The New Sudoku Players' Forum

[re'born]

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

Cec,

You've missed the point. Where do you put the first 1 in 0.000000000000001? Well, you put it in the 15th place after the decimal, which means you think 1-(.999...) = 10^{-15}, which is clearly not true. Of course that isn't what you meant anyway. What you meant was that the difference is 0.000...1 where there are an infinite number of 0's before the 1. But what does that even mean? If there is a 1 somewhere, in must be the k-th place for some natural number k, that is the difference is 10^{-k}. At this point you may refer to my above argument.

[udosuk]

It seems many people are tricked by the representation of "0.999..."...

It actually means "0." followed by an infinte number of "9"s... Numerically it's equal to 1, or 1.000...(infinite 0's)...0001, or 1.000...(infinite 0's)...000987654321...

It all depends on how well you understand the concept of infinity...

Suppose I ask you, what is 1 divided by "infinity"? The mathematical answer is 0, but if you think about the inverse of that relationship, how can 0 times anything equal to 1? That's the mystery about infinity...

rep'nA, your rigid proofs can be well-understood by anyone who studied higher-level maths, but I'm afraid it's not the best tool for explanation to the general public...

[re'born]

rep'nA, your rigid proofs can be well-understood by anyone who studied higher-level maths, but I'm afraid it's not the best tool for explanation to the general public...

Perhaps, but I did (and still) recommend they read your proof first. On the other hand, I've found the people on this forum to be extremely smart and good with dealing in abstractions. I don't think my rigid proofs are any harder than the ALS xz-rule or some of the Ultimate Fish deductions.

[re'born]

It seems many people are tricked by the representation of "0.999..."...

It actually means "0." followed by an infinte number of "9"s... Numerically it's equal to 1, or 1.000...(infinite 0's)...0001, or 1.000...(infinite 0's)...000987654321...

It all depends on how well you understand the concept of infinity...

The last two representations you give are not well defined. It doesn't make any sense to write an infinite number of anything and then a 1 (or 987654321).

[JPF]

Let x=0.999999...

10x=9.999999...

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

Therefore x=1.

(Q.E.D.)

To perform operations with x, x and the operations on x have to be defined...

(How do you define x, 10x, 9x and justify 9x=10x-x) ?

Proposed solution :
x=9x10^(-1)+9x10^(-2)+....
x=9r[1+ r+r^2+r^3+...+r^n+....] with r=10^(-1)=0.1
x=9r/(1-r)=9x0.1/0.9
x=1

JPF

[MCC]

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.

Semantically you are correct, what I meant was that no matter how many 9's you add to the end of 0.9999.... it never equals 1.

Suppose I ask you, what is 1 divided by "infinity"? The mathematical answer is 0.

What is mean by infinity

If 0.9999.... tends to infinity, then:

1.0000..../0.9999.... approaches 1.0 not zero.


MCC

[re'born]

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.

Semantically you are correct, what I meant was that no matter how many 9's you add to the end of 0.9999.... it never equals 1.

MCC

To be precise, you need to say that adding a finite number of 9's will never get you to 1. Repeating 9 infinitely will.

[udosuk]

What is mean by infinity

If 0.9999.... tends to infinity, then:

1.0000..../0.9999.... approaches 1.0 not zero.

"0.9999...." doesn't tend to infinity, but "9999...." does...

1.0000..../0.9999.... does approach 1 when you keep adding 0s and 9s to the end... And if you claim there are an infinite number of 0s and 9s after the decimal points, it is exactly 1...

We indeed need to take a lot of care on operations involving infinity... Consider this:

infinity+1=infinity=infinity+0

Therefore 1=0.

Of course this is invalid reasoning as the object "infinity+1" is not well-defined...

It seems many people are tricked by the representation of "0.999..."...

It actually means "0." followed by an infinte number of "9"s... Numerically it's equal to 1, or 1.000...(infinite 0's)...0001, or 1.000...(infinite 0's)...000987654321...

It all depends on how well you understand the concept of infinity...

The last two representations you give are not well defined. It doesn't make any sense to write an infinite number of anything and then a 1 (or 987654321).

You're probably right... I just wanted to stress the "fact" that once there are an infinite number of 0's in between, the end isn't significant anymore...



And thanks JPF for the alternative solution... If people find your lines more convincing, than I'll be glad too...

[daj95376]

The point is perspective and the realization that an infinitely repeating decimal value doesn't accurately represent anything. The question then becomes, what does an infinitely repeating decimal value come closest to representing.

For the case of 1/3, the infinitely repeating decimal value 0.333... is as close as you can come. Now, extending this logic, you get.

1 = 3 * 1/3 = 3 * 0.333... = 0.999...

Thus, an infinitely repeating sequence of 9s comes closest to representing 1.0 .

[re'born]

We indeed need to take a lot of care on operations involving infinity... Consider this:

infinity+1=infinity=infinity+0

Therefore 1=0.

Of course this is invalid reasoning as the object "infinity+1" is not well-defined...

Actually, each sum in ∞+ 1 = ∞ = ∞ + 0 is well defined and perfectly true. The reason you can't conclude that 0 = 1 is that subtraction of ∞ is not defined.

It seems many people are tricked by the representation of "0.999..."...

