Is 0.9 recurring equal to 1?

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Joram
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Re: Is 0.9 recurring equal to 1?

Post by Joram » July 2nd, 2010, 2:45 pm

Midnight_Carnival wrote:1 what?
1 apple? is 0.9999(rec,) or an apple the same as an apple, that is is an apple with an infinitely small bite taken out of it still and apple?

the value "1" is an abstraction. Is one part red 75000 parts blue still blue? it depends on how you define colours. Numbers and mathematics are made to be objective, this is becasue they are a tool used to measure things (mostly differences). You have to define "1" first. If you're dealing with countries, 0.999999(rec.) of a country could still be a country. Generally speaking, taken in isolation, numbers are (abstractions of) "pure" or "pepfect" values which don't exist in the real world. Don't try to apply "real-world logic" to numbers, it doesn't work. Mathematics is a codified system of conventions, you can work within the conventions or you can ignore them.

(guess which one I/we chose...)
The thing is that an infinitely small part equals zero, so 0.999(rec) of an apple is an apple.

Look at your example of colors; you have a finite amount of red vs a finite amount of blue. In this case, it isn't true blue because it is 1/75000 red. If it had one part out of 1,000,000,000 then it wouldn't be true blue, because it would be 1/1,000,000,000 red. But in the case of 0.999(rec), it is 1/oo parts something else. And 1/oo = 0 (for those who don't know, a sideways figure 8 represents infinity, and is easier to type, though I suppose this explanation wastes that).

Your example with a country: How would one take an infinitely small portion of a country? You can't. However small a portion you take, it will be a finite amount. If you take an infinitely small portion, that infinitely small portion will be zero because 1/oo = zero.

0.999(rec) = 1 is not an "approximation". Assuming that it truly is infinite, it is a statement of fact. 0.999(rec) of a country isn't a country by convention or because mathematicians say so (anymore than 1+1=2 is a convention), it is because it is the same thing as 1 country. The same holds true for apples.

The reason that "real-world logic" doesn't work on 0.999(rec) is that infinitely small portions don't exist in the real world, and they are hard to visualize. When people think of infinitely small portions, a lot don't realize what that means, and have in their head a finitely small portion that is just really, really, really small. But no matter how small, a finite portion will always mess up the question because if the portion is finite, then no matter how small, it will not equal 1.

Real world logic applies to things that exist in it (for the most part), but in the real world, everything is finite (with the possible exception of the universe).
Pentarctagon wrote:but if they converge at infinity, doesn't that mean that they never actually converge (or at least that's what my math teacher said)?
No, because it isn't a process. You don't have to take 0.9 and add a .09, then add a .009, then add a .0009, etc. That would take infinite time, and we'd never get them to converge perfectly (like your math teacher said). If you are banging 9s into your calculator, you'll never get it to converge perfectly to 1 (until your calculator rounds it), because you can't put in infinite 9s.

However, we know that the 9s go on infinitely, and so we can see what that number represents in an instant. We don't require infinite time. So in our calculations, we use the more useful representation of 1, since we have already established they are identical. The number 0.999(rec) equals 1 because it is infinite. Any attempts to write down/use 0.999(rec) on a computer or paper are doomed to failure because we cannot write down or process an infinitely long number.

So in a sense your math teacher is correct (in terms of using 0.999(rec) in the real world), but it is not true that it doesn't equal 1.
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Re: Is 0.9 recurring equal to 1?

Post by Velensk » July 2nd, 2010, 3:00 pm

I've always settled for the answer that (.999 recurring *1) has a limit of 1. As .999 recurring is an infinite number it cannot be defined by traditional means a limit is needed.
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Re: Is 0.9 recurring equal to 1?

Post by Midnight_Carnival » July 2nd, 2010, 3:19 pm

Joram:
With my country and apple as well as colour analogies, I tried to portray how ridiculous the argument was.

Is an apple with any size bite taken out of it still and apple?
If a country looses land, people, etc... is is still a country?
And do you see the same colour as blue that I see as blue?

