Simple mathematical problem (?)

[LuxFerum]

The mistake is right there. You can't have a leftmost digit when you write infinity, so you would have to put something like:

...9999999999999999...

Note: there are three dots at both ends.

And then the rest of your proof falls apart.

If I have an infinity line and divide it in half?
those halfs are infinity?


yes they are.

 

[ceptimus]

If I have an infinity line and divide it in half?
those halfs are infinity?


yes they are.

Yes, and they would both have three dots at either end...

 

[xouper]

LuxFerum:

10*0.999... = 9.999... = 9 + 0.999..., you can subtract 0.999... from that.

no you cant.

Yes you can.

look at this
infinity - infinity =0

No, infinity - infinity does not equal zero.

And whe know that
9+infinity = infinity

Agreed.

Then whe got
9+infinity-infinity=0
9=0

Which is why infinity-infinity is not zero.

You just cant work with infinity like that.

Agreed. But 0.999.... is NOT infinity, so we CAN work with it like that.

 

[LuxFerum]

• Find a real number N such that N is less than 1 and N is greater than 0.999...

• 
  If you say that
  0,999.. = sum 9*10(-n), n=1 to infinity
  
  That number is
  
  1-sum 99*10(-2n), n=1 to infinity

 

[LuxFerum]

Yes, and they would both have three dots at either end...

so what you are saying is that
0.999... don't have infinites nines in it?

 

[LuxFerum]

Agreed. But 0.999.... is NOT infinity, so we CAN work with it like that.

then 0,999... dont have infinity nines on it?

 

[xouper]

ceptimus: The mistake is right there. You can't have a leftmost digit when you write infinity, so you would have to put something like:

...9999999999999999...

Note: there are three dots at both ends.

And then the rest of your proof falls apart.

And the reason his proof falls apart is because ...999 is not a real number. Nor is 999...

 

[Walter Wayne]

It proves that sum 9*10^(-n) is not 1 in infinity.

That is what Im trying to prove there.

That sum just tend to 1.

If that sum is 1 in infinity then

sum 9*10^(-n) - sum 2^(-n) "is" zero.
but that is not true.
it tends to zero.

All you say is that you can think of a time where 0*Infinity = Infinity. It doesn't show 0.999... != 1.

Consider f(x)=g(x)*h(x)

Case 1:
g(x) = x
h(x) = 1/x

g(0) = 0
lim h(x) as x -> 0 = infinity

lim f(x) as x -> 0 = 1

Case 2:
g(x) = x²
h(x) = 1/x

g(0) = 0
lim h(x) as x -> 0 = infinity

lim f(x) as x -> 0 = 0

Case 3:
g(x) = x
h(x) = 1/x²

g(0) = 0
lim h(x) as x -> 0 = infinity

lim f(x) as x -> 0 = infinity

Three cases where g at a value is 0, and I multiply it by something which tends to infinity at that value and in each case I get a different answer. None of these showed that g(0) != 0.

Your proof does not show that (1-0.999...) != 0

Walt

 

[TillEulenspiegel]

this have the same mistake as this one

            999999999999999999999999...  This is infinity
            999999999999999999999999...  This is infinity too


When we multipliy something by ten, the numbers move one row to the left.
And when we multiply infinity by ten, the result is infinity.


           9999999999999999999999999...  This is infinity multiplyed by 10, and still infinite
            999999999999999999999999...  This is infinity 

When we subtry infinite from infinte the result is zero
sooo 


           9999999999999999999999999...  
           -999999999999999999999999...  
           ----------------------------
           9000000000000000000000000...

So by this result is clear that infinity is equal to zero.

Uhh no the results are any number subtracted from itself =0, in this case we have an number N plus an imaginary number X. The fact that X is an imaginary number that recurrs toward infinity has no bearing as we have established its qualities in the varible X.

so N + X = y y-X=N yes? the fact that X has certain properties is understood but negated by subtraction the result? 9.000... is NOT 0.. or did you not see the leading digit 9 in the sum of your calculation?

 

[LuxFerum]

Lets start again

Let 0,99... be sum 9.10^(-n) n=1... infinity
Let 0,99... be sum 2^(-n) n=1.. infinity

If we subtract one from another we have


sum [2^(-n)-9.10(-n) ] n=1... infinity

This tends to zero.
But it is not zero wen n is infinity.

And I can show that buy multiplying that sum buy 10^n
which leads the result to infinity.


which is not true is both series can't be 1 in infinity.
They just tend to one.

That is the point here.

when you say 0.999.... is one you are ignoring the fact that
it tends to 1 with some speed.

You simply can not ignore that speed when you put it side by side with another serie that tends to the same number with different speed.

And that is when 0.9999.. behave diferente from 1.

one is already there.

sum 9.10^(-n) n=1... infinity is going there.

 

[69dodge]

Originally posted by LuxFerum
So you agree that 0,999.. tends to 1 and is not equal to 1?

Well. This is kind of frustrating.

I thought I was being pretty clear. I guess not.

No, 0.999. . . doesn't tend to 1, because 0.999. . . is a number. A number doesn't tend to anything; it just is whatever number it is.

It happens to be that in order to determine what number "0.999. . ." represents, we need to find out what limit a certain sequence tends to. But once we've done that, what we're left with is a number. Just a plain, regular number. In this case, that number is 1.

