To be clear: you also need a²=0 and b²=0, otherwise you just get a²+b².
you get that from the fact that it's fora all a,b and and so a^2 = -a^2 --> a^2=0
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Ok, it wasn't clear that the statement was universally quantified.
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that's fair but i was running out of characters lol
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But also, in that case then, it has nothing to do with (a+b) being a binomial!
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that's right
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Wait, doesn't this come from a=-b, which makes a+b=0? What exactly is for all a and b? Please shed some light
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The relation x^2 =0 for all x, implies xy = -yx. In characteristic 2 (i.e. when -1 = +1), the converse is not true. So mathematicians (at least if they care about characteristic p) define the exterior algebra of a vector space V as the tensors of V mod v^2 = 0 for v \in V.
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