Answer to all 3: "if you say so".

the number "1" is an absolute value, hence you don't find "1"s in the real world. You wouldn't be able to tell a shopkeeper that your apple lost a cell and now it's not a whole apple and so you demand a price reduction, you'd probably get punched in the mouth if you start telling people "actually, that's not pure blue becasue there is 0.000003% red in there, do you want me to do a spectral anaysis?" and as for the country, well, it's best not to discuss politics too much. Numbers are created to make our lives easier: one coral reaf for example is a very complex notion which includes billions of organisms, as well as geographical features, etc... but we can say "if there is one coral reef in the see and another one grows there will be two coral reefs in the sea" and be understood.
I hope I have explained myself better here.

(summary: mathematics is rubbish, but convenient, Canada isn't a real contry and the French inventied blue.)
...apparenly we can't go with it or something.

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Re: Is 0.9 recurring equal to 1?

Post by chaoticwanderer » July 2nd, 2010, 6:40 pm

.9999.... and 1 are the same. If something is infintely approaching a value, it is that value; for example, we can see this kind of thing in calculus, where there are two accepted ways of writing an expression for the area under a curve. You have standard Leibnitz notation with integrals, or you can use Riemann sum notation where you take the sum of an infinite amount of infinitely small rectangles under the curve.

Now, realistically speaking, it's not possible to actually sum an infinite amount of rectangles, but when you use any sort of rectangular or trapezoidal approximation method (LRAM, MRAM, RRAM etc.) you'll see that as the degree of accuracy increases (based on how many rectangles are under the curve) it gets closer and closer to the actual value of the area under the curve which can be found through anti-differentiation. From a mathematical standpoint, we know that when we allow the number of rectangles to approach infinity the approximation becomes infinitely accurate.

Another way to look at it is if you consider .9999... as a geometric series, where the first term is .9 and r = .1

To find the sum of an infinite geometric series we can use the formula Si = a1/(1-r)

a1 = .9
r = .1
.9/(1-.1) = .9/.9 = 1
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Re: Is 0.9 recurring equal to 1?

Post by PsychoticKittens » July 2nd, 2010, 11:26 pm

Infinity is a principle. Principles are not numbers.

So 1.999endlessly is not a number. Therefore incapable of being equal to a number.

1.999xwhatever is a number, and still isn't equal to 2.

We just round because its really freaking close to 2, and simplifies things. Infinity doesn't really occur in day to day things, so you can't really say you'd ever have to worry about something literally being 1.999infinity, since eventually you would be able to say "oh hey, it stopped." And then it becomes a number, instead of a principle. Someones probly gonna bring up something, or divide, to prove me wrong. But who divides outside of school without rounding to the nearest something? Generally whole numbers.

Thats my thoughts on the matter anyway. No matter how many people talk about this we'll probably never get a conclusive answer due to how many different ways this is taught. Teachers would argue about it too.
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Re: Is 0.9 recurring equal to 1?

Post by Sapient » July 2nd, 2010, 11:59 pm

PsychoticKittens wrote:Infinity is a principle. Principles are not numbers.
The precise answer is that it depends on which model of mathematics you are using. For example, it is possible to create a model of geometry in which all lines must have a measurable width and thus the concept of an infinitely thin line (such as found in standard geometry) would not exist. This would change the way that you calculate the volume and surface area, for example.

Similarly, it is possible to create a model of mathematics different from the standard one.
For example, I could introduce a model of mathematics where there is no plain zero but instead only positive and negative numbers, thus +0 and -0.

Herein lies the confusion at the heart of the question. The fact that you would bother even writing "0.9 recurring" instead of "1" implies that you deliberately chose not to represent in a simplest form. Why would you choose not to express your answer in simplest form? Possibly because it does not recur infinitely, because you are using a computer program with limited storage and memory to express it (for example). This would imply that it reaches an end and does not recur infinitely, but only until the end of machine resources is reached.
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Re: Is 0.9 recurring equal to 1?

Post by Jetrel » July 3rd, 2010, 6:28 am

Alink wrote:This is a common pattern, people arguing that 0.999.. is not 1 usually rely on intuition (which is misleading here, as often when dealing with infinity) or vague "proof" where others have easily indicated the error. In the other hand, proofs that 0.999.. = 1 are often rigorous like:
1 = 9/9 = 9*0.111.. = 0.999..
And nobody argues against that.

Sorry, there is also people who accept the truth but are disturbed by its counter-intuitive aspects. To them, I suggest to check the underlying points used by their intuition (like "there is only one way to write a number")
The trick behind this counter-intuitive fact is the essential trick to calculus (especially, the core delta/epsilon proof). This whole business feels wrong, but it's a very profound and powerful fact.
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Re: Is 0.9 recurring equal to 1?

