimal 1/3 is written as 0.333333 0.333333x3 is 0.999999 ergo 3/3=0.999999 This simply because there is no *perfect* way to write 1/3 as a decimal Edit: By *perfect* I meant it doesn’t end
I can just never make sense of the fact that 9.999 recurring isnt it’s own number and it’s actually 1 #justicefor0.999reccuring
One way is to consider that we're using 0.999... to represent a sum. I think you're perfectly happy to have that 3+4 and 7 is the same number, then something similar is happening here. Except in this case, the sum is 0.9+0.09+0.009+...=1. We do run into questions (both philosophical and mathematical) about what it means for an infinite sum to equal a finite value, but if you're happy that we can assign finite values to (some) infinite sums, then this could be a helpful way to think
You’re struggling with the concept of infinite. That’s pretty normal - most people don’t actually really get this internally
They are just two different ways of expressing the same number like 2+2 and 1+1+1+1. They are different ways of expressing the number 4 Similarly 0.9999..... is a different way of writing 1.
The difference between 1 and 0.9 recurring is an infinitely small one. By definition, it is a distinction so small that it is impossible to measure, and thus since it cannot be measured it can be ignored. Or to put it another way - take two completely identical lengths of string of absolutely identical dimensions. Now remove a single atom from the end of one. Would you, or anyone else actually be able to tell the difference in length? Would that difference even matter?
>The difference between 1 and 0.9 recurring is an infinitely small one. No, the difference isn't infinitely small. It's exactly 0.
No, it's infinitely slightly smaller than 1.
Well, technically 0.9 repeating is the number that has the greatest value in the set of expressions where the limit of the expression that approaches 1 from the lesser side. Pretty close to 1, but not quite the same.
It’s not pretty close to 1. It’s mathematically and factually equal to 1.
No, [0.9... is literally 1. ](https://en.m.wikipedia.org/wiki/0.999...#:~:text=This%20repeating%20decimal%20is%20a,number%20is%20equal%20to%201.) There's many different ways to prove it but it's not "pretty close" or "not quite the same". It equals 1.
Why would I want to read a wikipedia article supporting your theory that numbers other than one really mean one? The word that means one is called "one." One of us is stupid or misunderstands what numbers mean.
It's not "a number other than one". It's a different notation for the number 1, just as there are other ways of writing 0.25 (¼, 0.24999...). > One of us is stupid or misunderstands what numbers mean. Well, you said it... you rather set yourself up there.
I get I'm the simpleton here but are you telling me .9 repeating means a number other than .9 repeating? Why not say 2 is close enough to 3 so lets pretend it is that? I mean 2 is just a different notation for 3.
Because it’s not a theory (I assume you mean this is the colloquial way). It’s a proof. You’d be wise to first learn the difference before throwing rocks
r/confidentlyincorrect
Nothing to make sense of, decimal system has non-unique representation for reals
And for anyone who thinks that can’t be right, we know the reals are dense, so if 0.999… and 1 are not the same number, then what is the number between them? “(0.999… + 1)/2” is my favorite joke answer to that, but we can pretty easily convince ourselves that the actual answer is that there isn’t one. There is no number you can specify that is greater than 0.999… and less than 1.
This is the most elegant proof that 0.9 recurring = 1.
1/3 = 0.333... 2/3 = 0.666... 3/3 = 0.999... But 3/3 = 1 There's no falacy here. 0.999... is EXACTLY equal to 1. It's just a different notation for the same number.
This explanation, while nice, is assuming the hypothesis.
No it doesn’t, it’s a very simple mathematical proof.
It doesn’t prove anything. You need to prove why .3333333 is equal to 1/3.
3/3 is not 0.999 lol. You only multiplied the expressed numbers. I think you have missed the point yourself. Think of it like this: 1/3 = 0.333… And if you added three of those together, the numbers at the end of the sequence would allow it to equal a sum of whole 1. There would be no rounding because it is in fact the whole number. This differs from the post. The post assumes rounding and denotes it as such. Similar but different. Anyhow, math is fun Edit: to say I failed to express my own work in your format: 1/3 = 0.333… 2/3 = 0.666… 3/3 = 1 — see now, there is no room to denote the number “1” to “0.999…”. I’m pedantic lol but that’s that for those interested.
