Let's discuss Mathematics

[Souron]

Shifting the radix works for finite decimals, but that doesn't mean it works for infinite ones.

No integer number of the digit 3 will ever make 0.33333... = 1/3.

More formally: no integer n will make the sum of {n>0|3*10-n} = 1/3.

Yes I am aware that this is an old debate and I've been on both sides. There was a point when I had your view point. That was before I took calculus.

 

[ParadigmShifter]

So explain how you define it for infinite decimals then.

Maybe you been reading this, this is the minimum for arguing nonstandard analysis

http://en.wikipedia.org/wiki/Nonstandard_analysis

 

[Souron]

Please explain what 0.999... times 10 is then. Good luck!

What do you mean explain? I can define such a number, and it is plainly an irrational number. Irrational numbers cannot generally be expressed except by restating their definition. For instance the square root of two cannot be explained as anything other than the number who's square is two. Simmilarly 10 * 0.999.. is the number that is the product of 10 and 0.999... .

 

[ParadigmShifter]

0.999.. is not an irrational number. EDIT: Since it is equal to 1

All recurring decimals are rational.

I laid down the challenge earlier... give me a recurring decimal and I will give you the rational number it represents.

No takers so far

 

[Souron]

Maybe you been reading this, this is the minimum for arguing nonstandard analysis

http://en.wikipedia.org/wiki/Nonstandard_analysis

Never heard of it before. My arguments stem from the definition of a limit.

The limit as n approaches infinity of the sum of the series {9*10^n} is 1. But the sum of the series {9*10^n} is not 1.

 

[ParadigmShifter]

Infinity is not a real number.

This is where your argument fails.

Yes it is (sum of series as n-> inf is 1). Again, infinity is not a real number and don't treat it as one.

 

[Souron]

0.999.. is not an irrational number. EDIT: Since it is equal to 1

All recurring decimals are rational.

Well if you define it that way then sure.

I laid down the challenge earlier... give me a recurring decimal and I will give you the rational number it represents.

No takers so far

No deal. I can do it too. But you have to assume that 1/9 == 0.1111.... You won't get a rational number for this though:

0.101001000100001...

Though it is a decimal, and something in it does repeat in an obvious pattern...

 

[ParadigmShifter]

Yep, that's a non-recurring decimal though. EDIT: Congrats you found an irrational number, there are infinitely many more of those than rational numbers though, so it's not too hard

As I said, all recurring decimals (in any base, by the way) are rational.

EDIT: I should add any integer base since obviously pi would be rational in base pi

 

[Souron]

Infinity is not a real number.

This is where your argument fails.

Yes it is (sum of series as n-> inf is 1). Again, infinity is not a real number and don't treat it as one.

How am I treating infinity as a real number?

In fact my point is that infinity is not an integer.

 

[Souron]

Yep, that's a non-recurring decimal though. EDIT: Congrats you found an irrational number, there are infinitely many more of those than rational numbers though, so it's not too hard

As I said, all recurring decimals (in any base, by the way) are rational.

This holds if and only if you define 1/(10-1)=0.1111...

 

[ParadigmShifter]

Firstly, you did a sneaky edit

Secondly, your argument for .999... x 10 obviously involves infinity, unless you are gonna say there are infinity minus 1 9's.

Thirdly prove [EDIT: Me WRONG] that all rationals can be expressed as a finite fraction or provide a counterexample

EDIT2: Good debate, lame subject though. You are wrong. You can do better in this thread.

 

[Souron]

Firstly, you did a sneaky edit

Well it's for the good, because what I had was wrong.

Secondly, your argument for .999... x 10 obviously involves infinity, unless you are gonna say there are infinity minus 1 9's.

Yes, but two infinities are not necessarily equal. If we relax the defintion of 0.999... so that it is not the sum of the particular sequence I named for it, then yes 10 * 0.999... does equal 9.999..., but then 0.9999... does not always equal 0.9999....

EDIT: in other words your definition fails at 9.999...-0.999... = 9.

