Werewolf - Priest of Lycan - Game Thread.

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Strikingwolf

New Member
Jul 29, 2019
3,709
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Let me demonstrate what happens if zero had a multiplicative inverse in a ring with 1.

Suppose there was an a with 0*a = 1.
Then 0*a = 0*a + 0 = 0*a + 0*a - (0*a) = (0+0)*a - (0*a) = 0*a - (0*a) = 0.
So we have 1 = 0.
For each b in the ring we have b = 1*b = 0*b = 0. In other words, all elements of the ring are zero.
So you can only divide by zero if nothing else than zero exists.
Actually not. You are forgetting some key things. Such as limits. If we take the limit of 1/x as x approaches 0 we get infinity and here is how the magic happens. You can't do arithmetic on infinite values. It just outright fails. Therefore since your theorem is entirely arithmetic it will fail. Whereas limits are not arithmetic so my theorem stands
 

Pyure

Not Totally Useless
Aug 14, 2013
8,334
7,191
383
Waterloo, Ontario
Actually not. You are forgetting some key things. Such as limits. If we take the limit of 1/x as x approaches 0 we get infinity and here is how the magic happens. You can't do arithmetic on infinite values. It just outright fails. Therefore since your theorem is entirely arithmetic it will fail. Whereas limits are not arithmetic so my theorem stands
.......
My cat's breath smells like catfood.
 

Strikingwolf

New Member
Jul 29, 2019
3,709
-26
1
Actually not. You are forgetting some key things. Such as limits. If we take the limit of 1/x as x approaches 0 we get infinity and here is how the magic happens. You can't do arithmetic on infinite values. It just outright fails. Therefore since your theorem is entirely arithmetic it will fail. Whereas limits are not arithmetic so my theorem stands
To go further it is kinda like i If we go by arithmetic there should be no value that when squared gives us -1 to show this.
a^2 = b
+- a = +- sqrt(b)
We get both positive and negative answers. Thus encompassing all the real numbers. However, we can define i as i^2 = -1
It is the same way for infinity. IIRC the way infinity is formally defined is. a/0 = infinity
 

Pyure

Not Totally Useless
Aug 14, 2013
8,334
7,191
383
Waterloo, Ontario
To go further it is kinda like i If we go by arithmetic there should be no value that when squared gives us -1 to show this.
a^2 = b
+- a = +- sqrt(b)
We get both positive and negative answers. Thus encompassing all the real numbers. However, we can define i as i^2 = -1
It is the same way for infinity. IIRC the way infinity is formally defined is. a/0 = infinity
.....
I bent my wookie :(
 

SpwnX

New Member
Jul 29, 2019
210
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0
I may be doing engineering, but i ain't claiming anything.
I just know that my scientific calculator gives me a nice "math error" error if i try "anything"/0. Should be enough evidence.
Also limit 1/x (x-> 0) does not actually divide 1/0 but a number infinitely small that approaches to 0.
 

Strikingwolf

New Member
Jul 29, 2019
3,709
-26
1
Also limit 1/x (x-> 0) does not actually divide 1/0 but a number infinitely small that approaches to 0.
Limits are used in all the higher branches of mathematics to do calculations like these that don't make sense. They are taken as fact because a number infinitely close to a that approaches a is...a
Bold = repeating
For example. .9 is a number infinitely close to 1. However, let me prove to you it is 1
x = .9
10x = 9.9
10x - x = 9.9 - x
9x = 9.9 - .9
9x = 9
x = 1