0.99~=1 - Page 5 - Electronic Circuits Projects Diagrams Free

bananasiong · 28th May 2007, 09:25 AM

Quote:

Originally Posted by tkbits

How about an alternate proof?

Code:

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

So what's the answer for this?
0.99...99 PLUS 0.00...09 numerator and 9 denomenator, so the answer is still 1.

Quote:

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

How about this:

Code:

c = 0.99...99
10c = 9.99...90
10c - c = 9.99...90 - 0.99...99
c = 0.99..99

Roff · 28th May 2007, 05:14 PM

Quote:

Originally Posted by bananasiong

So what's the answer for this?
0.99...99 PLUS 0.00...09 numerator and 9 denomenator, so the answer is still 1.

How about this:

Code:

c = 0.99...99
10c = 9.99...90
10c - c = 9.99...90 - 0.99...99
c = 0.99..99

I think you miss the point. The string of 9's is infinitely long. You can't multiply by 10 and have a 0 show up at the end. If you think this will happen, then you don't understand the concept of infinity (which, admittedly, is difficult to comprehend).

bananasiong · 29th May 2007, 01:13 AM

Quote:

Originally Posted by Ron H

I think you miss the point. The string of 9's is infinitely long. You can't multiply by 10 and have a 0 show up at the end. If you think this will happen, then you don't understand the concept of infinity (which, admittedly, is difficult to comprehend).

You're right too. But sometimes this can be done by imaging it, but not writing it out.
For example, does anyone know about Heaviside's rule of partial fraction? There is a step which is something like this:

Code:

2(infinity)/infinity = 2

But this cannot be written on the paper, and my lecturer call this as 'illegal job'

3iMaJ · 29th May 2007, 06:17 PM

infinity/infinity is an indeterimant form, IE you can't say 2*inf/inf = 2. The rules dont work like that.

bananasiong · 30th May 2007, 03:22 AM

Quote:

Originally Posted by 3iMaJ

infinity/infinity is an indeterimant form, IE you can't say 2*inf/inf = 2. The rules dont work like that.

Yes, you can't say 2*inf/inf = 2, but this is done when the Heaviside's rule of partial fraction is applied. Something like this:

Code:

A = lim(x-->inf)  [2x/x  +  (x+B)/x]
A = 3 edited

Straight away as shown above, but the infinity over infinity cannot be written.

3iMaJ · 30th May 2007, 05:52 AM

Thats completely different. You're thinking of l'Hospital's rule where you have an indeterimant form involving limits it goes as:

if lim g(x)/f(x) is either 0/0 or inf/inf then the following applies

lim g(x)/f(x) = lim g'(x)/f'(x) where g'(x) and f'(x) are the derivatives which is what you would apply in your problem for both sections.

Applying to your problem:

lim (x->inf) 2*x/x + (x+B)/x = inf/inf + inf/inf as above l'Hospital's rule applies so:
d/dx of above = 2 + 1/1 = 3

lim (x -> inf ) [2*x/x + (x+B)/x] = 3

bananasiong · 30th May 2007, 06:16 AM

Yes, you're right, that is 3. I didn't notice that. Why you call that as hospital's rule??

So, do you think 0.99...99 =1?

3iMaJ · 30th May 2007, 12:29 PM

Quote:

Originally Posted by bananasiong

So, do you think 0.99...99 =1?

Yep. I posted the proof above. I'll post it again here.

HudzonHawk · 3rd June 2007, 01:37 PM

Wait, people are still arguing over the whole 0.99.. = 1 equation despite the fact that a 5 minute search on the Internet will show thousands of credible sources which prove that the equation is true?

No offense to anyone, but do you guys also disprove that the Earth is round?

Roff · 3rd June 2007, 10:27 PM

Quote:

Originally Posted by HudzonHawk

Wait, people are still arguing over the whole 0.99.. = 1 equation despite the fact that a 5 minute search on the Internet will show thousands of credible sources which prove that the equation is true?

