### _Bunny's blog

By _Bunny, history, 2 weeks ago,

hi everyone!

I read a blog write about formula of a^b = a'b + ab' and i dont understand. Someone can explain it for me pls :((

(sorry i'm poor E).

• +7

 » 2 weeks ago, # |   0 Auto comment: topic has been updated by _Bunny (previous revision, new revision, compare).
 » 2 weeks ago, # |   -37 lol, that whole blog is shit. (a xor b) + (a xor c) = (a xor b) + (b xor c) is not true. U got (a xor b) on both sides, so basically ur saying that (a xor c) = (b xor c) which doesn't make sense. Simplest example you can prove it yourself is a = 1, b = 0, c = 1
•  » » 2 weeks ago, # ^ |   +8 something something dunning kruger
•  » » 2 weeks ago, # ^ |   +1 ah yes, (x+y)^2 = x^2 + y^2
•  » » 2 weeks ago, # ^ |   -8 I can't lie: I was gonna make this exact comment a few days ago, but I saw your comment, so I didn't end up commenting myself. I was even gonna preface my comment with a snarky little phrase, as you have.Thank God you made it before me so we all can laugh at you instead.
 » 2 weeks ago, # |   +37 It's not wrong. it's just that the blog is in the context of boolean algebra, where, in fact, a^b = a'b + ab'. Here, ' symbol represents the negation of the boolean value of the variables. In boolean algebra, a^b is true when a is false and b true OR a is true and b is false, predicate that is represented with a' and b or a and b' which is just written as a'b + ab'
•  » » 2 weeks ago, # ^ |   0 thanks u very much.
•  » » » 2 weeks ago, # ^ | ← Rev. 2 →   0 Also the definition of $+$ and $\times$ differs from our common one. In boolean algebra, “ $+$ ” basically means bitwise xor and “ $\times$ ” means bitwise and. One can verify that these operations forms a ring.