The whole 0.999...=1 argument (1 Viewer)

[BrotherBread]

For the first time in maths you can kinda argue that it is 'close enough', in layman's terms.

Yeah I understand all f what you said. I get that whole bit. I spose there are just two sides of the coin. Like everything in math, there are always little things that contradict. I also understand that there are more proofs that are far more elegant and shoot what I said down in flames. Personally I prefer the whole 'close enough is good enough' senario in math. Oftenmakes things simpler. Like with limits when you have an x term on the top and botton you can't siplyfy, and you divide by x, as x -> infinity one or maybe more of the numbers become so small it is discarded (x/infinity).

But yes I can see where everyone is coming from

 

[seremify007]

From memory, this is year10/year11 maths (depending on your school) so if you haven't learnt it yet, it'll come soon enough

 

[Forbidden.]

Then don't?

Just making my point in a democratic country.

 

[Templar]

I spose there are just two sides of the coin. Like everything in math, there are always little things that contradict. I also understand that there are more proofs that are far more elegant and shoot what I said down in flames. Personally I prefer the whole 'close enough is good enough' senario in math.

Mathematics at this level is consistent, so there are never two sides of the coin. One will always be correct. (And for anyone who wants to argue with me against that on the Godel theorems, don't even get started. I will debate you over the entirety of that topic.)

Close enough is good enough rarely works in pure. Take e^Pi*Sqrt[163]. Is it an integer?

 

[Shortbreads]

Wait a second! I have an avatar for this!

I'll leave as a pic though.

Sorry, it's not adding much to the discussion, but forgive me. It's not everyday something this specific becomes so utterly appropriate.

Oh, and I <3 xkcd. :-D

 

[KeypadSDM]

Uhh...

0.99999...= sum[n=1->inf][9/(10^n)] By definition.
So just take the limit of that sum - which is nearly trivially 1.

For a more esoteric view:

Given any recurring or terminating decimal beginning with 0. ... we find that it is either equal to or less than 0.99999.... as in the mathematical definition. (Either all digits are 9 or some digits are not, and thus are less than 0.999... by taking partial sums and comparing the differences). But noting that progressive terms in the series converge as: a_n+1/a_n > 0 strictly, we find that the series converges. But this series must converge to the supremum of the reals less than 1, hence 0.999... = 1.

I think that's right. Anwyay it's got the right ideas for a rigorous proof.