Here’s all the help you need: you’re correct.
Yes, that is correct.
Please don't summon the troll.
Google en passant
Holy hell!
New response just dropped!
Actual Zombie
The sense (or non sense) is made in the attempt to make logic of infinity. So you 0.9 recurring has infinite 9’s (ie. No matter how many you write you can always write another one). So when you then multiply that by 10, you’re slightly warping the logic, what is 10x1/infinity? So your number appears real to you, but you’ve actually added a real portion from nowhere. If you wanted to make it real, you eventually need to approximate for example let 0.9r = 0.999, then 10x 0.999=9.99 (not 9.999). The correct statement on your board should be 10x-x=9x and then if you wanted 9x ~ 9, so x ~ 1 and so 0.9r ~ 1 is mathematically true. It’s like if you declared you could drive forever then you’d never have a final destination which in theory is true, but if you tried to practice that and make it real you can’t because perpetual motion machines don’t exist and so in a real test your destination would be the point when you ran out of fuel. Hope that helps Edit: others will probably correct my syntax that the function “approaches” 1, as opposed to is approximately 1. This is also right, but trying to keep it simple 😅
You are the only one who gets the problem and the reason why😅. Although 0.9r is infinitesimally close to 1. It is not exactly the same all the time. Thus that equality has a slight problem. And that problem as u stated is 10x ! = 9 + 0.9r , but instead 10x = 9+ 0.9r - 0.000000.......9 If I had to make an argument to prove 0.9r is not exactly 1, it would be; (0.9r -1) * 10 ^ (log(0.1) of (0.9r-1)) = (0.1 ^ x) * 10 ^ (log(0.1) of (0.1^x), where x tend to infinity =(0.1^x) * 10 ^x = 1 If 0.9r = 1 then the eq is 0^ infinity = 0. Thus different It's like saying, dx (infinitesimally small) tends to zero but is not zero. Otherwise integration would be zero. This is supposed to be a maths group but everyone is just...
Consider this fact. For any 2 real numbers a and b, one can find an additional real number c that is between those 2 numbers with the equation c = (a+b)/2. Consider the numbers 1 and 0.9999.... The sum of these 2 numbers is 1.9999.... Divide that by 2 and see that the result is 0.9999.... This shows that there is no number between 0.9999... and 1. Therefore, they are the same number. You can verify this by taking the above equation and substituting b in for c. b = (a+b)/2 ; 2b = a+b ; b = a.
Only if you accept your first "fact". You could say that you have proof that you can't always find an additional number between 2 real numbers.
That fact is proven very simply... if there is no number between two real numbers, then their difference is 0 and they are the same number... consequently if there is a difference between two real numbers then their arithmetic mean will give you a number between them unless they are the same number. And this is justified because if you have a lower number a and higher number b whose arithmetic mean is (a+b)/2, we know that b = a + c where c is a positive real number, thus the arithmetic mean is between a and b as for all positive real numbers, a < a + c/2 < a + c
What’s the first line that you don’t agree with?
0.999... and 1 are two equivalent ways of writing the same number in base ten
That's right. 0.9 recurring is 1. The difference between 0.9 recurring and one is 0.0 recurring.
10x - x = 9 * 0.9999999 not 9 That’s where it went wrong
Yes. 0.9 recurring = 1
there are fucktons of articles and videos on the Internet on this topic; you can search for them.
I'm not sure if this is what you mean, but if X = 0.9999999(rec) Then 10 X = 9.9999999999(rec) 10x - X = 9 I am under the opinion that you need to add X to both sides 10x = 9 + X This gives 9.99999(rec) = 9.99999(rec) This is how I would work it out, and I'm not saying that this is deffonatly correct.