No offense to anyone, but do you guys also disprove that the Earth is round?

I've been trying to get the naysayers to rewrite the Wikipedia entry. No takers yet.

bananasiong · 10th June 2007, 08:49 AM

Quote:

Originally Posted by Ron H

I've been trying to get the naysayers to rewrite the Wikipedia entry. No takers yet.

So you can proof that 0.99...999 is not 1?

Roff · 10th June 2007, 03:08 PM

Quote:

Originally Posted by bananasiong

So you can proof that 0.99...999 is not 1?

No, my statement was meant as sarcasm. What I meant was, if the naysayers believe so strongly that 0.999... is not equal to one, they should correct the Wikipedia entry. In English, we have a saying: "Put your money where your mouth is". You probably know that saying, or have a similar one in your culture.

**Nigel Goodwin** · 10th June 2007, 10:36 PM

The most amazing thing about this entire thread is that I've not deleted or locked it yet!

Sceadwian · 11th June 2007, 07:22 AM

Please Nigel! PLEASE!

marplots · 13th June 2007, 03:01 AM

1 - .999... is zero. The reason for this is that between any two different rational or irrational numbers, there is an infinite number of other numbers. The only way that there can be no difference between two numbers is if they are, in fact, the same number.

Hence, there is no number between 3 and 3, but an infinite number of them between 3.141414... and 3.1399999...

Since, for the expression .999... and 1 there is no number 'between', they are, in fact, the same number. Any difference you come up with is an artifact of your calculator or truncation.

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29th May 2007, 06:17 PM	(permalink)
3iMaJ	infinity/infinity is an indeterimant form, IE you can't say 2*inf/inf = 2. The rules dont work like that.

30th May 2007, 05:52 AM	(permalink)
3iMaJ	Thats completely different. You're thinking of l'Hospital's rule where you have an indeterimant form involving limits it goes as: if lim g(x)/f(x) is either 0/0 or inf/inf then the following applies lim g(x)/f(x) = lim g'(x)/f'(x) where g'(x) and f'(x) are the derivatives which is what you would apply in your problem for both sections. Applying to your problem: lim (x->inf) 2x/x + (x+B)/x = inf/inf + inf/inf as above l'Hospital's rule applies so: d/dx of above = 2 + 1/1 = 3 lim (x -> inf ) [2x/x + (x+B)/x] = 3

30th May 2007, 06:16 AM	(permalink)
bananasiong	Yes, you're right, that is 3. I didn't notice that. Why you call that as hospital's rule?? So, do you think 0.99...99 =1? __________________ Superman returns..

3rd June 2007, 01:37 PM	(permalink)
HudzonHawk	Wait, people are still arguing over the whole 0.99.. = 1 equation despite the fact that a 5 minute search on the Internet will show thousands of credible sources which prove that the equation is true? No offense to anyone, but do you guys also disprove that the Earth is round?

10th June 2007, 10:36 PM	(permalink)
Nigel Goodwin	The most amazing thing about this entire thread is that I've not deleted or locked it yet! __________________ PIC programmer software, and PIC Tutorials at: http://www.winpicprog.co.uk

11th June 2007, 07:22 AM	(permalink)
Sceadwian	Please Nigel! PLEASE! __________________ "Because I be what I be. I would tell you what you want to know if I could, mum, but I be a cat, and no cat anywhere ever gave anyone a straight answer, har har."

13th June 2007, 03:01 AM	(permalink)
marplots	1 - .999... is zero. The reason for this is that between any two different rational or irrational numbers, there is an infinite number of other numbers. The only way that there can be no difference between two numbers is if they are, in fact, the same number. Hence, there is no number between 3 and 3, but an infinite number of them between 3.141414... and 3.1399999... Since, for the expression .999... and 1 there is no number 'between', they are, in fact, the same number. Any difference you come up with is an artifact of your calculator or truncation.