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Acmlm's Board - I2 Archive - Lost Section - The best number ever is approximately 8.53973422862967682 | | Thread closed
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Smallhacker

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Since: 03-15-04
From: Söderhamn, Sweden

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Posted on 11-13-04 02:48 AM Link
Originally posted by Cymoro
CYS MATH LESSON: Dividing By Zero

Step 1: You can't.


If you ask me for my opinion, I would say this:
X>0
Y<0
X/0 = Infinity
0/0 = 1
Y/0 = - Infinity. (An infinitly small number)
Alastor the Stylish
Hey! I made a cool game! It's called "I poisoned half the food, so if you eat you might die!" Have a taco.


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Since: 03-15-04
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Posted on 11-13-04 03:45 AM Link
Originally posted by Smallhacker
Originally posted by Hiryuu
And one is still the loneliest number.

Are you serious? Every single number can be divided by one. It's far from lonely!
Figures you wouldn't understand this, being Swedish. It's a reference to a song.
HyperLamer
<||bass> and this was the soloution i thought of that was guarinteed to piss off the greatest amount of people

Sesshomaru
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Since: 03-15-04
From: Canada, w00t!
LOL FAD

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Posted on 11-13-04 06:25 AM Link
Hey, speaking of geekiness numbers... 31337 is a prime number.

Squash: Seek mental help.

(edited by HyperHacker on 11-12-04 09:26 PM)
NSNick
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Posted on 11-13-04 02:43 PM Link
Originally posted by Smallhacker
0/0 = 1

0/0 = Undefined.



If 0/0 equaled 1, then...

0/0 = 1
0/0 * 2/1 = 1 * 2/1
0/0 = 2
1 = 2

And 1 does not equal 2.
Smallhacker

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Posted on 11-13-04 04:30 PM Link
Lame guess:
0/0 = The square root of 0-(10^-infinity)

Wait...

0/0 = 1
0/0 * 2/1 = 1 * 2/1
0/0 = 2
1 = 2

I can't see how that works...

Hmm...
0/0 = 1
0/0 * 2/1 = 1 * 2/1
X * Y = 1 * Y
...I can't understand how to make that mean 0/0=2...

(edited by Smallhacker on 11-13-04 07:37 AM)
Kitten Yiffer

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Vivent l'exp����¯�¿�½������©rience de signalisation d'amusement, ou bien !
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Posted on 11-13-04 06:23 PM Link
For being his 1337's post that was a nice one.

But well, I can sorta see why people think 0/0 = 1 since X/X is always 1. Still 0/0 may as well be an expection from that rule then... or?
NSNick
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Posted on 11-13-04 10:56 PM Link
Originally posted by Smallhacker
I can't see how that works...

Hmm...
0/0 = 1
0/0 * 2/1 = 1 * 2/1
X * Y = 1 * Y
...I can't understand how to make that mean 0/0=2...


   0  =  1
   0              (I start with your assumption that 0/0 = 1)

   0  =  1
   0     1        (1 is equal to 1/1)

2 * 0 = 1 * 2
1   0   1   1     (Multiply both sides by 2/1)

2 * 0 = 1 * 2
1 * 0   1 * 1     (Showing the fraction multiplication)

    0 = 2
    0   1         (Result of the multiplication)

    0 = 2
    0             (2 is equal to 2/1)

    1 = 2         (Substitute in from the first line)
Smallhacker

Green Birdo

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Posted on 11-14-04 03:28 AM Link
Ah. I see... What about 0/0=0, then? Or what if 0/0 is as real as -1^0.5?
HyperLamer
<||bass> and this was the soloution i thought of that was guarinteed to piss off the greatest amount of people

Sesshomaru
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From: Canada, w00t!
LOL FAD

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Posted on 11-14-04 09:00 AM Link
Hey, here's another fun number. I'm exactly six thousand six hundred and six days old today. And, this is my 0x800th post. Woo.
MathOnNapkins

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Posted on 11-15-04 09:46 AM Link
Hmm some ideas:

Let a/0 = a' existing in R'. a is from R is the real number field.

Some axioms:

1. a/0 * b/0 = (ab)/0^2 Which would be in R'', not R or R'. See how far this can go?
2. a/0 * b = (ab)/0. A better multiplication rule.
3. a/0 + (-a)/0 = 0/0 (additive inverse rule)
4. a/0 * 0 = a. Note this is an exception to rule 2. This sets the precedent for fractional multiplication with zero in the numerator.
5. thus, a/0 * 0/a = a/a = 1, a =/= 0 (multiplicative inverse)
6. 0/0 is the additive identify, 1 is the multiplicative identity
7. a * 0/0 = 0/0, for a =/= 0. a/0 * 0/0 = a/0 here a can be 0. Note that here we cannot say 0/0 = 1 in general because it acts differently upon the sets R and R'. Thus R and R' are still not trivial ( = {0} )

Miscellaneous:
8. a/0 + b = a/0 + b/1. And since division by zero is an operator that sends b -> b', we cannot use it to place b in the set R'. hence there can be no common denominator, and the sets R and R' are therefore orthogonal. a' + b is then the simplest expression.
9. For this to change you have to change normal notions that things like 0/2 = 0. Here such terms cannot be so easily eliminated. Also, 0*1 is not the same as 0 anymore.
10. From this we might infer that multipication by 0 also yields something like a field R(sub)', and R(sub)'', etc. with further multiplication. Hence 0*b might be b(sub)' in R(sub)'. I haven't explored that much though.

If you find any problems with this I wouldn't be surprised, since it was thrown together in like 20 min. If you don't find any problems, then I guess this is a way to divide by zero. Nyah.

(edited by MathOnNapkins on 11-15-04 12:48 AM)
Zem
You can be civil without being flowery, dipshits.
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Posted on 11-19-04 10:54 AM Link
I am genuinely curious. Did you write all that out on a napkin before you posted it? ~;o
MathOnNapkins

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Posted on 11-20-04 12:43 AM Link
YES, YES It's all true! *sob*












No actually I typed it from scratch into the reply box.
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