.999...= 1?

[papernuke]

in the wiki, it said that .999... is equal to 1. what did it mean? i mean isnt .999... still .000.......................................1 less than 1?

[Hugh]

in the wiki, it said that .999... is equal to 1. what did it mean? i mean isnt .999... still .000.......................................1 less than 1?

But the point may be that you never actually reach that .......1

Here is the page on Wikipedia that talks about this, and gives various proofs why they are equal. http://en.wikipedia.org/wiki/0.999...

[papernuke]

i read it but i didnt get it, especially the equations part.

[Nick]

Huh! I wouldn't believe it if not for the proof.

@Icon:
_ represents infinity
1/3 = .3_
3 * 1/3 = 1
therefore, .3_*3 = .9_ = 1

[PWrong]

in the wiki, it said that .999... is equal to 1. what did it mean? i mean isnt .999... still .000.......................................1 less than 1?

0.000....1 is equal to 0, because:

0.1 = 1/10
0.01 = 1/100
0.001 = 1/1000
...

The limit of this sequence is 0

[bo198214]

what is 0.0.........1?
An infinite number of zeros followed by a 1?
If there is any 1 in the sequence then before are only finitely many 0s.

[papernuke]

An infinite number of zeros followed by a 1?

yea

[jinydu]

An infinite number of zeros followed by a 1?

yea

By the usual definition of a decimal expansion, that is not possible. A decimal expansion assigns each integer to another integer from 0 to 9. 'Infinity' is not an integer.

[Super Hans]

I just always thought that if-

0.999... = x

10 x = 9.999...

9x = 10x - x = 9.999... - 0.999... = 9

(all 9's cancel except the one to the left of the decimal point)

9x = 9

x = 1

1 = 0.999...

[wendy]

The sets B_n describe those numbers included in {Z, C/n} ie if z is in B_n, so is z/n.

The set B_10 then designates those numbers that are exact decimals. So it includes, eg 0.125, but not 1/3. Such sets as B_10, or B_120 are class-two sets, which are infinitely dense and simply ordered. One can, therefore find unique representations, for which there is an ordered spelling, that represents all members of the set. Since these points project onto a line, one can approximate a point of the line by a series of decimal approximates, eg not exceeding the number.

So pi is 3, then 3.1, then 3.14, then 3.141, then 3.1415, etc. None of these are _equal_ to pi, but all are approximates to it.

Likewise, we could locate the point just under ' 1' . We get 0.999999999&c.

Since these progressions represent ordered points on a line, to ever increasing fineness, we see that ( a ), No number in expressable in decimals unless the represented number is part of B10, and ( b ), that it is legitimate to round a particular calculation to its approximate, so

0.9999999999&c = 1. [Zeno paradox]

[Keiji]

A bit late, but I'm moving this to General.