Are you familiar with long divisions?
divide 1 by 9;
9 does not fit in 1, but fits in 10, so 1/9 = 0.1 with remainder 0.1
9 does not fit in 0.1 but it does fit in 10, so 1/9 = 0.11 with remainder 0.01
No amtter how long i continue i will keep dividing 1 by 9, with each division reulting in a 1 one decimal place more to the right.
This is a constant process, the calculation never changing. Each step adding a new 1 to the row. Therefore we are allowed to say 1/9 = 0.111...
See, it's because the steps are the same every time that we can say that the answer is 0.111... 🙂
THe numbers 0.1, 0.11, 0.11, ... are only approximations of 1/9. If you use approximations to disprove our "claim" then you'll succeed every time :p
Things like this are the reason i wish to be a teacher in math in a year and a half 😀
Originally posted by TheMaster37maybe I'm the one not clear :-)
Are you familiar with long divisions?
divide 1 by 9;
9 does not fit in 1, but fits in 10, so 1/9 = 0.1 with remainder 0.1
9 does not fit in 0.1 but it does fit in 10, so 1/9 = 0.11 with remainder 0.01
No amtter how long i continue i will keep dividing 1 by 9, with each division reulting in a 1 one decimal place more to the right.
This is ...[text shortened]... ime :p
Things like this are the reason i wish to be a teacher in math in a year and a half 😀
I dont see the diffrence between the endless task of dividing 1
in nine parts and multiplying an endless line of 1:s with 9
Those tasks are exactly the same.
so from examining the results of these there are at least four
possible conclusions (remember, I'm talking logic here, not anything else):
1. 1=0.99999....
2a. 1/9 not equal to 0.111....
2b. 9*0.1111.. not equal to 0.999....
3. multiplication is not inverse method of division
4. 9/9 not equal to 1
Do you get my point?
(This is basic logic. Nothing fancy. Nothing difficult. I am just pointing out that any proof reliant on an assumption of its truth is not, in fact, a proof at all. Examine, for example, Descartes proof of the existance of God. In which he states "I think therefore I am". A wonderful example of running around in futile circles.)
Ah, I see, i think 🙂 You have problems with dividing 1 by 9, then multiplying it again by 9 to get 0.999... and not 1, the number you started out with?
The best i can answer to that is that dividing 1 by 9 results in a neverending sequence. No matter how long you go on, you can't write the result down accurately. Multiplying it again with 9 takes that little inaccuracy with it, resulting in the 0.999...
Division and multiplication ARE eachothers inverse (at least the multiplication and division we speak of here 😛). That fact is the foundation of the proof that 1 = 0.999...
You gave me an idea for another proof:
Given two functions m(x,a) = ax and d(x,a) = x/a defined on the real plane. You see that these function are eachothers inverses;
m(d(x,a),a) = d(m(x,a),a) = x
Now we look at a specific pair (x,a), namely (1,9)
1= m(d(1,9),9) because m and d are inverses of eachother. Also,
m(d(1,9),9) = m(0.111... , 9) = 0.999... wich proves that 1 = 0.999...
Look at it this way if you're still not believing 😉
the difference 1 - 0.999... is smaller then any positive number you can name. The only non-negative number wich has that property (being smaller then any positive number) is 0 (*). Having a difference 0 means that the two numbers are equal.
(*) PROOF: Let's say there is a second number with that property (non-negative and smaller then all positive numbers), say O. Then because of the ordening on R we have either O<0, O>0 or O=0.
CASE 1: O<0. X<0 means that X is a negative number, so O is a negative number, contradiction with the fact that O is a non-negative number.
CASE 2: O>0. Make a number A = O/2. Then 0<A<O, wich is a contradiction with the fact that O is smaller then any positive number.
The only possibility left is CASE 3; 0=O wich must be true because we must have case 1,2 or 3.
QED.
Originally posted by TheMaster37Great proof! (the bottom part)
Ah, I see, i think 🙂 You have problems with dividing 1 by 9, then multiplying it again by 9 to get 0.999... and not 1, the number you started out with?
The best i can answer to that is that dividing 1 by 9 results in a neverending sequence. No matter how long you go on, you can't write the result down accurately. Multiplying it again with 9 takes th ...[text shortened]... ly possibility left is CASE 3; 0=O wich must be true because we must have case 1,2 or 3.
QED.
It proves 1=0.999... in a complete way, and thus
it proves 1/9=0.111... by induction.
It is still true however that assuming 1/9=0.111... in an effort
to prove 0.999...=1, is a logical fallacy.
This is because of the tiny inaccuracy you mention.
A solid proof has no room for inaccuracies.
Concluding this from my part:
I started reading this thread because I did not belive 1=0.999...
and have learned its so. I think we are on the same page here (well
almost :-) ) and that we're (or I am anyway) splitting hairs over the
1/9=0.111... issue. This might be because I love these kinds of arguments and the things they teach me.