It actually means "0." followed by an infinte number of "9"s... Numerically it's equal to 1, or 1.000...(infinite 0's)...0001, or 1.000...(infinite 0's)...000987654321...

It all depends on how well you understand the concept of infinity...

The last two representations you give are not well defined. It doesn't make any sense to write an infinite number of anything and then a 1 (or 987654321).

You're probably right... I just wanted to stress the "fact" that once there are an infinite number of 0's in between, the end isn't significant anymore...

It's even better than that. With an infinite number of zeros, that is the 'end', which is significant as it can get.

[Bigtone53]

For the case of 1/3, the infinitely repeating decimal value 0.333... is as close as you can come. Now, extending this logic, you get.

1 = 3 * 1/3 = 3 * 0.333... = 0.999...

Thus, an infinitely repeating sequence of 9s comes closest to representing 1.0 .

What makes the infinitely repeating decimal value 0.333 ... equal one third, any more than 0.999... equals 1? This sounds a bit circular to me. The clearest and most proper explanation is the one given by udosuk on the 20th.

[JPF]

A recurring or repeating decimal is a number which when expressed as a decimal has a set of "final" digits which repeat an infinite number of times.
....
a recurring decimal is a rational number
...
A shortcut
If the repeating decimal is between 0.1 and 1, and the repeating block is n digits long occurring right at the decimal point, then the fraction (not necessarily reduced) will be the n-digit block over n digits of 9. For example,
• 0.444444... = 4/9 since the repeating block is 4 (a 1-digit block),
• 0.565656... = 56/99 since the repeating block is 56 (a 2-digit block),
• 0.789789... = 789/999 since the repeating block is 789 (a 3-digit block), etc.

So :
0.333333...=3/9=1/3
0.999999...=9/9=1
0.777777...=7/9

etc...

JPF

[Cec]

Cec,

You've missed the point. Where do you put the first 1 in 0.000000000000001? Well, you put it in the 15th place after the decimal, which means you think 1-(.999...) = 10^{-15}, which is clearly not true........"

Hi rep'nA. Pardon my ignorance of maths terminology but I can't follow your explanation. What does 10^{-15} mean?.

Looking at my previous explanation another way what I was trying to explain was this....
1 minus 0.9 = 0.1 No problem here so let's take it further..
1 minus 0.99 = 0.01 Taking this further gives...
1 minus 0.999 = 0.001 and so on.
In my previous post I randomly chose the fifteenth step to show that 1 minus 0.999999999999999 still produces an answer which is greater than zero.
Continuing this exercise will always give an answer albeit decreasing in value but never reaching zero because 1 will always be higher than 0.9999repeating.

Cec

[re'born]

Cec,

You've missed the point. Where do you put the first 1 in 0.000000000000001? Well, you put it in the 15th place after the decimal, which means you think 1-(.999...) = 10^{-15}, which is clearly not true........"

Hi rep'nA. Pardon my ignorance of maths terminology but I can't follow your explanation. What does 10^{-15} mean?.

10^{-15} = .000000000000001

Looking at my previous explanation another way what I was trying to explain was this....
1 minus 0.9 = 0.1 No problem here so let's take it further..
1 minus 0.99 = 0.01 Taking this further gives...
1 minus 0.999 = 0.001 and so on.
In my previous post I randomly chose the fifteenth step to show that 1 minus 0.999999999999999 still produces an answer which is greater than zero.
Continuing this exercise will always give an answer albeit decreasing in value but never reaching zero because 1 will always be higher than 0.9999repeating.

Cec

I tried to anticipate this response when I wrote:

Of course that isn't what you meant anyway. What you meant was that the difference is 0.000...1 where there are an infinite number of 0's before the 1. But what does that even mean? If there is a 1 somewhere, in must be the k-th place for some natural number k, that is the difference is 10^{-k}.

The key point in your argument is when you said, "continuing this exercise...". For your argument to be correct, you are only allowed to continue the exercise a finite number of times, placing a finite number of 9's after the decimal point. As soon as you allow an infinite number of 9's, all bets are off.

Early in the thread, Bigtone53 referred to Zeno. His argument (used to prove that motion is impossible) goes similar to your argument. To get from point A to point B, you must first travel half of the distance. To get from the 1/2-way point to point B, you again must first travel half of the distance. To get from the 3/4-way point to point B, you must (once again) travel half of the distance...So you proceed along the numbers 0, 1/2, 3/4, 7/8, 15/16, 31/32,... You can get as close to 1 as you like, but never actually get there. The best you can do is (2^n-1)/2^n for arbitrarily large values of n. Consequently, you can never move from point A to point B (and so motion is impossible). Zeno's paradox fails for the same reason your argument is incorrect, our intuition about the finite is just plain wrong when we talk about the infinite.

[underquark]

Code: Select all

                    ......
1/7 = 0.142857142857142857
When I type that into my calculator and add it up seven times it equals one but 0.142857 x 7 = 0.999999.

Alternatively, 0.99999... as an asymptotic curve tends to 1.0.
1.0 as a straight line can be considered a tangent to this curve.
They meet at a point ergo they are equal.

Alternatively, who cares? We made numbers in order to try and explain the universe. It must be recognised that these numbers are not the be-all and end-all of everything and are necessarily limited in their scope. If we could expand them in complexity to fully explain the universe then they would be larger and more complex than the universe that they try to explain.