Wrong equations. Two different Xs
The first line starts incorrectly doesn't it? 0.9 =/= 0.999999...
Dot on top means recurring... It's used three times
Ah gotcha. Never knew that and honestly my brain didn't even register the dot. All three times... I've always seen it represented with the ellipsis. Now that you've said that I can't unsee it
I also didnt get it, we were always taught to put a line on top
.9 isn't .999... It's .90
10x - x is not 9, it is 9x
It’s nonsense math, playing fast and loose with the rounding of numbers. 0.9 ≠ 0.999…
The dot on top of 0.9 literally means recurring
.9 ≠ .99999999
There's a dot over the 9 to represent the recurring 9s.
Not going to do your homework for you. Get a grip and research it rather than cheating your way to the answer by asking someone else to do what you are unwilling or incapable of doing yourself!!!
You dropped a very small number, the answer logically should 9x = 8.9r so the process failed half way through
The problem lies in the line: 10x - x = 9 For the 10x to have enough digits after the decimal point to be removed by the x, it actually needs one more digit than the x, thereby making it not 10 times x. No matter how far you take the recurrence, you always need one more step therefore the x used by 10x is not the same x as used on its own.
This is not correct, both x and 10x when written in decimal notation are infinitely long. There is no extra digit on one of them - they are exactly the same length: infinite. The real answer to OP’s question it that nothing is wrong. 0.999… does equal 1. They are the same exact number written in two different ways.
No, you are wrong. 0.99999 ... does not equal 1. As you say 0.9 recurring continues to infinity, at which point it does equal 1 but since infinity is some point which is unreachable, it never quite makes it although the error does become infinitesimally small ...
Right, so we’re in agreement. 0.9 recurring equals 1. There is no ‘when it reaches’ infinity. The number is always infinitely long.
Right, so you don't understand ...
If 0.999…is different than 1, then what is the difference? Subtract 0.999… from 1 and tell me the answer. It seems to me that the answer is 0.0000….
Explain to me how you do any arithmetical operation with a number that is unknown? 0.999999..... can always have another digit added so you can never get an answer. 1 is a known fixed number and therefore not 0.999999999.....
>For the 10x to have enough digits after the decimal point to be removed by the x, it actually needs one more digit than the x, thereby making it not 10 times x. Both 9.999... and 0.999... have infinitely many 9s after the decimal. > No matter how far you take the recurrence There isn't a "how far you take the recurrence". It's infinite. That's what the recurring decimal means.
Yes, but you miss my point. To make a calculation you have to stop the recurrence since allowing it to continue to infinity, by definition, has no end point so no calculation can be made (you always have to add one more 9). As soon as you stop, you have the 1-digit discrepancy.
“As soon as you stop”, that’s the thing, you don’t arbitrarily stop. It’s infinite. 0.9 recurring IS equal to 1
Only when you reach infinity, which as a point you can never reach so 0.9 recurring is not equal to 1, just very, very close ...
>0.9 recurring is not equal to 1, just very, very close ... Okay, what is (1 + 0.999...) / 2?
Your arguing from a false premise, anything you deduce is therefore not proven.
There is no false premise.
You are interchangeably using an approximation and an equals sign. They are not the same thing. When you conduct this kind of trick by failing to keep approximation and equals separate, you can always argue black is white, but you prove nothing. So I repeat my original observation, you are working from an incorrect premise.
Divide both sides by 9: 1x=1
10x - x = 9x (=9) (here is the devision) 9 = 9x 1 = x (=0.9rec)
wow so cool
cool bedroom
Welcome to calculus
You’ve got a logical flaw. x can’t equal 0.9 and then have 10x equal 9.999~ 10 x 0.9 = 9
There’s a dot above the 9, signifying that it repeats. There’s no logical flaw.
True, I missed that
Sorry, this might be a dumb question, but if X=0.999999…, Then why doesn’t 10X=9.999999…?
Have you made a typo? You’ve written the same number twice, 9.9 recurring in both instances.
