ambreenjava Posted September 15, 2003 Share Posted September 15, 2003 Just to clear a confusion. Among the identities, there is one x + xy = (x+y)(x+z) I cannot understand it. Quote Link to comment Share on other sites More sharing options...
rafi_dery Posted September 15, 2003 Share Posted September 15, 2003 Neither do I. take x=F y=T z=T x + xy = F (x+y)(x+z) = T Any typos in the identity? where did you get it from? Quote Link to comment Share on other sites More sharing options...
wood Posted September 15, 2003 Share Posted September 15, 2003 ambreenjava, using Distributive Laws, x + xy should be the same as (x+x)(x+y), which is in turn x(x+y). Are you sure you copied it right? Quote Link to comment Share on other sites More sharing options...
karteek Posted September 16, 2003 Share Posted September 16, 2003 Hi all... i guess it should be x + yz = (x+y)(x+z). the easiest way to check this will be by constructing a truth table. regards karteek Quote Link to comment Share on other sites More sharing options...
wood Posted September 16, 2003 Share Posted September 16, 2003 I believe you're right karteek. That's what ambreenjava must have meant! Quote Link to comment Share on other sites More sharing options...
ambreenjava Posted September 16, 2003 Author Share Posted September 16, 2003 I read it in Computer System Architecture By M.Morris Mano (3rd Ed.) Quote Link to comment Share on other sites More sharing options...
crackit Posted September 16, 2003 Share Posted September 16, 2003 hi karteek, it is x+yz=(x+y)(x+z) (x+y)(x+z)=x+xz+yx+yz =x(1+z)+yx+yz =x+xy+yz =x(1+y)+yz =x+yz hope it is useful Quote Link to comment Share on other sites More sharing options...
karteek Posted September 19, 2003 Share Posted September 19, 2003 Originally posted by crackit hi karteek, it is x+yz=(x+y)(x+z) (x+y)(x+z)=x+xz+yx+yz =x(1+z)+yx+yz =x+xy+yz =x(1+y)+yz =x+yz hope it is useful very elegant solution.... Much better than the boring truth table stuff!! karteek Quote Link to comment Share on other sites More sharing options...
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