Implications of 0.9999.... = 1

[mathscurious1]

Hi there,
I'm no expert on math but I am curious about the implications of 0.999... = 1. I've read that this statement is true (please let me know if I am incorrect).

Here's a few thoughts I've had and wondering whether a) these equations are valid and b) what the answers and implications may be.

• 1 x 1 ^Infinity = 1. So what is 0.999.... x 0.9999...^Infinity equals to? Is it all 1? Usually any numbers that are 0 followed by a decimal followed by digits become smaller and smaller when multipled by themselves and progress towards zero.
• If 1 - 1/3 - 1/3 - 1/3 = 0, is 0.999... - 1 = a negative fraction? Or is 0.999... - 1 = 0 in all cases? I understand that the initial point I made is 0.999... = 1 and, of course, 1 = 1 but I also see clearly that 0.999... is in some way distinct from 1 since it is not 'technically' 1. Or am I misunderstanding something here?

What implications of 0.999.... = 1 can you see? Are there any implications that are not obvious at first glance? Or do you think that 0.999... = 1 is a false assumption to start with?

Please be nice, I am new and just wanted to see what those who know far more than I do have to say about the implications of 0.999... = 1 because I think there could be something interesting conclusions.

 

[mathscurious1]

Just had a fresh thought.

So 1 x 1 ^Infinity = 1, however, 0.999... x 0.999...^infinity may not equals 1 for the following reasoning:

0.99 x 0.99 = 0.9801
0.999 x 0.999 = 0.998001
0.9999 x 0.9999 = 0.99980001

As these equations show, every 9 you add to both numbers being multiplied, the 8 digit is pushed along and the number of 0s that follows it increases and increases yet a 1 is left totally on its own.

If you kept going following this logic, are we left to conclude that 0.9999.... x 0.999... ^infinity must equal to an answer that includes an 8 and an infinite number of zeros?

 

[Dr.Peterson]

I also see clearly that 0.999... is in some way distinct from 1 since it is not 'technically' 1. Or am I misunderstanding something here?

You are misunderstanding. It is true that 0.999... = 1; it is not true that it is a different number.

All you have here is a different way to write the number 1. Just as 1/1, 2/2, 1.000, and 1 all represent the same number, so does 0.999... .

The notation 0.999... means the infinite sum 0.9 + 0.09 + 0.009 + ... = 9/10 + 9/100 + 9/1000 + ... . This is a geometric series; if you stop the sum anywhere, you have not yet reached 1, but as you add more terms, the sum approaches closer and closer to 1. The value of 0.999... is not any of those partial sums, but the limit they approach, which is exactly 1.

And since it is the same number, there are no implications.

If you kept going following this logic, are we left to conclude that 0.9999.... x 0.999... infinity must equal to an answer that includes an 8 and an infinite number of zeros?

No. The numbers in your sequence approach closer and close to 1; the most important digits are on the left, not the right, and they all become 9 eventually. So the result is still 1.

But to clarify, it appears that what you are calculating is not 0.999... x 0.999...^infinity but just 0.999...² (which, again, is really just 1^2 = 1).

 

[mathscurious1]

You are misunderstanding. It is true that 0.999... = 1; it is not true that it is a different number.

All you have here is a different way to write the number 1. Just as 1/1, 2/2, 1.000, and 1 all represent the same number, so does 0.999... .

The notation 0.999... means the infinite sum 0.9 + 0.09 + 0.009 + ... = 9/10 + 9/100 + 9/1000 + ... . This is a geometric series; if you stop the sum anywhere, you have not yet reached 1, but as you add more terms, the sum approaches closer and closer to 1. The value of 0.999... is not any of those partial sums, but the limit they approach, which is exactly 1.

And since it is the same number, there are no implications.


No. The numbers in your sequence approach closer and close to 1; the most important digits are on the left, not the right, and they all become 9 eventually. So the result is still 1.

But to clarify, it appears that what you are calculating is not 0.999... x 0.999...^infinity but just 0.999...² (which, again, is really just 1^2 = 1).