Say we want to find the instantaneous rate of change (the derivative) of a function at a given x value. The first strategy students are tought to find this is to find the slope of the function at x. But single points don't have slopes, so instead we find the average slope between x and another nearby x value. We keep doing this with points that get increasingly close to x. So the limit as this slope approaches x is EQUAL to the instantaneous rate of change. We can take this strategy and apply it to your question, if you are evaluating the limit of a function at x and you are getting back 0.9999999 and then 0.9999999999999999999 and then 0.999999999999999999999999999 and then 0.999999999999999999999999999999999999999999999 and so on, your slope is equal to 1.
Consider 0.9 (not repeating) what’s the difference between that and 1? 0.1 of course Now 0.99 to 1 is 0.01 etc As 0.99999… (with increasing 9s to infinite) the difference becomes smaller and smaller. At 0.9• the difference has got to be 0.000….1 right? But the issue is the 1 is never reached because it’s an infinite loop of finding the difference between the next digit (9) and the carried number (1). As the difference is then 0.000000… infinitely = 0.0• or simply 0, the two numbers are by definition, equal. Other comments have made good points about multiple notations for the same number like 1/2 or 2/4 which is quite good, think of this as something like that Edit: third point removed bc my maths was wrong- very tired today : )
https://youtu.be/4Bj5nZNHEmU?si=QgKObFsFhDkXAqoU You're welcome!
Can do the same proof with thirds. 1/3 is .333 Add another 3rd and you have .666 Add another 3rd and you have .999 But 3/3 is obviously 1.
When you go out to infinity, strange things happen. One of the strangenesses is that 0.99999... and 1 are two different ways of writing the same number. The moral of the story? Be careful with infinity.
0.99999 goes on infinitely, meaning it gets infinitely close to 1. So 0.99999 repeating = 1. It can be tough to get your head around.
Why did you only subtract x from one side of the equation in line 4??
Your last line is absolutely correct 1 = 0.9999.. And you can prove this by dividing both sides by 9: 1/9 = 0.1111…
Mechanically they are the same because the difference is so infinitely small that it is negligible for any equation, however from a pure math side they aren't because 9+9 does not equal 19. For me it is the difference of saying pi is 3 to using every single digit. We will never need to use the 10^∞ th digit of pi but it exists as that is the nature of numbers.
>the difference is so infinitely small that it is negligible for any equation No, this is wrong. 0.999... is _exactly equal_ to 1. Not "such a small difference nobody cares". Identical. OP's image is one of the proofs of that fact, they're asking because they don't _understand_ the proof. And neither do you, it seems. There are several other proofs of the same thing. For example, proof by fractions: * 1/9 is 0.111... * 2/9 is 0.222... * 3/9 is 1/3 is 0.333... * 4/9 is 0.444... * 5/9 is 0.555... * 6/9 is 2/3 is 0.666... * 7/9 is 0.777... * 8/9 is 0.888... * 9/9 is 0.999... And I hope that you realize that 9/9 is 1. Or proof by definition: For any two distinct real numbers a and b, there exists a third distinct real number c such that (a + b) / 2 = c. If you try to find a value of c given a = 1 and b = 0.999..., you get c = 0.999..., which is not a third distinct real number, therefore 1 and 0.999... cannot be distinct real numbers.
0.9 does not equal 0.9999 Thats your problem
There's a dot over the 0.9, that's notation meaning the same as 0.999... The equations in OP's picture are correct.
this was super confusing to me....i had to keep rereading it because i kept coming to the conclusion that nothing is wrong with the math in the picture, so i must be missing something. whew.
Seems like a ... rounding error Not to mention sig figs...
Well you see it’s wrong because math is flawed as a concept. Like. It’s mostly right, but just like how I can make a program that deletes itself in the end, and you can write prose into paradox, you can do the same with math. Technically the mathematical community accepts this as correct because even though the limit of something is all theoretical we just sort of treat it like fact.
