What is .99 repeating as a fraction?
Quote: WizardIt shows several explanations why 0.9...=1. For what it is worth, I've never known anyone with a strong mathematical background to claim otherwise.
Mathematics as we know it for well over a thousand years says that 0.999 ... = 1 . But the philosophical debate goes back almost 3000 years. Zeno of Elea was trying to prove mathematically that motion of two bodies always involved the sum of an infinite series. By his argument the sum of an infinite series could never equal a finite number. The point of the discussion was that the motion of two bodies was an illusion, and we were left with a universe that was actually "all is one". He was defending a school of thought championed by Parmenides that we associate nowadays with Zen Buddhism.
But most of the dissenting posts on this thread are not sounding so rational as to be defending a philosophical train of thought.
Somewhere in high school the problem is usually 'solved' as letting S be the sum..
S=.99999 ....
S= 9/10+ 9/10^2+ 9/10^3+ 9/10^4.... (equation #1) definition of S
1/10*S= 9/10^2+ 9/10^3+ 9/10^4.... (equation #2) multiplying (equation #1) by 1/10
S can be replaced by 10/10*S in equation #1
9/10*S = 9/10 (equation #3) gotten by subtracting (equation #2) from (equation #1)
S=1 (equation #4) gotten by multiplying both sides by 10/9
Quote: WizardI didn't read all of it, but agree it is a good mathematical treatment of the concept. It shows several explanations why 0.9...=1. For what it is worth, I've never known anyone with a strong mathematical background to claim otherwise.
Here is the link again.
My favorite line?
Fascinating thread. I sure am glad I didn't bet mustangsally that 9/9 is a fractional expression of .999... I think I would have lost :)
This statement that .999... is equal to zero, though, has a little wording problem. You see, they didn't prove that .999... is equal to 1. That would mean, of course, that .999... is always equal to 1. That's not correct. What they proved is that certain fractions are equal to both .999... and also to 1. For instance, 9/9 is a fractional expression for .999... and also 1.
You see, other fractions, such as 2/2, for instance, only equal 1, they never equal .999... The proof, from the website above, would fail for this fraction.
So, to just say .999... is equal to 1 is misleading. It should be stated that .999... is equal to 1 for certain fractions. ex 9/9. See, 1/1 =1, 2/2=1, 3/3 = .999... or 1 There's 2 valid expressions for this particular fraction. However, it can't be both. You can interpret 9/9 to be either .999... or 1. If you have 2 different people one might interpret 9/9 as being 1, another might interpret it as .999..., both would be valid.
Fascinating Stuff.
The website from above did inspire me to try to take a stab at proving that 1 is not equal to .999... mathematically instead of logically.
.999... - .999... = 0
1 - .999... = a number >0
or
solve for x
.999... + x = 1
1 + x = 1.x
The interest thing is of course these fractions with two possible interpretations.
1 - 1 = 0
1 - 2/2 = 0
1 - 9/9 = a number >0 , or just 0 It has two valid answers.
This might make for a nice prop bet out here in Vegas:)
Quote: DocThanks for not exposing us to that drivel twice.
Wow. Dude, chill out.
YDR11: I think that you and I have had polite exchanges on this topic, even though we disagree. I think that you believe in your position, even though I contend that your sources have errors in logic: I have pointed out the circular reasoning and the attempts to discredit the .999... = 1 position by disregarding/refuting the standard definitions that are used by those holding that position. Even though we disagree, I have found no reason not to maintain a polite dialog.Quote: YoDiceRoll11Wow. Dude, chill out.
In contrast, JyBrd0403 has made numerous posts in this thread that make no sense at all. On several occasions yesterday, he noted that he was drunk, and I mostly disregarded his posts, allowing that these may have just been the typical results of intoxication. If that was the case, it's time either to sober up or to spend more time away from this forum. Earlier, I made a comment about sometimes forming the opinion that someone either is an idiot or is putting me on. I suspect (don't really know) that JyBrd0403 is putting us on. I am not convinced that he has considered this topic sufficiently to have actually formulated an opinion. I think he just likes to jerk folks around by posting stuff that I think I complimented by calling it drivel. He may even be another one of the trolls who show up here far too often just trying to provoke arguments by posting inane comments/opinions/claims. He has been a member of the forum for 10 days and has made 54 posts in 3 threads. I don't think I have noticed a post yet that really made much sense.