Post by pauxlo » July 3rd, 2010, 12:05 pm

PsychoticKittens wrote:Infinity is a principle. Principles are not numbers.
OK.
PsychoticKittens wrote: So 1.999endlessly is not a number. Therefore incapable of being equal to a number.
What is your definition of "number"? Only "finite decimal fractions"?

The word "number" has no precise meaning, you should qualify which kind of numbers you mean. There are, for example, natural (ℕ), integer (ℤ), rational (ℚ), real (ℝ), and complex (ℂ) numbers, to speak of the standard number sets. There are some other interesting sub- and supersets of these, but I know of none where 1.99995 is a number and 1.9999... is not one.

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Re: Is 0.9 recurring equal to 1?

Post by Tonepoet » July 3rd, 2010, 2:13 pm

First of all, I'd like to say I'm terrible at math.

Secondly, I'd like to suggest that without context .999999... may not be approaching a value but that instead, it may be a value in and of itself. I say this because we have another infinitely procedural calculation being scientifically considered its own distinctive value all on its own. It's known as pi, which is defined as 3.1459... it's not 3.2, it's not 3.3, it's not 4 and it's certainly not infinity but every digit known thereafter can be considered closer to infinity than how we know it now. As it's linearly impossible to approach one number without approaching the others in the same direction, which I'll submit a ruler and a pencil as proof of, I don't think this presumption should have too much difficulty in being accepted.

Thirdly, I'd suggest our mathematical proofs are inherently flawed because they rely upon converting fractions to decimals. I'd say that this is an inherently flawed process of converting abstract concepts into concrete values. I say this because not every fraction is a denominator or factor of ten which leads to an obvious problem when we attempt to represent thirds in tenths.

One third is a quantity defined as an amount which if multiplied three times, makes a complete whole. On a scale of one to ten, we have no such number because the proper amount should fall between three and four, which isn't allowed by our given degree of granularity. This means in order to represent a third most accurately on this one to ten scale, we have to use the number three since it is the very closest number on our scale. However in doing so we lose quantity as evidenced by the fact that three times three, is nine which is less than our whole of ten with a remainder of one, meaning 3/10 is not one third but rather 3/10ths and one third of a whole is.

In our decimal system we attempt to resolve this issue by circularly asking "How do we represent one third on a scale of 10?" In this fashion we end up continuously compromising the truth with our closest estimates to arrive to the nearest answer, which is how we come up with .333... to begin with. Now this is awfully convenient to use it as an estimate, as it's as close as we can get to one third without actually being one third. However it can never actually be one third. We know this because if it could, that would mean that we could repeatedly perform the exact same process with no variables whatsoever and inexplicitly get different results somewhere along the line. Didn't somebody once say that was the definition of insanity?

One could say that since we've infinitely postponed the event where we lose quantity, it may as well not exist but I'd suggest that it may be more accurate to say that we've infinitely postponed the event where we've actually solved the problem, so we may as well not have solved it. Edit: This would mean we never get to .333 = 1/3rd, which is the basis of this logic, I presume?
But you already accepted that 0.000... equals 0, why not accept that 0.999... equals 1. If there is a rule like "infinite sequence of 0 can be rounded to the lower integer", why not have a rule like "infinite sequence of 9 can be rounded to the upper integer"?
Zero has no value so it doesn't add or subtract from anything: X +/- 0 = X, always. It'll should only occur when we have no need for offset or change. Other numbers represent values of their own; 7-3 is 4 but 7+3 is 10. So what we're saying with .3333 recurring is 3 tenths of 3 tenths of 3 tenths of 3 tenths, when what we're actually trying to show each time is 3 tenths and 1/3rd of a tenth. I'd surmise this may even we lose quantity every step of the way but this is kinda contradictory to the fact that each decimal point we add makes the number larger, since .33 is more than .3

However what happens when we try to combine .9 recurring with another number, like ...001.75? 1+...001.7500000... should be ...02.750... but .9 recurring should be ...02.749... on the same scale?
Last edited by Tonepoet on July 3rd, 2010, 5:31 pm, edited 6 times in total.
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Re: Is 0.9 recurring equal to 1?