Thanks for your thoughts. I understand what you mean. So as you keep going, the 8 and 0s just vanish, per se? I understand that there cannot be a point at which this happens either, since you cannot 'jump from the finite to the infinite'. However, that part that confuses me a little but I understand that infinity is a different world to the finite. I guess I'm trying to understand where the 8 and 0s go, if they do not disappear at any one point but are not present once you reach 0.999... x 0.999...

 

[Cubist]

Thanks for your thoughts. I understand what you mean. So as you keep going, the 8 and 0s just vanish, per se? I understand that there cannot be a point at which this happens either, since you cannot 'jump from the finite to the infinite'. However, that part that confuses me a little but I understand that infinity is a different world to the finite. I guess I'm trying to understand where the 8 and 0s go, if they do not disappear at any one point but are not present once you reach 0.999... x 0.999...

The ellipsis are "evaluated" before anything else in this case, since they are an integral part of a number. Implied brackets are like this...

0.9999...^∞ = ( 0.9999... ) ^∞ = 1^∞

You could formulate an entirely different problem like [imath]f(x) = (1 - 10^{-x})^x[/imath]

f(1) = 0.9^1
f(2) = 0.99^2
f(3) = 0.999^3

But in the limit as x tends towards infinity f(x) still tends towards 1

--

If you're wondering how quickly the exponent would need to grow in order to change this, try evaluation of the limits of [imath]g(x) = (1 - 10^{-x})^{10^x}[/imath] and [imath]h(x) = (1 - 10^{-x})^{11^x}[/imath] as x tends towards infinity

 

[Deleted member 4993]

Hi there,
I'm no expert on math but I am curious about the implications of 0.999... = 1. I've read that this statement is true (please let me know if I am incorrect).

Here's a few thoughts I've had and wondering whether a) these equations are valid and b) what the answers and implications may be.

• 1 x 1 ^Infinity = 1. So what is 0.999.... x 0.9999...^Infinity equals to? Is it all 1? Usually any numbers that are 0 followed by a decimal followed by digits become smaller and smaller when multipled by themselves and progress towards zero.
• If 1 - 1/3 - 1/3 - 1/3 = 0, is 0.999... - 1 = a negative fraction? Or is 0.999... - 1 = 0 in all cases? I understand that the initial point I made is 0.999... = 1 and, of course, 1 = 1 but I also see clearly that 0.999... is in some way distinct from 1 since it is not 'technically' 1. Or am I misunderstanding something here?

What implications of 0.999.... = 1 can you see? Are there any implications that are not obvious at first glance? Or do you think that 0.999... = 1 is a false assumption to start with?

Please be nice, I am new and just wanted to see what those who know far more than I do have to say about the implications of 0.999... = 1 because I think there could be something interesting conclusions.

You are correct - 0.999999.... = 9/9 = 1 ....... like 0.33333.... = 3/9 = 1/3 or

0.6666... = 6/9 = 2/3

What did you expect?

 

[Steven G]

I sometimes think of this as follows.

Either .999...<1, .999....=1 or ,999...>1.

I will assume that you agree that .999.... is not greater that 1.

If you think that .999... is less than 1, that is fine by me. However, you need to give me a number in between .999.... and 1. (note that if a<b then there is a number between a and b (for example, (a+b)/2 works))

 

[pka]

}Hi there,
I'm no expert on math but I am curious about the implications of 0.999... = 1. I've read that this statement is true (please let me know if I am incorrect).

Suppose that [imath]D=0.\overline{\;9\;}[/imath]. Now multiply by ten to 'move' the decimal one place to the right.
We now have [imath]10D=9.\overline{\;9\;}[/imath] Both [imath]D~\&~10D[/imath] have the same repeating decimal to the right of decimal point.
Thus [imath]9D={\color{blue}(10D)}-{\color{red}(D)}={\color{blue}(9.\overline{\;9\;})}-{\color{red}(.\overline{\;9\;})}=9[/imath]
So that it must be that [imath]\large\bf D=1[/imath]
[imath][/imath]