Rf wrote:
here lies subzero wrote:I guess in this scenario zero doesn't count as a real number. Sad!
I guess in this scenario division by zero counts as a defined operation. Sad!
Um, I'm standing right here.
Rf wrote:
here lies subzero wrote:I guess in this scenario zero doesn't count as a real number. Sad!
I guess in this scenario division by zero counts as a defined operation. Sad!
Um, I'm standing right here.
Bad Wigins wrote:
not only can't you do that in general, it's not equivalent for any a and b in the set of real numbers.
Suppose a/(a+b) = 1 + a/b
then
ab/(a+b) = b + a
ab = (b+a)^2
ab = b^2 + ab + a^2
b^2 + a^2 = 0
This can be extended to show that there aren't even any fields where a/(a+b)=1+a/b for any nonzero a and b (being lazy and writing x*y^-1 as x/y).
Assume for the sake of contradiction that F is a field where the property holds. Then by BW's argument aa=-bb for any nonzero a and b in F. But then ab=(a+b)^2=-b^2=a^2, so a=b=-b for any nonzero a,b in F. But then a+b=0, so (a+b)^-1 is not defined, a contradiction.
In can be interesting to ask when there are a and b in some structure such that this property holds at all. For example, one can show that if V is an R vector space there exist a and b in V such that (a+b)^-1=a^-1+b^-1 iff there is an operator T on V such that T^2=-I for I the identity map. I.e if the vector space is some way resembles the complex numbers.
Rf wrote:
Bad Wigins wrote:not only can't you do that in general, it's not equivalent for any a and b in the set of real numbers.
Suppose a/(a+b) = 1 + a/b
then
ab/(a+b) = b + a
ab = (b+a)^2
ab = b^2 + ab + a^2
b^2 + a^2 = 0
This can be extended to show that there aren't even any fields where a/(a+b)=1+a/b for any nonzero a and b (being lazy and writing x*y^-1 as x/y).
Assume for the sake of contradiction that F is a field where the property holds. Then by BW's argument aa=-bb for any nonzero a and b in F. But then ab=(a+b)^2=-b^2=a^2, so a=b=-b for any nonzero a,b in F. But then a+b=0, so (a+b)^-1 is not defined, a contradiction.
In can be interesting to ask when there are a and b in some structure such that this property holds at all. For example, one can show that if V is an R vector space there exist a and b in V such that (a+b)^-1=a^-1+b^-1 iff there is an operator T on V such that T^2=-I for I the identity map. I.e if the vector space is some way resembles the complex numbers.
well the joke's on you because actually if
ab = (b+a)^2 then
ab = b^2 + 2ab + a^2
a^2 + ab + b^2 = 0
still no real solutions, but it has complex roots, as did the previous. Why were you saying that one didn't?
Bad Wigins wrote:
well the joke's on you because actually if
ab = (b+a)^2 then
ab = b^2 + 2ab + a^2
a^2 + ab + b^2 = 0
still no real solutions, but it has complex roots, as did the previous. Why were you saying that one didn't?
I showed there's no field where that holds for every a,b, not that there can't exist specific instances where it holds. Things like the complex numbers fall into the category of R vector spaces I mentioned later. In fact operators such that T^2=-I are called complex structures precisely because they look like negative numbers.
Oh, maybe replace "for any" with "for all" and the proof makes more sense. Words are trickeir than just drawing an upside down A.
Obvious troll question was answered by the first response, yet people keep chiming in. I think giving troll ratings are silly but this was more successful than it should have been.
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