I have really enjoyed this discussion. I have learned lots, trained my logic and hopefully provided entertainment for someone else. Maybe I even got my point across regarding the solidity of logical proofs.
Thanks again :-)
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Originally posted by TheMaster37Unless the difference is not a real number. I think we could say the difference d is infinitessimal but non-zero (and non real) and still construct some self consistent mathematics.
Look at it this way if you're still not believing 😉
the difference 1 - 0.999... is smaller then any positive number you can name. The only non-negative number wich has that property (being smaller then any positive number) is 0 (*). Having a difference 0 means that the two numbers are equal.
(*) PROOF: Let's say there is a second number with t ...[text shortened]... ly possibility left is CASE 3; 0=O wich must be true because we must have case 1,2 or 3.
QED.
Originally posted by iamatigerThat's the word I was looking for. In any case, they're a viable extension of the real numbers, but one which coventional mathematics avoids.
I think we can declare 1-0.9999... to be infinitesimal (non zero) and we still have workable mathematicics. See http://mathworld.wolfram.com/NonstandardAnalysis.html
Here's a question: is this extension of the reals (including infinite and infinitesimal numbers) generated as a field by R and d, where d = 1 - 0.999... ?
Theorem: Between any two rational numbers there are infinitely many more rational numbers.
Proof: Let a and b be different rational numbers, with a<b. Then (a+b)/2 is also a rational number, and between a and b.
Suppose that (strictly) between a and b there are only finitely many many rationals. Then there will be a largest such - let us call it M. Then (M+b)/2 is a rational number lying (strictly) between M and b, hence (strictly) between a and b. Bur (M+b)/2 is greater than the suppose greatest such rational, M. This contradiction shows that the assumption of only a finite number of rationals lying betwwen a nd b cannot be true. thus there are infinitely many.
Now 0.999..(recurring) and 1 are both rational numbers, so if they were different there must be infinitely many other rational numbers between them. Since there aren't any rational numbers between them, they must be the same.
Well there is one mistake in your algebraic approach. You subtract .999999... from one side and x from the other, and although x is .999999... you are not using true algebra.
Saying .9999 is equal to one is like saying in the Olympics someone who got .000001 of a second less tied with the other.
Look, I don't see what everybody's arguing about. (That sentence is true, but I haven't been following the nonstandard analysis part of this thread although it's probably cool and worth reading, so I'm in no way trying to 'have the last word'.)
We can take a few approaches to this:
1. First:
0.9 rec = 9/10 + 9/100 + 9/1000 + ... +9/10^n + ....
= 9(1 + 1/10 + 1/100 + 1/1000 + ... + 10^-n+...) - 9
= 9(1-10^-1)^-1 - 9
= 1
The first equality is really the definition of 0.9 rec, motivated by the basis representation theorem. The second follows from the fact that multiplication distributes over addition. I've included a 10^0 term, and compensated, because the third equality (which is the sum of a geometric series-this one converges since 1/10<1) is nicer that way. The last bit is just 10 - 9 = 1.
2. We can also think of 0.9 rec as the limit of the sequence defined by (10^n - 1)10^-n as n goes to positive infinity. Neatening things up, we get 1 - 10^-n, which approaches 1 as n approaches infinity.
There are a few other ways to do it. I think the most comfortable way to define 0.9 rec is by the series method, and a critical person could point out that 1 and 2 seem to be proving that two different things are equal to 1. However, note that:
9(1/10 + ... 10^-(n-1)) = 9(10^-n - 1)/(10^-1 - 1)
= 1 - 10^-n
The infinite series definition of .9 rec is just the limit of the left as n --> infinity, while the final expression is just the bit from 2, so they're equivalent, and the number written as 0.9 rec in positional decimal notation is just 1.
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Originally posted by SiskinYes, 0.999... and 1.000... are equal in their rational parts, but they have different infinitessimal parts. The difference between them is infinitessimal (non-rational) see?
Theorem: Between any two rational numbers there are infinitely many more rational numbers.
Proof: Let a and b be different rational numbers, with a<b. Then (a+b)/2 is also a rational number, and between a and b.
Suppose that (strictly) between a and b there are only finitely many many rationals. Then there will be a largest such - let us call it M. Then ...[text shortened]... bers between them. Since there aren't any rational numbers between them, they must be the same.
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Originally posted by AcolyteNo, there have to be an infinite number of different infinitessimals for it all to work. If b is an infitessimal number then 2.b is a larger infinitessimal number.
That's the word I was looking for. In any case, they're a viable extension of the real numbers, but one which coventional mathematics avoids.
Here's a question: is this extension of the reals (including infinite and infinitesimal numbers) generated as a field by R and d, where d = 1 - 0.999... ?