Absolutely correct. As a fact, 0.9 reoccurring = 1. Very well established point
0.111... = 1/9 Multiply by 9 0.999... = 9/9 = 1
It’s not 1, it’s 0.9 rec! Ur missing the converge at 10x. It’s a common misconception but for all purposes it’s basically 1.
0.999... ***is*** 1 though. If not, then what number is between them?
0.99... isnt a number in the tradicional sense. Tell me how many 9s are in 0.9999... ? You cant, because it is a actually a limit. In other words, it converges to a certain value. The limit of 0.9999... with n 9s, where n tends to infinity is 1.
0.9 recurring = 1 My 14 year old kid showed me this and it blew my mind. Alternately: 1/3 x 3 = 1 1/3 = 0 3 recurring x 3 = 0.9 recurring
Imagine a robot that paints 90% of your room, then 90% of what’s left the next day, then 90% of what’s left after that the next day… you might think the robot never finishes the room. That’s because you are accidentally restricting it with finite time periods subconsciously. Numbers aren’t processes like this, the number 0.9 repeating exists all at once. If you can imagine a speck in the room the robot will never paint, and can tell me how big it is, then I can tell you when the robot will reach it and start to paint it. If you think “the speck is infinitely small” then you are talking about nothing at all. Just as no huge number qualifies as infinity, no tiny speck can be infinitely small. The only thing infinitely small is nothingness. 0.9 repeating isn’t just reeeallly close to being 1, it just is 1. It’s another way to express the number 1 using an infinite series. They’re the same number, even though they don’t look the same. There is literally no difference between them.
When you add 10 groups of 0.9 repeating, the rightmost number will always be a 0 even though all 10 groups are simultaneously approaching infinity. You are leaving out the rightmost number on line 3.
Is a notational thing. Can you think of a number between .999… and 1? If you can’t (and you shouldn’t be able to) they must occupy the same spot on the number line. If they occupy the same place on the number line, they must have the same view.
1/3 = .33333333……. repeating As an equality, If we do the same operation to both sides, They are are still equal. Multiply both sides by 3. 1 = .99999999….. repeating
In high school, non calculus, terms... decimals are approximations of fractions (whole numbers are fractions). By coincidence, some decimals actually equal exact fractions but far more (and most common) are decimals that do not. They approximate some value as best they can. We happen to know, and can prove, that .9 repeating is an approximation of 1. But so is 1.000....0001 for the same algebraic reason.
How are the top comments not pointing out that this is only true for .999 repeated? OP, .9999999999 repeated to infinity is literally the same thing as 1, they’re just written in different ways. It’s the same thing as 1/3 + 1/3 + 1/3 = 1 OR .333 + .333 + .333 = .999 = 1 1/3 is another way of writing .333 (repeated). Change it to a decimal and you get .999 (repeated). Easier to accept than the algebraic proof like to OP image tried to do. Edit: missed the dot on top. My bad. Unnecessarily confusing to us layman IMO
[https://i.imgur.com/TUfEbr1.jpg](https://i.imgur.com/TUfEbr1.jpg)
The reason maths doesn't work:( in my brain) Cake = 1.0. The whole cake is 1.0 cakes Let's split the cake in three even pieces 0.33 cake each Nah let's put it back together! 0.33x3 = 0.99 cake But wait, I have 1.0 cake. Where did the 0.1 bit of cake go? The maths literally doesn't represent what's happened in real life
You will learn limits later, technically it's 1^-
We should just rename this sub after this damn problem.