Post by Deusite » July 3rd, 2010, 2:28 pm

Jetrel wrote:The trick behind this counter-intuitive fact is the essential trick to calculus (especially, the core delta/epsilon proof). This whole business feels wrong, but it's a very profound and powerful fact.
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If 0.9 recurring =/= 1 then calculus is wrong, which it definitely isn't. It's equal and unequal at the same time, isn't that nice? Whether it's equal or not depends on what kind of maths you're dealing with. Or, basically, this debate will never have a definite answer, just like 0.9 recurring.
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Re: Is 0.9 recurring equal to 1?

Post by Tonepoet » July 3rd, 2010, 3:27 pm

Why can't calculus be inaccurate? Are there no margins of error in its applications? Also, can you extrapolate upon how something can be equal and unequal at the same time?

Also just to better demonstrate my thoughts on recombination:

1+1=2
.9+.9=1.8
.99+.99 = 1.98
.999+999=1.998
.9...+.9... = 1.9...8?
1.9...8 + ...2 = 2

We get the ...8 from the fact that we have to add 9+9 at the logical start of the process (not that we'll ever find it normally). Meaning 1.9...8 isn't 2, since you can add to it to achieve two. This in turn means you get different results from adding the two numbers with themselves, meaning these equations aren't equivalent, which they should be if all of the numbers held within are equivalent.

Also, if we can even have infinitely recurrent nines, then we should be able to achieve results that would mathematically combine with that ...8 by utilizing other recurrent numbers, unless the digits can somehow become maligned.

Just to go over this one more time though, I'm suggesting that one nineth isn't .1111...; which is the basis of our presumption. It's the point just immediately after .1111... that cannot be represented on the decimal scale, so we get a system error in the form of recursion when we attempt to make the conversion until the memory buffer overloads and you have to press Alt Control Delete to kill the program. =P This problem would not occur in this form on the larger scale of 90, where one ninth is 10 precisely.
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Re: Is 0.9 recurring equal to 1?

Post by Joram » July 3rd, 2010, 6:30 pm

Tonepoet wrote:Thirdly, I'd suggest our mathematical proofs are inherently flawed because they rely upon converting fractions to decimals. I'd say that this is an inherently flawed process of converting abstract concepts into concrete values. I say this because not every fraction is a denominator or factor of ten which leads to an obvious problem when we attempt to represent thirds in tenths.
So far, only one proof of at least 3 has relied on this. The algebraic proof and the proof using convergent series did not.

Tonepoet wrote:In our decimal system we attempt to resolve this issue by circularly asking "How do we represent one third on a scale of 10?" In this fashion we end up continuously compromising the truth with our closest estimates to arrive to the nearest answer, which is how we come up with .333... to begin with. Now this is awfully convenient to use it as an estimate, as it's as close as we can get to one third without actually being one third. However it can never actually be one third. We know this because if it could, that would mean that we could repeatedly perform the exact same process with no variables whatsoever and inexplicitly get different results somewhere along the line.
I don't follow you. You are in essence saying that the only way 0.3333(rec) could equal one third is if it ends? That is not true. 0.333(rec) can equal one third. In fact, it does equal one third. The reason that it does so is because it is infinite. If it was ever finite, then it would not.

Because it is infinite, your italicized statement is fallacious. We don't do the same thing and get different results, because we keep doing the same thing infinitely.

As I said before, we will never be able to do a calculation with 0.333(rec) and have it equal to 1/3, because the calculation would take infinite time. But 0.333(rec) does equal 1/3.

Tonepoet wrote:One could say that since we've infinitely postponed the event where we lose quantity, it may as well not exist but I'd suggest that it may be more accurate to say that we've infinitely postponed the event where we've actually solved the problem, so we may as well not have solved it. Edit: This would mean we never get to .333 = 1/3rd, which is the basis of this logic, I presume?
Again, your argument rests on the assumption that we cannot have an infinitely long number. You keep thinking in terms of an "end" or "solution where no more decimal places are added" to the point where you appear to be arguing that because the number has no end, it cannot equal 1/3. Such is not necessary in the least, unless you wish to have a practical application (though any application that requires precision down to the thousandth decimal place is rather impractical).
Tonepoet wrote:However what happens when we try to combine .9 recurring with another number, like ...001.75? 1+...001.7500000... should be ...02.750... but .9 recurring should be ...02.749... on the same scale?
If 1 = 0.999(rec), what is wrong with 2.75 = 2.74999(rec)? All of the proofs used on the former could be used just as easily with the latter (and only one of the proofs relied on fraction to decimal equivalence).


