Manuel
1=.999
i am going to give 20 fri$ points to the one who answers this.
i am going to give 20 fri$ points to the one who answers this.

1=.999....why? 20 frih$Manuel
1=.999
i am going to give 20 fri$ points to the one who answers this. ocalhoun
1=.999
Round 1=1 Surely that's too simple to be the correct answer, though... Manuel
Nope, that΄s not the answer
mathiaus
We did this in maths ages ago (23 years) when we started decimals and fractions.
1/3 = .333 recurring 2/3 = .666 recurring so add them together 1/3 + 2/3 = 3/3 = 1 .333 (R) + .666 (R) = .999 (R) So 1 = .999 recurring (theoretically) Of course many people say it is wrong. I disagree because I don't think 3 should be used (or any odd number for that fact). 2*(1/2 = .5) = 2/2 = 1 I think people will argue over this forever though. When you split things into three, you ALWAYS have one piece slightly larger than the other two! Manuel
correct. i am going to pay you your 20 points
Arnie
Sorry, that is not a good mathematical proof. This one comes a bit closer. I don't care about the frih$ though, you can keep that...
x = 0.999... 10x = 9.999... 10x  x = 9.999...  0.999... 9x = 9 x = 1 Manuel
That one is cool too! Anyway, both end in the same result and you were the second to answer it correctly. Sorry, I would give you the points if you had been the first one. imagefree
It seems like the contest is over. Yet i want to say that the solution provided is not correct. You yourself said that .333 (R) + .666 (R) = .999 (R) , but in the next step you replaced .333 (R) + .666 (R) and wrote 1 at its place. It seems that you are trying to proof that .333 (R) + .666 (R) = 1 because you replaced the first for the second, instead of proving .999(R) = 1 . Morever you said So 1 = .999 recurring (theoretically). Actually if you are rounding off your figures, you have to round off all the figures under consideration to maintain the quality of your results. So, i think the solution is not correct. rohan2kool
he replaced .333 by 1/3 and .666 by 2/3, the sum of whose is obviously 1. Note that it is not .333 but .333... [digits continue to repeat till infinity]. He just skipped a step and replaced their sum by 1, while keeping the LHS unmodified, by which we can prove that .999 = 1. although personally, i think that the second solution was way more cool. mathiaus
You misunderstood what I wrote. Please allow me to clarify... Left 1/3 + 2/3 = 3/3 = 1 (This is where the 1 comes from on the LEFT) Right .333 (R) + .666 (R) = .999 (R) (This is where the .999 (R) comes from on the RIGHT) So 1 = .999 recurring (theoretically) Arnie's way was better, I just mentioned the way I knew it (and got in first ) Guelila
I love this answer! I never thought about it this way Did u figure this out by yourself or did you learn this from somewhere else? Arnie
No, I didn't figure it out myself
hunnyhiteshseth
Well, I used to do such stuff a lot in junior classes, but now i have found an inherit flaw in all these types of proofs.
Here your assumption that 1/3 = 0.333.... is wrong. Remember they are not equal, 0.333... only gives the best approximation of 1/3 Since this basic step is flawed, rest of proof stand null and void. Arnie
The proof I posted does not use 1/3 = 0.333... Have you read the entire topic before posting? I also already pointed out that the 1/3 proof is not mathematically correct.
hunnyhiteshseth
oh yes, about your proof. Actually I was also going to post what your faulty assumption was , but then forgot.
Anyway, your assumption that if x = 0.999... , then 10x = 9.999...... is wrong. We assume that because, we consider that multiplying by 10 takes one 9 off from infinite 9s. So, practically, that would not make any difference. But when you are considering such type of proofs then you have to consider them. So, 1 > 0.999... by 0.000.........(infinite times)01 Hope that makes it clear. Related topics