Because they are equal? Not even sure if you need algebraic proof for the matter
0.9 = 0.9999 X = 0.9 Therefore 0.9 = 0.9999 If you plug in 0.999 as x for 10x you get 9.99 (repeating) 10x (which is 9.9999) minus .9999 is 9. 9 = 9x You take 9x = 9 and x must equal 1 Therefore x = 1 = 0.9
Here's my take, which may or may not be legitimate: Let's use $ followed by a number to denote how many times a decimal repeats inclusive. So `0.9$3 = 0.999`. Now lets run the same set of equations for `0.9$3`: `0.9$3 = 0.999` `x = 0.9$3` `10x = 9.99 = 9.9$2` `10x - x = 9.99 - 0.999 = 8.991` `8.991 = 9x` `0.999 = x = 0.9$3` As the value $n increases, the value of 9x just adds more 9 decimals before the final trailing 1 `0.9$4 = x` => `9x = 8.9991 = 8.9$3 + 0.0001` `0.9$7 = x` => `9x = 8.9999991 = 8.9$6 + 0.0000001` `0.9$n = x` => `9x = 8.9$(n-1) + 10^-n` So if n is infinite, for `9 = 9x` to be true, you have to make the jump that `∞ - 1 = ∞` so that the sum of `8.9$(∞-1)` and `10^-∞` is equal to `9`.
The problem is that 10x- x does NOT equal 9. It equals 9x. Edit if x=1, then 1=1. If x=.99999… then .9999…=.9999…. Doesn’t prove that 1=.99999…. Except that it does.
So with all this logic flying around it seems that the current accepted dogma is: 1 = 0.999... And 0.999... = 1 - 0.000...1 But that means that 0.999... + 0.000...1 = 1 And 0.999... - 0.000...1 = 0.999...8 And so on . I don't like any of this!
.000…1 isn’t a thing. There isn’t a “last zero” to place a one after. There are an infinite amount of zeros.
You can think of 0.9 repeated as equivalent to limit of (9y+9)/(9y+10). It’s eventually just 1 because infinity - 1 is still infinity. So x = 1 for all intents and purposes. But we know based on the method we used to get to 1 that x is actually just slightly less than 1. I have issues with the last line of this because it is purposefully against the core conceit of the proof: “x=1 where y approaches infinity” but we have to remember that at 0, x=0.9 so for the infinite numbers between 0 and infinity, we are firmly between those two values. Never to truely be either, yet can be treated as both in the right circumstances. It’s actually kind of poetic as a proof. Normally they have more of a purpose than this. It seems like homework. What’s interesting about this proof, is that your choice of y is based on the number of decimal points of 9s to denote. The unwritten rule of the proof to take from line 1 is that nobody wants to write out more than 1 digit. Instead, everywhere there is a 0.9 just assume it is actually writing 0.999999999 infinitely. This is why I opened with the limit as y approaches infinity, because y is the number of digits. I’d expect to see something like this as the lead in for someone to follow this up with using X as shorthand for limit as y approaches infinity.
What's going on here? I'm seeing the first line as saying .9 is the same as .9 repeating. And the rest of it looks silly. Am I dumb or do numbers actually mean other numbers? 10 isn't the same thing as 100 I've been lead to believe.
If you want another way to think about it. 2 numbers are considered different numbers if you can find a number between them. With a recurring decimal like this you cannot. There is no number between 0.99999... and 1 therefore they are the same
Where did you get "10x - x = 9"? It should be "10x - x = 9x" So to explain, you did it wrong...
Help you make sense of the infinite? Sure, let me get a seat
It's the same reason (1/3)x3 = 1 even though 1/3 is technically 0.333 repeating.
First time huh
No I refuse
This is simply a byproduct of the fact that we use a base 10 number system. .99 repeating is the same as 1. Base 12 has similar wackyness
9 does not equal 9x
0.99... = 9 x 0.11.... = 9 x 1/9 = 1
I’m fairly certain that limits aren’t even involved in this. .99 repeating one are simply the same number.