Tonepoet wrote:Why can't calculus be inaccurate? Are there no margins of error in its applications? Also, can you extrapolate upon how something can be equal and unequal at the same time?

Also just to better demonstrate my thoughts on recombination:

1+1=2
.9+.9=1.8
.99+.99 = 1.98
.999+999=1.998
.9...+.9... = 1.9...8?
1.9...8 + ...2 = 2

We get the ...8 from the fact that we have to add 9+9 at the logical start of the process (not that we'll ever find it normally). Meaning 1.9...8 isn't 2, since you can add to it to achieve two. This in turn means you get different results from adding the two numbers with themselves, meaning these equations aren't equivalent, which they should be if all of the numbers held within are equivalent.
As has been stated before, finite mathematics has no bearing on infinite mathematics. 0.999(rec) does not have an end, so any demonstrations with finite numbers will have no bearing whatsoever on the issue.

If 0.999(rec) ever had an end, it would not equal 1. Since it never has an end, it does. Showing that if it had an end, it wouldn't be equal to 1 therefore has no bearing on whether it equals 1 if it doesn't have an end.

Tonepoet wrote:This problem would not occur in this form on the larger scale of 90, where one ninth is 10 precisely. (On that note, .111 recurring, multiplied by ten would be 1.11.... wouldn't it? Nowhere near one tenth) It's an inherently faulty methodology, in other words.
You lost me, and I suspect the inherent fault is in your logic rather than the methodology.

If you are assuming that "1" represents 90 objects (which it appears you are doing), then what you have makes no sense with the current rules of multiplication. 1*1 would equal 1, which would mean you could multiply 90 objects times 90 objects and get 90 objects. In every numbering system in use, you have a 1's column. By removing it, you are changing the very nature of mathematics. So don't complain if you get faulty methodology.

If you mean base 90, then the math still works.
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Re: Is 0.9 recurring equal to 1?

Post by Tonepoet » July 3rd, 2010, 8:35 pm

I don't follow you. You are in essence saying that the only way 0.3333(rec) could equal one third is if it ends? That is not true. 0.333(rec) can equal one third. In fact, it does equal one third. The reason that it does so is because it is infinite. If it was ever finite, then it would not.
Why would it make a difference if it ends or not? In converting 1/3rd into counting base ten, it would seem to me that the reason 3 recurs indefinitely is because we always have a remainder left over. If we have a remainder we cannot eliminate, then that means there's a difference which does not equate to the concept we're trying to convey.
Again, your argument rests on the assumption that we cannot have an infinitely long number. You keep thinking in terms of an "end" or "solution where no more decimal places are added" to the point where you appear to be arguing that because the number has no end, it cannot equal 1/3. Such is not necessary in the least, unless you wish to have a practical application (though any application that requires precision down to the thousandth decimal place is rather impractical).
No, we can have infinitely long numbers, I never argued against that. Just not remainders. I am thinking in terms of an end to infinity but only because the process is made upon the basis of searching for those ends, which are inherently indefinite due to there being no degree of granuality which would allow it.
As has been stated before, finite mathematics has no bearing on infinite mathematics. 0.999(rec) does not have an end, so any demonstrations with finite numbers will have no bearing whatsoever on the issue.

If 0.999(rec) ever had an end, it would not equal 1. Since it never has an end, it does. Showing that if it had an end, it wouldn't be equal to 1 therefore has no bearing on whether it equals 1 if it doesn't have an end.
Hmm. I'd guess that technically there's no difference between the two, since we always have an infinite number of zeroes before and after every number. Also, in theory we should be able to manipulate this theoretical remainder if we can match it somehow. Couldn't we do that with another infinitely long number?

I have no idea why somebody would want to actually work with infinitely long numbers but just humor me for a moment and say that there may be an unforseen potentially useful application, like shifting the universal topography to create a warp drive engine? What would .999+Pi be? 4.14295...? It should be if .999 is equal to one. We may not be able to come to an answer without knowing all of the digits in pi, which is impossible. However I may reason that the answer may very well be different since we're not working from the last digit up, like the process of addition demands. How would all of the nines roll numbers up, if we could start this properly?
You lost me, and I suspect the inherent fault is in your logic rather than the methodology.

