agurkie

how many mathematicians does it take to change a lightbulb? 0.999...

There will always be another 9. The whole idea with 0.9 recurring is that it approaches 1 but never gets there. This is also the whole idea behind a limit. Why do people want to make it equal to 1? What is the mathematical use of making 0.9 recurring = 1?

Don't you think this statement is a bit condescending? Anyway, moving on.... The definition of a limit states that if lim x->c f(x) = L, then for any epsilom > 0 there exists a delta > 0 such that 0 < |x-c| < delta and |f(x) - L| < epsilom. Meaning, give me any arbritrary value, and i will find a value for x which will make the distance between f(x) and L closer than that arbritrary value without it being 0. But it never says f( c ) = L. f( c ) does not even have to be defined. Of course this works for a function of x. So if you are looking to use the limit argument on 0.9 recurring, then you are saying 0.9 recurring is some variable function. In which case, by the very definition of a limit I can prove to you that there is a value closer to 1. Plus i thought i gave a pretty decent proof of why 0.9 recurring is not equal to 1 in my previous post. Partial sum of an infinite geometric series: a(r-r^n)/(1-r), in this case 9( 1/10 - (1/10)^n)/ (1- 1/10) To stretch this to the sum for an infinite series, we must assume that (1/10)^n = 0 as n => infinity, so that the formula becomes 9(1/10) / (9/10), which is equal to 1. But for which n is (1/10)^n ever going to be 0? I dont quite understand how the sum of geometric series proves to some people that 0.9 recurring = 1, while to me it proves exactly the opposite.

The sum of finite geometric series is a(r-r^n)/(1-r). Now, to derive the sum of infinite series from the sum of finite series above, we are using the fact that r^n => 0 for n => infinity, if |r| < 1. Then the formula becomes ar/(1-r). Great, and this is then how you prove using the sum of infinite geometric series, that 0.9 recurring = 1. But wait a moment! r^n will never ever be 0 unless r=0. We just threw it away and used the result to prove what we wanted it to be. That r^n that we threw away is in fact the difference between 0.9 recurring and 1, (if you divide it by (1-r)). This actually proves that we can never accurately determine the sum of that series unless we know what n is, which also highlights the error in the 10x = 9.9 recurring theory. So once again, it gets very close, but it just isnt. I think where the cofusion comes in, is around whether or not 0.9 recurring is a constant number or a number that converges. Lets take (x-1)/(x+1). We could easily prove using L'Hospitals rule that this converges to 1 as x => infinity. The reason is because the value of this expression actually changes as x gets larger and larger. But what happens to 0.9 recurring as x goes to infinity. Well, nothing. 0.9 recurring is not dependant on any variable. So, in fact you could argue that it does not converge to anything at all. The limit of 0.9 recurring is 0.9 recurring. It just feels to me that saying 0.9 recurring = 1 is turning mathematics into some form of fancy trickery. Look I can prove that this hat contains a rabbit, as long as nobody checks my sleeve.

But 1/3 is not equal to 0.3 recurring. 0.3 recurring approaches 1/3, but never gets there. same situation: x=0.3333 10x=3.3333 9x=3 x=1/3 and in my opinion the same problem in line 2. the series proof proves convergence. but just because something converges to something else does not mean it is actually equal to that something does it?

I know this a topic which has been discussed to infinity, but I have a problem with this theory and would like other peoples' opinion on it. The 'proof' states that: x = 0.9999 10x = 9.9999 9x = 9 x = 1 I believe the problem lies in the second line already. Is it possible to do an arithmetic operation on a number with an infinitely repeating fraction? Let's take a finite number. x = 0.9999, then 10x = 9.9990. We have to know there are 4 digits after the decimal point, so each one moves one to the left, and the fourth digit after the decimal is replaced by a 0. I suppose the argument is with regard to the theoritical interpretation, but in practice it is not possible to multiply a number with a repeating fraction. Looking also at the irrational number PI, any computer or calculator actually stores PI to about 40 significant digits. So to any computer or calculator, PI is a number with a finite number of digits after the decimal point. If you think about it, how could you expect a computer to use PI if it continued forever? So my opinion is, yes I agree that 0.9 recurring approaches 1, but we do not have the mathematics to prove that it is in fact 1 and until then it just...well....isnt.