@WassimBerbar said in #13:
> Imagine you cut an invisible pizza into nothing, the answer is confusing: Is there one invisible pizza left so it's 1, or is there one invisible pizza but it's not a pizza so it's 0?
imagine you divide a pizza into -4 slices, actually that's pretty hard to imagine
> Imagine you cut an invisible pizza into nothing, the answer is confusing: Is there one invisible pizza left so it's 1, or is there one invisible pizza but it's not a pizza so it's 0?
imagine you divide a pizza into -4 slices, actually that's pretty hard to imagine
@ST4RSCR34M said in #32:
> imagine you divide a pizza into -4 slices, actually that's pretty hard to imagine
It could work in an imaginary world. Pun intended.
> imagine you divide a pizza into -4 slices, actually that's pretty hard to imagine
It could work in an imaginary world. Pun intended.
Suppose we have a \in F ( where F is a field with usual sum and multiplication), 0 = e_+ (ie, identity of sum) and 1 = e_*(multiplication identity.
a * e_+ = a * 0 \implies 0 * a since F is a field
a * 0 = 0 * a \implies a * 0 * a^{-1} = 0 * a * a^{-1} since \forall a \in F, \exists a^{-1} \in F st a*a^{-1}=a^{-1}*a=e_* (Same for +)
a * 0 * a^{-1} = 0 * e_* \to a * 0 * a^{-1} = 0 * 1 \implies a * 0 * a^{-1} = 0 since a*e_*=e_**a=a, \forall a \in F
a * 0 * a^{-1} = 0 \to a * (a'+(-a')) * a^{-1} = 0 \to (a*a'+a*(-a'))*a^{-1} = 0 since a+e_+=e_++a=a \implies e_+ = a + (-a)
(a''+(-a''))*a^{-1} = 0 \to 0 * a'''' = 0 \implies 0 * 0 ^{-1} = a, a \in F. ie 0/0 can be any value...
Running in circles in abstract algebra just to find the classic 0/0
edit: Pls add Latex symbols here plsss
a * e_+ = a * 0 \implies 0 * a since F is a field
a * 0 = 0 * a \implies a * 0 * a^{-1} = 0 * a * a^{-1} since \forall a \in F, \exists a^{-1} \in F st a*a^{-1}=a^{-1}*a=e_* (Same for +)
a * 0 * a^{-1} = 0 * e_* \to a * 0 * a^{-1} = 0 * 1 \implies a * 0 * a^{-1} = 0 since a*e_*=e_**a=a, \forall a \in F
a * 0 * a^{-1} = 0 \to a * (a'+(-a')) * a^{-1} = 0 \to (a*a'+a*(-a'))*a^{-1} = 0 since a+e_+=e_++a=a \implies e_+ = a + (-a)
(a''+(-a''))*a^{-1} = 0 \to 0 * a'''' = 0 \implies 0 * 0 ^{-1} = a, a \in F. ie 0/0 can be any value...
Running in circles in abstract algebra just to find the classic 0/0
edit: Pls add Latex symbols here plsss
Nothing... divided by nothing... leaves nothing
You gotta have something... if you want to be with me - :]
You gotta have something... if you want to be with me - :]
How many nothings must be arranged in a row to equal nothing. One nothing will work. But won't two nothings work, too? Or three nothings? Where does one stop? It's like having a six pack of beer. Or a frisky and lovely wife. There is no right answer.
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