_Bunny's blog

formula of a ^ b
By _Bunny, history, 5 months ago, In English

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).

link: https://www.homeworklib.com/qaa/1383353/2prove-that-a-xor-b-a-xor-c-a-xor-b-b-xor

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5 months ago, # |
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Auto comment: topic has been updated by _Bunny (previous revision, new revision, compare).

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5 months ago, # |
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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

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    5 months ago, # ^ |
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    something something dunning kruger

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    5 months ago, # ^ |
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    ah yes, (x+y)^2 = x^2 + y^2

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    5 months ago, # ^ |
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    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.

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5 months ago, # |
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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'

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    5 months ago, # ^ |
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    thanks u very much.

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      5 months ago, # ^ |
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      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.

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