The New Sudoku Players' Forum

[Chessmaster]

does .999 repeating equal 1 excatly?

[Bigtone53]

does .999 repeating equal 1 excatly?

Zeno says no.

[r.e.s.]

does .999 repeating equal 1 excatly?

See http://en.wikipedia.org/wiki/0.999...

For adic-tional fun, don't miss the part about what integer might be represented as "...999" (repeating to the left).

[Chessmaster]

does .999 repeating equal 1 excatly?

Zeno says no.

who is Zeno?

[Bigtone53]

Chessmaster, I am not sufficiently adept to give you links to Zeno, but if you google Zeno and paradox, especially the paradox of Achilles and the Tortoise, I think that you will see the relevance

[Chessmaster]

Ok, I guess.

[MCC]

Here's a wiki link to Zeno:

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

And Zeno's paradoxes:

http://en.wikipedia.org/wiki/Zeno%27s_paradoxes


MCC

[wapati]

does .999 repeating equal 1 excatly?

Yes. One third is .333 repeating.

One third plus one third plus one third = 1, exactly.

.333 repeating plus .333 repeating plus .333 repeating = .999 repeating,
which is one.

[Smythe Dakota]

Or, to look at it another way:

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

Bill Smythe

[udosuk]

Let x=0.999999...

10x=9.999999...

9x=10x-x=9.999999...-0.999999...=9

Therefore x=1.

(Q.E.D.)

[Chessmaster]

Or, to look at it another way:

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

Bill Smythe

Thanks i suppose that proves they are equal at least to me.

[MCC]

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

However large 0.9999.... gets, it never equals 1.


Pedantically.
MCC

[lunababy_moonchild]

However large 0.9999.... gets, it never equals 1.

I'm inclined to agree with this. However small the difference is, there will always remain a difference, therefore they are not equal.

Just because you don't get enough change from £2 having paid £1.99 for your item to make a difference to your finances doesn't mean that the amounts are mathematically equal - I think !

Luna

[re'born]

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. If you prefer a variation on Bill's argument, consider the following: I define two numbers to be equal if and only if their difference is 0, i.e., a = b if and only if a-b = 0. An immediate consequence of this is that if a<b, then b-a = ε for some positive real number ε.

Assume that .999... ≠ 1. Then 1-(.999...)=ε for some positive real number ε. For any real number x, there exists a unique integer k such that 10^k ≤ x < 10^{k+1}. Fix such a k for our ε. Since 10^k is no bigger than ε, adding it to .999... cannot get you bigger than 1. However, it is easy to see that .999...+10^k > 1, a contradiction. For example, if k = -3, then .99999999...+10^{-3} = .9999999...+.001=1.000999999....

However large 0.9999.... gets, it never equals 1.

.999... is a fixed number, it can't get larger, it just is what it is, a real number equal to 1. Your statement reminds me of the joke 2+2=5 for large values of 2.

[Cec]

This is certainly an interesting scenario but I have to conclude 0.999repeating will never equal 1. Here's my logic.....
If 0.9999repeating equals 1 then the difference between these two numbers would be zero but this can never be because...

1.000000000000000 etc. minus
0.999999999999999 etc.
=0.000000000000001 and so on but never becomes zero.

Cec