## .99999(repeating)=1 ?

Anything goes, but keep it seemly...
Myth Jellies wrote:
Trying another approach

0.9999...

equals

.9*(1) + .9*(.1) + .9*(.01) + .9*(.001) +...

equals

.9*(.1)**0 + .9*(.1)**1 + .9*(.1)**2 + .9*(.1)**3 +...

which is an infinite geometric series.

a geometric series has the form

ar**0 + ar**1 + ar**2 + ... + ar**(n-1)

the sum of which, for any finite n, equals a(1 - r**n) / (1 - r)

if -1 < r < 1 and we let n go to infinity, then r**n goes to 0 and the sum is

a / (1 - r)

in our .9999... case we have

a = 0.9, r = 0.1, and n => infinity

therefore the sum

.9*(.1)**0 + .9*(.1)**1 + .9*(.1)**2 + .9*(.1)**3 +...

equals

0.9 / (1 - 0.1) = 0.9/0.9 = 1

Surely that is the same as my approach, except that it is a much longer way of expressing the same idea?
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999_Springs

Posts: 468
Joined: 27 January 2007
Location: In the toilet, flushing down springs, one by one.

999_Springs wrote:Surely that is the same as my approach, except that it is a much longer way of expressing the same idea?

It is pretty much the same. I included the extra finite series to infinite series step because some people were not seeing how an infinite series could result in a mundane sum such as 1.
Myth Jellies

Posts: 593
Joined: 19 September 2005

daj95376 wrote:Face it. For some people, 0.999(repeating) is never going to be equal to 1 because they will always hang onto the belief that a *difference* exists. Talking about geometric progressions, infinite sequences converging, limits going to zero, and sums of fractions that are infinitely repeating decimal sequences, isn't going to change their minds. It's time to let it go.
but it is never time to stop trying to prove it to them. your proofs make me see how it is equal to 1. there is a wikipeida article that has all the arguments that people use against it and it counters them at least i think so. you should keep trying to show them how it equals 1. Do you know of one that uses basic math.
Chessmaster

Posts: 191
Joined: 21 December 2005

Do you know of one that uses basic math.

Chessmaster, Without wanting to sound patronising in any way, once you move into the mathematical concept of infinity, basic math goes out of the window. You have some 'basic' proofs of the proposition earlier in the chain but I think that you have to accept that there are some heavy duty mathematical concepts that you need to master before you can get a 'real' (ie strict) answer to your question.

Please do not take this the wrong way. I am trying to help. Out of interest, what is driving your interest in this?
Bigtone53

Posts: 413
Joined: 19 September 2005

the only way the notion of .99999 repeating equals one is relative to the context the question is asked and the mathmatical form they are initally expressed in.

in math matical equations a series of numbers converted from one form to another
i.e 1/3
the matmatical decimal form of the number is represented by .33333 reating.

if u are summizing 1/3 + 1/3 + 1/3 you get 3/3 = 1, and conversly in this application. in decmial form the sum is .9999999 reapting and since the application delte with fractions initally in basic form the product must equal as 1.

if you are not working in decimal form initally then y would you round up?
i can see rounding to convert to a diffrent form.
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