 

[Walter Wayne]

Lux, one more nit-pick of you big proof, I ignored it in the previous post since I was rebutting you conclusion about finding something that tends to infinity faster.

You multiplied different fractions of a whole by different numbers. Lets take two different sequences that equal 1 and subtract them.

0.8 + 0.2 = 1
0.7 + 0.3 = 1

You multiplied the n-th terms in your two series, (0.9, 0.09, ...) and (0.5, 0.25, ...) by 10ⁿ. If I do this in my two series I get

80 + 200 =280
70 + 300 = 370
(80 -70) + (200 - 300) = -90

By your reasoning I have somehow that my first to series are not equal. You abused the communative property of multiplication. In your case in series 1 (0.9, 0.09, ...) you ended up multiplying 90% of the sum by 10, 9% by 100 and so on. In the second case you multiplied 50% of the sum by 10, 25% by 100 and so on. So you should not be surprised when you result is gibberish.

Walt

 

[xouper]

xouper: Find a real number N such that N is less than 1 and N is greater than 0.999...

LuxFerum: If you say that
0,999.. = sum 9*10(-n), n=1 to infinity

That number is

1-sum 99*10(-2n), n=1 to infinity

So you are claiming that N=0. Excellent. Then you also agree that 0.999... equals 1.

xouper: Agreed. But 0.999.... is NOT infinity, so we CAN work with it like that.

LuxFerum: then 0,999... dont have infinity nines on it?

Are you being intentionally obtuse? Having an infinite number of digits after the decimal point does not mean the number equals infinity.

 

[LuxFerum]

It happens to be that in order to determine what number "0.999. . ." represents, we need to find out what limit a certain sequence tends to. But once we've done that, what we're left with is a number. Just a plain, regular number. In this case, that number is 1.

the limit is 1 then.

 

[Walter Wayne]

And I can show that buy multiplying that sum buy 10^n
which leads the result to infinity.

You are not multiplying the sum by 10^n, you are multiplying the terms by 10^n. This is important. See what I did above. Two finite sums which are equal but with different terms, and multiplying those terms by 10^n, and look they are no longer equal.

Walt

 

[LuxFerum]

80 + 200 =280
70 + 300 = 370
(80 -70) + (200 - 300) = -90

80-70=10
200-300=-100
10-100=-90
That is still correct.

But when we change sum 9.10^(-n) n=1... infinity buy 1
and sum 2^(-n) n=1... infinity
whe got this

Sum [(1 -1)*10^n] n=1.. infinity

And that is zero.

the result is different from infinity wich whe know that is the correct answer.

sum 9.10(-n) n=1... infinity tends to 1. and is not 1.

 

[LuxFerum]

So you are claiming that N=0. Excellent. Then you also agree that 0.999... equals 1.

Are you being intentionally obtuse? Having an infinite number of digits after the decimal point does not mean the number equals infinity.

no I claim that N tends to zero
and I agree that 0,999.... tends to 1.

no Im not.
having a infinity nimbers of digits shows that you cant work with that digits the same way you do with non infinity string of numbers.

 

[xouper]

LuxFerum: when you say 0.999.... is one you are ignoring the fact that it tends to 1 with some speed. You simply can not ignore that speed when you put it side by side with another serie that tends to the same number with different speed.

Yes you can ignore the "speed", since there is no "speed" in the definition of the notation 0.999..., since all the nines are there simultaneously.

And that is when 0.9999.. behave diferente from 1.

one is already there.

0.999... is also "already there". That is the definition of the notation. I am getting tired of your attempts to redefin the meaning of the notation. Stop doing it.

If we were having this conversation on the newsgroup sci.math, you would have been dismissed as a crank a long time ago. The poeple here on this forum are being quite patient with you and trying to help you understand the flaws in your math. Please do yourself a favor and do not abuse that patience by continuing to insist you are right and we are wrong.

 

[69dodge]

(Previewing this, I see that Walter Wayne has already said the same thing.)

Originally posted by LuxFerum
Lets start again

Let 0,99... be sum 9.10^(-n) n=1... infinity
Let 0,99... be sum 2^(-n) n=1.. infinity

If we subtract one from another we have


sum [2^(-n)-9.10(-n) ] n=1... infinity

This tends to zero.
But it is not zero wen n is infinity.

And I can show that buy multiplying that sum buy 10^n
which leads the result to infinity.

You're not multiplying the sum by 10ⁿ. You can't multiply the sum by 10ⁿ, because 10ⁿ is not a single number: n is the index variable of the summation. What you're doing is multiplying each term of the series by 10ⁿ. In other words, you're multiplying the first term by 10, the second term by 100, the third term by 1000, etc. That results in a completely different series, which naturally might have a different value when summed. So what?

 

[TillEulenspiegel]

I'm not sure I understand what your trying to say...Maybe its my holoween hangover , or maybe a language problem but ..........

"And that is when 0.9999.. behave diferente from 1."

Of course it is , ask any chaos theorist or physicist
as far as translating that to practicle terms it's like the old problem of approching a wall at 1/2 the distance you are away from it, then 1/2 of that distance and so on ad infinitum ...the practicle is that you never get there, the theoretical is somewhere along the line you will bump in to a wall alright, like the planck length which will allow no further ( at least mesurable ) movement, but you still havent reached total convergence. But this kind of mental meandering seems fruitless as the issue ( in basic mathmatical terms ) can be addressed by other means then convoluted logic and bad math.