Quote: JyBrd0403mathematically instead of logically.
The interest thing is of course these fractions with two possible interpretations.
It has two valid answers
This statement that .999... is equal to zero, though, has a little wording problem.
The sum of the vanishingly small 9s approaches 1 as they go toward infinity. That apparent paradox is built into the system that we use to express the numbers.
Quote: JyBrd0403
1 - 2/2 = 0
this is great :-)))))))))))))
Quote: DocYDR11: Even though we disagree, I have found no reason not to maintain a polite dialog.
In contrast, JyBrd0403 has made numerous posts in this thread that make no sense at all. On several occasions yesterday, he noted that he was drunk.....
LOL, ok fair enough mate.
I have a hard enough time remembering ten characters.Quote: MathExtremistAs a postscript, I'm reading "The Information", the latest book by James Gleick. In it he discusses the cuneiform writing of the ancient Babylonians. Their tablets were full of accounting records, things like receipts for barley and livestock, and they were recorded in sexagesimal: base 60.
Drinking again tonight, Doc. Just in case, I say something out of line that I have to take back the next day, you'd know why. I thought it would be fair to mention the fact.
Babylonian numerals
Quote: JyBrd0403Well, doc, it may be drivel, but you're going to have to prove that. Now, the proof that was offered for .999... equals 1 was fascinating in and of itself, but as I stated it didn't prove that .999... is equal to 1. The proof fails for fractions such as 1/1, 2/2, 4/4, 8/8, 10/10 etc.
The proof doesn't involve fractions at all.
1) Let's assume you believe that 0.999... - 0.999... = 0
2) Let's also assume you believe that 9 + 0.999... - 0.999... = 9.
3) Let's also assume you believe that 0.999... is properly a decimal number, and as such, can be multiplied by 10 and in doing so merely shift the decimal point to the right one space, as is true for multiplication by 10 of any other decimal number.
4) Therefore, 0.999... * 10 = 9.999...
5) Let x = 0.999...
6) 10*x = 9.999...
7) 10*x - x = 9.999... - 0.999...
8) 9*x = 9
9) x = 1
Therefore, 0.999... = 1, QED.
I still question the use of infinity.
Quote: MathExtremistThe proof doesn't involve fractions at all.
3) Let's also assume you believe that 0.999... is properly a decimal number, and as such, can be multiplied by 10 and in doing so merely shift the decimal point to the right one space, as is true for mltiplication by 10 of any other decimal number.
.
Point, # 3 is where you run into problems with this proof. Assuming .999... is a proper decimal number. It's not. It runs to infinity. It hasn't been defined except to say that it is < 1. So you can't properly multiply it by 10. That would be like multiplying 10*n. If it wasn't for 3). I would agree with you.
Quote: JyBrd0403Point, # 3 is where you run into problems with this proof. Assuming .999... is a proper decimal number. It's not. It runs to infinity. It hasn't been defined except to say that it is < 1. So you can't properly multiply it by 10. That would be like multiplying 10*n. If it wasn't for 3). I would agree with you.
So, by your logic, pi, sqrt(2) and 1/3 arent' "proper decimal numbers" either? Would you be stumbled if you ever had to compute circumference of a circle given a radius (of 10)?
But, to answer your question, like i told someone else on this thread count to infinity and you may prove your point. I'm sure we'll all enjoy the results. Of course, if you ever report back we'll all know your full of sh**.
Quote: JyBrd0403Point, # 3 is where you run into problems with this proof. Assuming .999... is a proper decimal number. It's not. It runs to infinity. It hasn't been defined except to say that it is < 1. So you can't properly multiply it by 10. That would be like multiplying 10*n. If it wasn't for 3). I would agree with you.
Incorrect. 0.999... is simply a notational shortcut for the sum of the series 9*0.1^n for all n>1. That is a converging series and has a limit. What is that limit?
Do you not think that 0.333... is a proper decimal number either and a proper representation for the fraction 1/3? In short, do you dispute the notational convention? Because if you do, we're at the same definitional impasse as previously in this thread.
