0.99~=1

arod · Apr 18, 2007

ljcox said:
0.9999 approaches 1 as the number of decimal places approaches infinity.

0.3333 approaches 1/3 as the number of decimal places approaches infinity.

End of story.

0.33... = 1/3. You are trying to quantify infinity in your definition.

dknguyen · Apr 18, 2007

0.999...~= 1? Yes!
0.999... = 1? No.

End of story. If you do write down 1 in your calculations, it's because the difference is so small it doesn't matter (same reason why we round numbers).

Pommie · Apr 19, 2007

gramo said:
... And the result is lol?

(1+1/∞)^∞

(1+0)^∞

(1)^∞

1?

The result is e (2.71828). Try setting infinity to 1 then 10, 100, 1000, 10,000 etc. The graph levels out at e.

Mike.

ljcox · Apr 19, 2007

arod said:
0.33... = 1/3. You are trying to quantify infinity in your definition.

I did not say that 0.33... = 1/3. I said it approaches 1/3 as the number of decimal places approaches infinity.

No, I'm not trying to quantify infinity, I'm simply stating a mathmetical fact.

You can't quantify infinity, you can only approach it.

ljcox · Apr 19, 2007

dknguyen said:
0.999...~= 1? Yes!
0.999 = 1? No.

End of story. If you do write down 1 in your calculations, it's because the difference is so small it doesn't matter (same reason why we round numbers).

If you read the original question it was "0.99 -continuious- = 1" and the answer is no. Rounding up is a different issue.

ljcox · Apr 19, 2007

gramo said:
... And the result is lol?

(1+1/∞)^∞

(1+0)^∞

(1)^∞

1?

You can't divide by infinity or raise a number to the power of infinity.

If x = 1/n then x approaches 0 as n approaches infinity.

If x = 1^n then x =1 for all n. But n = ∞ is not valid. n can only approach ∞.

Pommie · Apr 19, 2007

And, as X approaches infinity then (1+1/X)^X approaches e.

Mike.

arod · Apr 19, 2007

ljcox said:
I did not say that 0.33... = 1/3. I said it approaches 1/3 as the number of decimal places approaches infinity.

No, I'm not trying to quantify infinity, I'm simply stating a mathmetical fact.

You can't quantify infinity, you can only approach it.

0.33~ = 1/3 => That is mathematical fact. What could it possibly be "approaching?" Here is one of many proofs:

c = 0.999~
10c = 9.999~
10c - c = 9.9999~ - 0.999~
9c = 9
c = 1

Hard to argue with that. Whether you want to believe it or not, this is accepted by all mathematicians. If you still don't want to believe me, I will show you the convergence of the infinite geometric series when I get home later today.

gramo · Apr 19, 2007

Pommie said:
And, as X approaches infinity then (1+1/X)^X approaches e.

Mike.

umm, where did e come from?

Pommie · Apr 19, 2007

The natural constant e is the amount in billions of dollars that google was valued at when it floated on the NYSE. It is also the base of natural logarithms. See wiki.

In fact they define e as,
**broken link removed**

Mike.

dknguyen · Apr 19, 2007

ljcox said:
If you read the original question it was "0.99 -continuious- = 1" and the answer is no. Rounding up is a different issue.

0.999 fixed to 0.999...

ljcox · Apr 20, 2007

arod said:
0.33~ = 1/3 => That is mathematical fact. What could it possibly be "approaching?" Here is one of many proofs:

c = 0.999~
10c = 9.999~
10c - c = 9.9999~ - 0.999~
9c = 9
c = 1

Hard to argue with that. Whether you want to believe it or not, this is accepted by all mathematicians. If you still don't want to believe me, I will show you the convergence of the infinite geometric series when I get home later today.

Correct me if I'm wrong, but I don't see how you can subtract 2 numbers that have an infinite number of decimal places. To do a subtraction, you start at the last digit. eg. 7.89 - 3.45, you start with 9 - 5.

But with an infinite number of decimal places, there is no last digit.

I will be interested to see your convergence of an infinite geometric series.

Optikon · Apr 20, 2007

arod said:
0.33~ = 1/3 => That is mathematical fact. What could it possibly be "approaching?" Here is one of many proofs:

c = 0.999~
10c = 9.999~
10c - c = 9.9999~ - 0.999~
9c = 9
c = 1

Hard to argue with that. Whether you want to believe it or not, this is accepted by all mathematicians. If you still don't want to believe me, I will show you the convergence of the infinite geometric series when I get home later today.

It is not hard to argue with because it is not mathematically correct.

Your first & last statement say that C = 0.999~ and also that C =1 implying that 1 = 0.999~ which is NOT true.

It is true however that Lim (n->inf) {0.999~} = 1 where n is number of repeating 9's. So your statements as they stand are NOT correct and the convergence of the geometric series is not the issue here technically. The geometric series convergence depends on the existence of its limit. You are showing no limits which is why it is wrong.

rumiam · Apr 20, 2007

ok if 1 does not = 0.99999~ then explain to me;
why is it that
1/3 = 0.333~
2/3 = 0.666~
3/3 = 0.999~

3/3= 1
so
1= 0.999~

Prove this to be wrong

ljcox · Apr 20, 2007

rumiam said:
ok if 1 does not = 0.99999~ then explain to me;
why is it that
1/3 = 0.333~
2/3 = 0.666~
3/3 = 0.999~

3/3= 1
so
1= 0.999~

Prove this to be wrong

Your error is in adding the decimal values for 1/3 and 2/3

These have an infinite number of decimal places and so I don't see how you can add them.

ljcox · Apr 20, 2007

Pommie said:
The natural constant e is the amount in billions of dollars that google was valued at when it floated on the NYSE. It is also the base of natural logarithms. See wiki.

In fact they define e as,
**broken link removed**

Mike.

I was intrigued by this so I did the maths to prove it. See attachment. What I forgot to write is that the first 3 terms in the last line are the first in the infinite series for e. The terms after 1/n go to 0 as n approaches infinity.

ljcox · Apr 20, 2007

I think the answer to this issue is in whether the attached series converges to 1 or to 0.9999~.

It is many years since I studied maths that I don't recall how to do the test for convergence of this series and don't have the time to revise it.

So if any knows how to do it, please enlighten us.

Sceadwian · Apr 20, 2007

I don't believe this post is still going.

arod · Apr 20, 2007

Optikon said:
It is not hard to argue with because it is not mathematically correct.

Your first & last statement say that C = 0.999~ and also that C =1 implying that 1 = 0.999~ which is NOT true.

Yes, isnt that the point of the proof? Showing that .999~ = 1...
I don't think you understand. When I type 0.999~, it means that the 9s repeat to infinity.

Instead of me reinvinting the wheel, here is a page showing the convergence of an infinite geometric series proving that 0.999~ = 1

tkbits · Apr 20, 2007

How about an alternate proof?

Code:

    0.9999....
   -----------
9 ) 9.0000....
    8 1
    ---
      90
      81
      --
       90
       81
       --
        90
        81
        --
         9....

0.99~=1

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