The New Sudoku Players' Forum

[Chessmaster]

0.99999... is clearly equal to 0.9+0.09+0.009+0.0009+0.00009...

This is a GP with first term 0.9 and common ratio 0.1 so the sum to infinity (i.e. the sum of all the terms) is

(first term)/(1-(common ratio)) when -1<common ratio<1

which is 0.9/(1-0.1)=1

Therefore, 0.99999... =1

Seemes reasonable to me.

[re'born]

0.99999... is clearly equal to 0.9+0.09+0.009+0.0009+0.00009...

This is a GP with first term 0.9 and common ratio 0.1 so the sum to infinity (i.e. the sum of all the terms) is

(first term)/(1-(common ratio)) when -1<common ratio<1

which is 0.9/(1-0.1)=1

Therefore, 0.99999... =1

Of course, this hides the tricky part of the whole thing which is "Why?". To prove the formula for the sum of a geometric progression, one needs the notion of a limit. The problem might be restated in the following way: Just what do we mean my adding up an infinite number of numbers. While 999_Springs writes that ".999... is clearly equal to .9 + .09 + .009 +...", I believe that it is intuitively true, but that there is actually nothing clear about it.

[Cec]

"..... so the sum to infinity (i.e. the sum of all the terms) is ..... =1

Therefore, 0.99999... =1

Seemes reasonable to me.

Hmm? This therefore concludes that "infinity" can be determined which seems contrary to the Oxford dictionary which defines "infinity" as "boundless number or extent" or "infinite quantity".

The dictionary also defines "infinite" as "inumerable" and "greater than any assignable quantity or magnitude...that may be continued indefinitely without ever coming to an end."

I still continue to be intrigued with this dilemma in not being convinced that .99999(repeating) = 1

Cec

[re'born]

"..... so the sum to infinity (i.e. the sum of all the terms) is ..... =1

Therefore, 0.99999... =1

Seemes reasonable to me.

Hmm? This therefore concludes that "infinity" can be determined which seems contrary to the Oxford dictionary which defines "infinity" as "boundless number or extent" or "infinite quantity".

The dictionary also defines "infinite" as "inumerable" and "greater than any assignable quantity or magnitude...that may be continued indefinitely without ever coming to an end."

I still continue to be intrigued with this dilemma in not being convinced that .99999(repeating) = 1

Cec

Part of the problem is that for anybody to be completely convinced, there is no alternative to writing down precise mathematical definitions of everything about which we've written. If you can get through the jargon, it is not a difficult fact to accept, but...there is still the jargon.

So lets go back to Bill's argument on page 1. If .999...=1, then you're convinced and we're all happy. If .999...≠1, then I ask you find me a number in between .999... and 1. As long as you believe that between any two distinct numbers there is another number (for instance, the average M = (x+y)/2 of x and y satisfies x < M < y), then not being able to find a number in between .999... and 1, should convince you that they must in fact be equal.

[Cec]

"...So lets go back to Bill's argument on page 1. If .999...=1, then you're convinced and we're all happy...."

No, I had (and still have) a problem accepting this argument and the following subsequent posts suggest I wasn't the only one (the bolding is mine)

Cec

Or, to look at it another way:

I challenge anybody to come up with a number larger than .9999.... but less than 1.

Doesn't that mean that 0.9999.... and 1 are seperate numbers

Suppose I said, can anybody come up with an integer (whole number) greater than 1 but less than 2, with the implication that if they can't then
1 = 2

You're right that Bill's argument is not a proof, however, you should not neglect the perfectly valid proof udosuk gave....."

Yes, I've previously acknowledged the two particular posts from udosuk and wapati which presented good arguments why .99999(repeating) =1.

However, I continue to be unsure about this but if I was required to give a definite "yes" or "no" answer I'd still lean towards the answer never reaching 1 unless infinity can be reached which doesn't seem possible.

Cec

[Bigtone53]

Cec wrote

This therefore concludes that "infinity" can be determined which seems contrary to the Oxford dictionary which defines "infinity" as "boundless number or extent" or "infinite quantity".

The dictionary also defines "infinite" as "inumerable" and "greater than any assignable quantity or magnitude...that may be continued indefinitely without ever coming to an end."

Unfortunately, infinity is a bit like Animal Farm's animals. All infinities are infinite, but some are more infinite than others.

Blame Geog Cantor and his infinity of infinities

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

[Mauricio]

However, I continue to be unsure about this but if I was required to give a definite "yes" or "no" answer I'd still lean towards the answer never reaching 1 unless infinity can be reached which doesn't seem possible.

You (Cec) have to understand that 0.999(repeating) is NOT a sequence of numbers, it is a real number equal to 1, so the expresion "0.999... reaching 1" is nonsense.

In the past times, people did not accept that negative integer numbers "existed", because we can not "touch" it; we can "touch" 3, ie, we can touch a set of 3 elements, for instance, of 3 oranges, but we can not touch -3 oranges.

Irrational numbers did not exist too, because they can not have the form a/b, where a and b are integers.

Then it came complex numbers, infinities and so on.

Unfortunately, infinity is a bit like Animal Farm's animals. All infinities are infinite, but some are more infinite than others.

Blame Geog Cantor and his infinity of infinities
http://en.wikipedia.org/wiki/Georg_Cantor

This might be too abstract for some people here. But for the records, there are a infinite number of "different" infinites.

[Cec]

Hi Mauricio....Could I suggest you edit your above post to show the reference to Animal Farm's animals, etc. was written by Bigtone53 and not me.

Cec

[Myth Jellies]

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

[daj95376]

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.

[Cec]

Myth Jellies, an interesting approach but daj95376 has touched a nerve ... I'm hoisting a white flag.

Cec

[udosuk]

Very well put Mauricio and daj95376... Advanced mathematics is a pill not that easy to swallow for everybody...

[daj95376]

Okay, I'm going to ignore my own advice. Sometimes, the best way to answer a question is to ask a question.

Code: Select all

1/9 = 0.11111(repeating)
2/9 = 0.22222(repeating)
3/9 = 0.33333(repeating)
4/9 = 0.44444(repeating)
5/9 = 0.55555(repeating)
6/9 = 0.66666(repeating)
7/9 = 0.77777(repeating)
8/9 = 0.88888(repeating)

9/9 = ?


[Cec]

I'm now waiving the white flag vigorously. Peace at last Well done daj95376.

Cec

[JPF]

I'm now waiving the white flag vigorously. Peace at last Well done daj95376.

Thanks daj,

daj.95376 (repeating) = daj + 95376/99999



JPF