Do you actually think that no rational numbers have valid decimal representations other than those with solely factors of 2 and 5 in the denominator? Because if you do, that's virtually all rational numbers, so you're basically throwing out the usefulness of the decimal system altogether. On what grounds do you dispute the notational conventions in use by virtually every mathematician on the planet?
Read these links if you need some education:
Decimal representation on Wikipedia
Repeating decimal on Wikipedia
You forgot to define the answer of what 0.999... ^n equals. You answered it with n. Bad answer. It basically, means 0.999... ex 0.999^x^y^z^a^b^c = 0.999... Matter of fact, that's what we we're talking about on this thread, you must have missed it.
Yes, in case you missed it, I don't think 0.333... is a proper decimal number either, however, I do think it is a proper representation for the fraction 1/3. Drinking tonight, but to try to make sense, I think 0.333... is > 0.332... and < 0.334... However, it has not been properly defined.
Don't have a clue about the rest of the stuff you're talking about, not to be harsh, but I don't think you're making sense.
On what grounds do I dispute the notational conventions in use by virtually every mathematician on the planet?
You should check the post with the website that was trying to prove that 0.999... is equal to 1. They didn't prove that, but they sure as hell, at least in my opinion, proved that virtuauly every notational convention in use by virtually every mathematician on the planet is incorrect! Ex 9/9, 3,3, etc. I shouldn't say every notation, and I guess I didn't, so there you go, but, don't blame me, blame the Wiz and the other guy who posted that link. I had no idea.
To reply to a point made before, what is the limit on a converging series? Maybe, this has been agreed upon before, if so, I had no idea. They must have left me out.
But, I'm still curious, as to what the limit is.
Quote: MathExtremistThe proof doesn't involve fractions at all.
1) Let's assume you believe that 0.999... - 0.999... = 0
2) Let's also assume you believe that 9 + 0.999... - 0.999... = 9.
3) Let's also assume you believe that 0.999... is properly a decimal number, and as such, can be multiplied by 10 and in doing so merely shift the decimal point to the right one space, as is true for multiplication by 10 of any other decimal number.
4) Therefore, 0.999... * 10 = 9.999...
5) Let x = 0.999...
6) 10*x = 9.999...
7) 10*x - x = 9.999... - 0.999...
8) 9*x = 9
9) x = 1
Therefore, 0.999... = 1, QED.
Sorry, I looked at this proof again. This fails at 8) 9*x = 8.999... 9) x= 0.999...
Quote: JyBrd0403Sorry, I looked at this proof again. This fails at 8) 9*x = 8.999... 9) x= 0.999...
ME is saying that (9.999...) minus the (.999...) part is 9. It can't be 8.999... (repeating), otherwise .999... =/= .999...
And anyhow, you don't understand that you are arguing against definitions of notation and processes, not absolutes. Whatever you think, it is wrong, unless you decide to come up with a new way of doing arithmetic and it supplants the current one worldwide.
Quote: MoscaME is saying that (9.999...) minus the (.999...) part is 9. It can't be 8.999... .
Maybe I have the notations and processes wrong, but I think that you stated at 8) 9*x= 9. That would be 9*.999... = 8.999... . That would also be a proof that .999... and 1 are not equal, by the way.
Oh, by the way, doc, that would make my other posts that .333... couldn't be multiplied by 10, drivel. Superbowl Sunday, what can I say. lol
Quote: JyBrd0403Maybe I have the notations and processes wrong, but I think that you stated at 8) 9*x= 9. That would be 9*.999... = 8.999... . That would also be a proof that .999... and 1 are not equal, by the way.
It follows statement 7, as a simplification of terms.
Quote: MoscaIt follows statement 7, as a simplification of terms.
Okay, but you stated at 5) let x = .999... then at 8) 9*x = 9 that would be incorrect. The answer should be 8.999... not 9.
Quote: JyBrd0403Correction, I do believe .333... is an appropriate expression of 1/3. I just don't think that .333... x 3 is equal to 1. Took me a minute to think about that but, I think that's right.
Okay, we'll go with this. You accept that 0.333... = 1/3.
In doing so, you accept that 0.333... is both properly a real number and is equal to the fraction 1/3, which is also a real number.
A fundamental rule of algebraic manipulation is that if a = b, then a*c = b*c, for all real numbers a, b, c.