Thirdly prove [EDIT: Me WRONG] that all rationals can be expressed as a finite fraction or provide a counterexample

So a point of contention between us is if the notation 0.11111... is should be considered equal to 1/9. It we take it to be so, then we'd agree. And it is a useful definition.

Perhalps it's my turn for a proof:
1/9
= 0 remainder of 1
= 0.1 rem of 1
= 0.11 rem of 1
....

repeating this process will always leave a remainder of 1. This means that no number of digits can ever be enough to reach 1/9.
EDIT:I admit that it's not a very satisfactory proof.

EDIT2: Good debate, lame subject though. You are wrong. You can do better in this thread.

Age old debates naturally tend have large amounts of contention.

 

[ParadigmShifter]

Indeed. You are still wrong though, and I'd rather this thread discussed serious mathematical stuff than nonsense, sorry

Open a non-serious maths thread if you want to talk spam

Nevertheless I admire your interest in arguing and added you to my buddies but not in this thread, OK?

 

[Souron]

Since you have failed to provide a satisfactory proof, you should be challenging your beliefs, not taking things on faith. In the formal language of math there is little room for disagreement. Either there is a proof, or you must resort to definitions.

I agree, however, that this discussion is best served in it's own thread.

 

[ParadigmShifter]

I'm too drunk to provide a proof

The 0.999... thing really needs another thread though.

EDIT: I think I gave a satisfactory proof. It's only you saying 10 x .999... - .999... != 9

 

[sanabas]

I'm too drunk

Interested in the proof though.

EDIT: Especially cos it wasn't mentioned on mathworld or wiki, so it's either homework or a new discovery

I think it's merely something interesting that I came up with the first time I met this series. But I couldn't see how it helped me to prove the series. I'll drop an even bigger hint.

Any odd number can be written as (p.2^n) - 1, where p is odd. Correct? What happens when you iterate from that?

Well I don't think 0.999 recurring is that obvious unless you are as familiar with maths as you say you are. you are parading urself in this thread I told it to my friends and only 1 believed me! And no I don't have 1 friend! have to be careful when posting you.

This just shows you need better friends. If it wasn't obvious, I wouldn't believe you or not, I'd sit down and try to work it out myself.

 

[cameltoe]

Perhalps it's my turn for a proof:
1/9
= 0 remainder of 1
= 0.1 rem of 1
= 0.11 rem of 1
....

repeating this process will always leave a remainder of 1. This means that no number of digits can ever be enough to reach 1/9.
EDIT:I admit that it's not a very satisfactory proof.

You have proven that for any finite number of 1's, 0.1...1 < 1/9. Induction does not extend to "infinity".

Shifting the radix works for finite decimals, but that doesn't mean it works for infinite ones.

No integer number of the digit 3 will ever make 0.33333... = 1/3.

More formally: no integer n will make the sum of {n>0|3*10-n} = 1/3.

Of course not, but we're not talking about finite sums.

If 0.999... = 1 is a question of notation. To be useful, we can define it as a limit, in which case it is true. However, if we do not define it as a limit, then the equality is false. For instance we can define it as "the sum of {for all N|9*10^-N}". Since this definition is not a limit, this infinite sum does not quite reach 1. Each part in the sum is one step closer to 1, but at no point does it quite reach 1. You can't even say that at infinity the sum reaches 1, because all talk about something happening at infinity implicitly means it's limiting behavior; infinity isn't a real number and it is only for convenience that we sometimes treat it like one.

This is where you are wrong. 0.999.. is precisely the infinite sum you have defined, and infinite sums are precisely defined as the limit of the sequence of its partial sums. There is no ambiguity here, and that limit is exactly 1.

 

[ainwood]

Moderator Action: quackers:
This thread is for serious discussions of mathematics. Please keep it to serious questions, not demands for 'proofs' such as you have made.

 

[ParadigmShifter]

Hey! Non-serious discussions of mathematics are OK by me

But you need to have a clue if you want to go down that route

 

[Riffraff]

I thought this whole 0.999... problematic was usually resolved in high school mathematics. Apart from the neat (10-1) proof you could also consider any possible numbers lying between 1 and 0.999.