If you are assuming that "1" represents 90 objects (which it appears you are doing), then what you have makes no sense with the current rules of multiplication. 1*1 would equal 1, which would mean you could multiply 90 objects times 90 objects and get 90 objects. In every numbering system in use, you have a 1's column. By removing it, you are changing the very nature of mathematics. So don't complain if you get faulty methodology.

If you mean base 90, then the math still works.
I was talking about counting base 90, in a sense.

I had second thoughts about the parenthical statement, which only arose as a tangical thought, so I tried to retract the statement before anybody saw it. This is why it doesn't show up in my seemingly unedited post. So long as I've made somebody curious, I may as well share what I was trying to do:

I was attempting a little thought experiment by attempting to multiply the denominator, the numerator and the sum all unilaterally to see if I got the expected result.

One ninth of 90 is 10, so I thought 1.111... couldn't equal one ninth because it would be 1.111... not 10. Again, I'm not sure if this is correct because again, I'm terrible at actual math. Maybe since I was thinking about counting base ninety, I should've converted 1.1111... into counting base 90? I don't know; I'm sure I made some mistake otherwise I wouldn't have attempted to retract the statement so stealthily, so have compassion upon the poor uneducated Rocklobber. :p
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Re: Is 0.9 recurring equal to 1?

Post by pauxlo » July 3rd, 2010, 10:18 pm

Tonepoet wrote:
I don't follow you. You are in essence saying that the only way 0.3333(rec) could equal one third is if it ends? That is not true. 0.333(rec) can equal one third. In fact, it does equal one third. The reason that it does so is because it is infinite. If it was ever finite, then it would not.
Why would it make a difference if it ends or not? In converting 1/3rd into counting base ten, it would seem to me that the reason 3 recurs indefinitely is because we always have a remainder left over. If we have a remainder we cannot eliminate, then that means there's a difference which does not equate to the concept we're trying to convey.
You try to get philosophical with mathematics, and this does not work.

Any real number (including the rational ones) firstly is a concept, not anything to write down. If we want to write it down, we have to use some convention how to to encode them. Since humans have ten fingers, we usually are using the decimal system, but the same thing works in any positional number system.

Given any base X ∈ ℕ, X > 1, we can write any nonnegative real number r as a infinite sum:
r = ∑ a · X⁻ⁱ, with a ∈ { 0, ... X-1 } for all i.

Then we call the (infinite) sequence
a[-k], a[-k+1], ..., a[-1], a[0], a[1], a[2], ...
a "base-X-respresentation of r", and write a "decimal" point after a[0].
(For negative numbers we have the same with an additional minus sign prefixed.)

For most real numbers r and bases X the base-X-representation of r is unique, but for each base X there are infinitely many numbers (including the natural ones) with exactly two representations - one which has a 0-sequence at the end, and one who has a (X-1)-sequence at the end (and they differ in the last digit before this by 1).

In the usual handling of mathematics, we write the decimal representation (= base-10-representation) for the number, and say "this is the number".

For some numbers, the rational ones, there are also representations as fractions (a/b, with a an integer, b a natural number), and we take those representations as "the number", too.

So, we have 1/3 = 3/9 = 2/6 = 0.3333... (in base 10) = 0.1000... (in base 3) = 0.022222... (in base 3) = 0.2000... (in base 6) = 0.155555... (in base 6) = 0.11111... (in base 4) = 0.0101010101... (in base 2), for example.

All these are representations of the same number - and all the positional representations have infinitely many digits, but as they are periodic, we only need to write the some digits at the beginning until the period is clear. (The rational numbers are exactly those whose base-X-representations are periodic (for every X), the irrational numbers have only non-periodic representations.)

All of mathematics is about "taking representations for the real thing" (since the real thing only is something abstract as a number).

Tonepoet wrote: I have no idea why somebody would want to actually work with infinitely long numbers but just humor me for a moment and say that there may be an unforseen potentially useful application, like shifting the universal topography to create a warp drive engine? What would .999+Pi be? 4.14295...? It should be if .999 is equal to one.