Now then, do you accept that 1/3 * 3 = 1?
If so, we have
1) 0.333... = 1/3.
2) Therefore, 0.333... * 3 = 1/3 * 3.
3) Since 1/3 * 3 = 1, then 0.333... * 3 = 1. QED.
Quote: JyBrd0403Okay, but you stated at 5) let x = .999... then at 8) 9*x = 9 that would be incorrect. The answer should be 8.999... not 9.
Erm, this is what that proof sets out to show that 8.9999.... IS 9 (and hence 0.9999.... is 1). The proof is all but done by step 8, you are just tidying up in the next two steps.
Which step in the proof is incorrect and does not follow from its immediate predecessor? That's how proofs work.
Quote: JyBrd0403Okay, but you stated at 5) let x = .999... then at 8) 9*x = 9 that would be incorrect. The answer should be 8.999... not 9.
Assuming arguendo that your logic is correct, which it's not:
Suppose x = 0.999..., as was given, and that 9*x = 8.999... as you state.
1) 9*x = 8*x + x
2) 8.999... = 8.0 + 0.999...
3) Therefore, 8*x + x = 8.0 + 0.999...
4) Since x = 0.999..., we'll subtract out those terms from #3 and arrive at 8*x = 8.0.
5) Therefore, x = 8/8, which also = 1. Therefore, x = 0.999... = 8/8 = 1. QED.
Quote: JyBrd0403Okay, but you stated at 5) let x = .999... then at 8) 9*x = 9 that would be incorrect. The answer should be 8.999... not 9.
let x = .999...
Then 10x = 9.999..., right?
10x - x is 9.999... - .999..., right?
And, 9.999... - .999... is 9, right?
And, 10x - x is also 9x, right?
So 9x is 9, right?
And therefore x is 1, right?
And that is the nature of infinity, when applied to infinitesimals.
x = 1
QED
Can we agree on this?
I just ask, because you seem to reject the other basic principles of math and algebra
1/3 * 3 could validly be interpreted as equaling 1, or also as validly be interpreted as equaling .999...
It's those weird fractions that they proved have 2 interpretations. Either .999... or 1. Again, try doing the same thing with a fraction such as 1/1, or 2/2, or 4/4, or 8/8, or 10/10. They can't be interpreted two different ways.
Quote: MathExtremistAssuming arguendo that your logic is correct, which it's not:
Suppose x = 0.999..., as was given, and that 9*x = 8.999... as you state.
1) 9*x = 8*x + x
2) 8.999... = 8.0 + 0.999...
3) Therefore, 8*x + x = 8.0 + 0.999...
4) Since x = 0.999..., we'll subtract out those terms from #3 and arrive at 8*x = 8.0.
5) Therefore, x = 8/8, which also = 1. Therefore, x = 0.999... = 8/8 = 1. QED.
If you subtract those terms from 3) (I assume that means we're taking the x's out of the equation) that would leave 8 = 8 + 0.999... That would equal 8.999... So, you're not really making sense.
Quote: JyBrd0403If you subtract those terms from 3) (I assume that means we're taking the x's out of the equation) that would leave 8 = 8 + 0.999... That would equal 8.999... So, you're not really making sense.
There's the problem. You don't know algebra. Subtracting terms means simplifying the equation. ME is subtracting x from the left side, and also its defined equivalent .999... from the right side.
I know that you are trolling here, but whatever, it's more fun than playing Free Cell.
Quote: JyBrd0403... Unfortunately, you expressed 1/3 in decimal form as .333... so there can be no misinterpretation, so .333 * 3 = .999... and can never be misinterpreted for equaling 1.
The obvious problem here is you believe that the value of something is not the same if written as a decimal vs written as a ratio.
If you agree that (1/3) = .333...
and you agree that (1/3) * 3 = 1
Then using simple algebra, it can be proved that .333... * 3 = 1.
Quote: JyBrd0403Here's the problem. 3) then 0.333... *3 = 1 . It doesn't equal 1 it equals 0.999... 1/3 * 3 = a number >1 or .999... Unfortunately, you expressed 1/3 in decimal form as .333... so there can be no misinterpretation, so .333 * 3 = .999... and can never be misinterpreted for equaling 1.