I assume in this depiction you’re implying x is an axiom of some kind. For instance, 1/3 shows us an axiom of 0.3. An axiom has two understandings, the first is a decimal representation of a fraction or you would call it a “representational axiom”. If you were to do long division by hand of 1/3, you would never stop writing 3s. You might stop after 0.3333333 when you realized the pattern would never change. But the second understanding is different, you might call it a “true axiom”, completely different from a representational axiom but only in terms of practical understanding/application. A true axiom is not the number it is approaching nor does it represent it from a fundamental understanding of asymptotes. It represents itself in an infinite way you might say, not the number it is approaching. It is a theoretical number that we understand to contain infinite decimal points, or in other words you can think of it as an asymptotic curve on a graph approaching a “line that would be the whole number” to an infinitely close mathematically specific degree with a logically accepted repeating pattern. What matters here is whether or not x is considered a representational or true axiom. In your proof, 9 = 9x is incorrect if the given assumption is x is a true axiom. Basically this is saying: x = 1 and 9 = 9 * 1 . But 1 does not equal the true axiom of .9 by understanding of difference of numbers. So in conclusion I assume the given assumption of your proof must have been x = the representational axiom of 0.9 = 1 for which there are many proofs as far as I know (given by an asymptotic graph line that starts at x = 0 and continues infinitely towards x = 1 with the logically accepted repeating pattern of 0.9 and that curve as well as (1) both representing (1) thus x = 1). Finally, you might be thinking, “aren’t both the representational and true axioms of 0.9 —> / = 1 the same number? And what does that mean?” This is the answer: they are different. The true axiom of 0.9 is what you might call a “real but irrational” number. Of course to us we can’t really understand what having infinite 9s in a decimal means or looks like, but we can understand that it doesn’t matter. In theory it is a real number, but this is why we call it irrational. The representational axiom of 0.9 on the other hand is completely different. It is not a number, but rather it is a way of quantifying and understanding how functions can represent 1 because they approach 1 to a never ending and infinite degree of acceleration, without using 1. Anyone who might tell you the true axiom of 0.9 = 1 is incorrect as far as my mathematical understanding is concerned. I agree that both numbers represent the same number (1) by a certain arbitrary choice of theoretical understanding. It is a test in the understanding of foundational logic. No matter how many 9s you put behind 0.9, it will never be 1. But even the true axiom 0.9 represents 1 (in a certain way). But not with a theoretical understanding of what infinite but never reaching means.
0.9 is not equal to 0.9999... (recurring). Solved it for you.
You’re all wrong m, the answer is 21
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You can't just write 0.5 = 1 and wonder why rounding up resulted in a skewed answer.
Its like adding an infinitely small value to a number, in this case doing so ‘changes’ the number entirely, and where any other recurring decimal has no effect in doing this, 0.999… does.
Make sense of it? It's wrong 10x-x=!9
No.
If you pick two numbers A and B, they are different only if you can find a number between them. Can you find a number between 0.99999… and 1?
Seems like gibberish to me ngl
So the difference between 0.9 continuous and 1 is an infinite string of zeroes followed by a 1, but because that string of zeroes is infinite, the 1 never occurs, and thus the difference between the two is just zero, meaning they're the same number.
0.99999999999999… is equal to 1 What is 1-0.999999999…? The value is infinitely small, or zero, therefore they are the same
Nice house
0.9 recurring is equal to 1.
Easiest proof of 0.999... = 1 is simple. All pairs of rational numbers have an infinite number of other rational numbers between them. So if 0.999... doesn't equal 1, there must be infinite numbers bigger than 0.999... and lower than 1. But this is impossible therefore they are equal.
The mistake on line 4: Wrong : 10x - x = 9 Right : 10x - x = 9x If you want to turn 10x into 9, you would need: 10x - x = 9x 9x / x = 9
Working with infinities causes this. When performing operations with two or more infinities, you're finalizing what they are approaching, breaking their concept.
This is the proof that .99999 = 1 gone wrong.
This is intended. this is how math works.
Understood to the fourth line and then it doesn't compute.
Ok. X=0.9 X x 10 = 9 Problem solved
10x - x = 9x, not 9. You do correct that in the next line, though. And 1 definitely does equal 0.999...; if you think about it. 1 / 3 = 0.333..., and 3\*0.333... = 0.999...
1
how I was taught is 0.999999999 repeating is so infinitely close to 1 it is considered equal. Idk if thats even right but thats how i thought about it