Jes, it is equal. For 1+pi (as for every irrational number) there is only one decimal representation (or base-X-representation for every base), not two.

Tonepoet wrote: We may not be able to come to an answer without knowing all of the digits in pi, which is impossible. However I may reason that the answer may very well be different since we're not working from the last digit up, like the process of addition demands. How would all of the nines roll up if we could start this properly?

The sum of real numbers is not dependent on anybody in reality doing the calculation - remember, the decimal representation is only a representation of the number.

Tonepoet wrote: I was attempting a little thought experiment by attempting to multiply the denominator, the numerator and the sum all unilaterally to see if I got the expected result.

One ninth of 90 is 10, so I thought .111... couldn't equal one ninth because it would be 1.111... not 10. Again, I'm not sure if this is correct because again, I'm terrible at actual math. Maybe since I was thinking about counting base ninety, I should've converted 11.1111... into counting base 90? I don't know. Have compassion upon the poor uneducated Rocklobber. :P

If you switch bases, you also should transform your numbers to those bases. Here, for you:

0.1111... (base 90) = 1/89 = 0.01123595505617977528089887640449438202247191 01123595505617977528089887640449438202247191 ... (in base 10) (yes, that's a 44 digit period).

0.11111... (base 10) = 1/9 = 0.[10][00][00][00][00][00][00].... (base 90) = 0.[09][89][89][89][89]... (base 90) (I'm writing the base-90-digits here as [..].)

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Zarel
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Re: Is 0.9 recurring equal to 1?

Post by Zarel » July 4th, 2010, 3:51 am

Here, I will use 0.9~ to denote "0.9 recurring".
Sapient wrote:I have a very simple proof that 0.9 recurring is not the same as 1

On your next math test, when the answer is 1 you should write instead "0.9 recurring"
I have a very simple proof that 54+3 is not the same as 57

On your next math test, when the answer is 57, you should write instead "54+3"

(This works especially well if the question is "What is 54+3?" - you would probably get zero credit for writing "54+3")

P.S. If I did write "0.9 recurring" on my next math test, my math prof would be like "that's a stupid joke" and give me full credit anyway.
Alink wrote:If you think that 0.999.. is not 1, you must then show the error in all these various proofs.
This is correct, and it reminds me of something a math teacher once told me...

"There is no such thing as a counterproof."

In math, if you've proven something to be false, and someone else has proven the same thing to be true, that means one of the two proofs has an error in it.
Velensk wrote:I've always settled for the answer that (.999 recurring *1) has a limit of 1. As .999 recurring is an infinite number it cannot be defined by traditional means a limit is needed.
That's a silly thing to say.

0.9~ does not have a limit of 1, because 0.9~ is not a sequence. 0.9~ is a number. {0.9, 0.99, 0.999, ...} is a sequence, and 0.9~ is traditionally defined to be the limit of that sequence.

In other words, the following are true statements:

"{0.9, 0.99, 0.999, ...} has a limit of 1"
"0.9~ is equal to 1"

And the following are false statements:

"{0.9, 0.99, 0.999, ...} is equal to 1" (false because {0.9, ...} is not a number)
"0.9~ has a limit of 1" (false because 0.9~ is not a sequence)
Deusite wrote:It's equal and unequal at the same time, isn't that nice? Whether it's equal or not depends on what kind of maths you're dealing with. Or, basically, this debate will never have a definite answer, just like 0.9 recurring.
This is wrong. There is no universally recognized branch of mathematics for which the notation 0.9~ ≠ 1 is true. The definite answer is that 0.9~ = 1, no matter what.

P.S. Unlike Gambit, I have been an actual math teacher. :P

P.P.S. One of the things I've learned from being a math teacher is that there are quite a few math teachers who have no idea what they're talking about. So if your math teacher says that 0.9~ ≠ 1, that doesn't mean we're wrong, it means he/she is wrong.

P.P.P.S. Other things I've heard from math teachers and read in math textbooks that are wrong:

"You can't take the square root of a negative number."
"pi is exactly 3.14"
"0*1 = 1, 0*2 = 2, ..."

Sometimes, they lie to you because they don't know better. Sometimes, they lie to you because teaching you the truth would take too long (like when they say that it's impossible to take the square root of a negative number - lots of fifth grade math teachers just don't want to explain what imaginary numbers are).
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