1/3 * 3 could validly be interpreted as equaling 1, or also as validly be interpreted as equaling .999...
It's those weird fractions that they proved have 2 interpretations.
Nope. I know you're trolling here too, but just to put a stake in it, equality is transitive. That means if a=b and b=c then a=c. Here, you agree that 1/3 = 0.333... and that 0.333... * 3 = 0.999... and that 1/3 * 3 = 1. Since by algebra, if x = y then x*z = y*z for all real x, y, z, we have
a = 1
b = 1/3 * 3
c = 0.333... * 3
d = 0.999...
You admit that
a = b,
b = c, and
c = d
Therefore by transitivity, a = d, that is, 1 = 0.999...
QED again.
Quote: MathExtremist
If so, we have
1) 0.333... = 1/3.
2) Therefore, 0.333... * 3 = 1/3 * 3.
3) Since 1/3 * 3 = 1, then 0.333... * 3 = 1. QED.
ME has you there. This is why I don't accept .333... = 1/3. It comes extremely close to it. The use of infinity while trying to pass .333... as a "real" number is absurd.
Quote: YoDiceRoll11ME has you there. This is why I don't accept .333... = 1/3. It comes extremely close to it. The use of infinity while trying to pass .333... as a "real" number is absurd.
YDR, I am curious, what is the highest level of math you completed in high school or college? I am in no way trying to be condescending, but the views you have espoused in this thread are interesting, so I am just curious what your background is.
Quote: YoDiceRoll11ME has you there. This is why I don't accept .333... = 1/3. It comes extremely close to it. The use of infinity while trying to pass .333... as a "real" number is absurd.
This is the part I don't get. What do you think is the limit of the series 3 * 0.1^n for n>=1? If it's not 1/3, what is it?
Quote: JyBrd0403Correction, I do believe .333... is an appropriate expression of 1/3. I just don't think that .333... x 3 is equal to 1. Took me a minute to think about that but, I think that's right.
So if you believe
1/3 = 0.3333333333333333.... (equation #1)
1/3 = 0.3333333333333333.... (equation #2)
1/3 = 0.3333333333333333.... (equation #3)
then
1 = 0.99999999999999999.... (equation #4) And if you add the three equations
To do the left side I Assume you believe 1/3+1/3+1/3 = 3/3 = 1
It isn't rocket science.
Quote: YoDiceRoll11Oh absolutely, thank you for asking. I have completed calculus(B-), statistics (A), criminal justice statistics (A), and business math(A) while in college. Nothing higher than that I'm afraid.
So, you don't believe in Pi?
Quote: weaselmanSo, you don't believe in Pi?
:), I believe in pi in its obvious practical applications in engineering and construction. I don't think you can ever fully notate it, but I think for its purposes, notating it to just 6 decimal points is enough to be accurate. Pi is one thing, using infinity to extrapolate a string to the next "real" number, is nothing but a theory, and holds no real practical application. Anyways, it's college basketball season guys. Take it easy.
Quote: YoDiceRoll11:), I believe in pi in its obvious practical applications in engineering and construction. I don't think you can ever fully notate it, but I think for its purposes, notating it to just 6 decimal points is enough to be accurate.
Who cares about the number of digits, that is "enough"? If you believe that Pi is a "real" number, then it should not be news to you that some "real" numbers require infinite number of digits to be expressed in the base-10 notation (in fact, there are uncountably more numbers like that, than those that can be expressed with a terminating fraction). There is nothing "absurd" about it or about performing arithmetic operations on those numbers.
(side note: and yes, you actually can fully notate pi, and you only need two characters for that. Hint: first character you need is "p" :)
just like you can fully note one-third using only 6 characters: 0.3... You see, "fully notate" does not necessarily mean that you have to spell out every single digit. Simply providing a method to determine the digit at any given position is sufficient).
Quote:Pi is one thing, using infinity to extrapolate a string to the next "real" number,
Nobody is using infinity to extrapolate anything. What do you mean by "extrapolate" by the way? I suspect, like with "infinity" (you can't really "use" it for anything) or "number" (you can't "extrapolate" anything "to" it), you imply some meaning to this term, that is different from what the rest of us think.
Quote: MathExtremistNope. I know you're trolling here too, but just to put a stake in it, equality is transitive. That means if a=b and b=c then a=c. Here, you agree that 1/3 = 0.333... and that 0.333... * 3 = 0.999... and that 1/3 * 3 = 1. Since by algebra, if x = y then x*z = y*z for all real x, y, z, we have
a = 1
b = 1/3 * 3
c = 0.333... * 3
d = 0.999...
You admit that
a = b,
b = c, and
c = d
Therefore by transitivity, a = d, that is, 1 = 0.999...
QED again.
I understand that. Here's what I was trying to explain, that this proof is only good for certain fractions. If you do the proof for 2/2 it would be.
a=1
b= 1/2 *2
c= .5 *2
You see 2/2 doesn't have a representation for .999... and therefore, can never = .999.... The proof would have to work for all fractions in order to state that .999... is always equal to 1. What you're proving is that certain fractions ex. 9/9 can be represented as either .999... or 1, or 9/9 =1 or .999... To prove that .999... is always equal to 1. You would have to be able to substitute b=1/3*3 with b=1/2*2 and get the same answer, otherwise the proof doesn't hold up for the fraction 2/2.
Quote: JyBrd0403What you're proving is that certain fractions ex. 9/9 can be represented as either .999... or 1, or 9/9 =1 or .999... To prove that .999... is always equal to 1. You would have to be able to substitute b=1/3*3 with b=1/2*2 and get the same answer, otherwise the proof doesn't hold up for the fraction 2/2.
Nope. Once you establish that 0.999... is equal to 3/3, and that 3/3 = 1, then you've also established that 0.999... is equal to all other representations of 1. That's what transitivity of equality means. You can't have one form of 1 (e.g. 3/3 or 1/3 * 3) be equal to 0.999... but not another form (e.g. 2/2 or 5/5 or cos(2pi) or combin(52,52) or 10^0) without violating that feature or otherwise at some point you'd reduce it to saying "1 doesn't equal 1 for some values of 1". To use your example:
9/9 = 0.999... = 1 = 2/2. There you go: 0.999... = 2/2.
And 2/2 = 2+cos(pi) = 11^2-5! = log(10) and so forth. Everything there is equal to 1.
Common sense dictates that .999... should represent the latter, but it doesn't. It's even that .(9)>.9+.9/10+...+.9/10^n for any n including infinity.
In simple terms, while the latter sequence increases with n in analysis, the digits of .999... are not a sequence and can only be considered simultaneously, in which they are just an alternate notation for 1.
Quote: MathExtremistNope. Once you establish that 0.999... is equal to 3/3, and that 3/3 = 1, then you've also established that 0.999... is equal to all other representations of 1. That's what transitivity of equality means. You can't have one form of 1 (e.g. 3/3 or 1/3 * 3) be equal to 0.999... but not another form (e.g. 2/2 or 5/5 or cos(2pi) or combin(52,52) or 10^0) without violating that feature or otherwise at some point you'd reduce it to saying "1 doesn't equal 1 for some values of 1". To use your example:
9/9 = 0.999... = 1 = 2/2. There you go: 0.999... = 2/2.
And 2/2 = 2+cos(pi) = 11^2-5! = log(10) and so forth. Everything there is equal to 1.
I completely disagree with you here. That's like solving an equation for x, let's say solving that x=2, then saying for every other equation x=2, because we've solved for x on this one particular equation. See, x mathematically speaking means unknown. From there when you solve x for a particular equation and state x=2, it means x=2 for that particular equation, however, mathematically speaking x still = unknown. You're trying to say that x= unknown = 2. That's incorrect. X is still a mathematical expression of an unknown, to prove that x = 2 you would have to prove that it works for all equations, just like x = unknown works for all equations.
Another way of pointing this out is 1=1/1= 2/2= 3/3= 4/4, etc The proof works for all these fractions. 2/2=1 , 3/3=1 , 4/4=1. The proof would, of course, fail for 5/4, or 3/2 etc. That would prove that those fractions do not equal 1. That's what I'm saying here, that .999... does not equal 1 for certain fractions. .999...= 1 only works with certain fractions. The proof fails for ex. 2/2. Which proves that .999... does not equal 1 for all fractions. It still blows my mind that it .999... does equal 1 for certain fractions.
