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  • Boolean Algebra problems

    SheldonCooper

    Member

    Updated: Oct 8, 2024
    Views: 3.4K
    1) If z=x*y then z*x is?
    Options-
    1)x
    2)0
    3)y
    4)1

    2) How many functions are possible with 'n' binary variables?

    3) How many functions are possible with 2 Boolean variables?

    Don't give only answers.I need explanations too.
    0
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Replies

  • nareshkumar6539

    Member • Nov 21, 2011

    1)If z=x*y then z*x is? if * is EX-OR operation then answer is y.because when both the symbols are same then it will give 0.
    0 EX-OR with Y gives y.(EX-OR follows associative law so we can change the order of input)
    2)with n variables if we construct a truth table then it will contain 2^n entries(rows).In a function that entry may be include or may not be include(we have 2 possibilities)total no.of functions possible with n variables is 2^2^n.
    3) if we apply n=2 in the above formula you will get answer 16.
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  • SheldonCooper

    Member • Nov 22, 2011

    nareshkumar6539
    1)If z=x*y then z*x is? if * is EX-OR operation then answer is y.because when both the symbols are same then it will give 0.
    0 EX-OR with Y gives y.(EX-OR follows associative law so we can change the order of input)
    But how you considered x=0? Is it an assumption? Then what if I take y=0?
    By the way I like the answer of second question.thanks.
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  • nareshkumar6539

    Member • Nov 22, 2011

    I am not assuming x=0. EX-OR follows the associate law means even if we change the order of operation result will be never change that i am using there
    z=x*y then z*x=?
    so i replaced z value in the euation
    x*y*x=>(x*x)*y[i apllied associate law ]
    EX-OR gives output 1 when both inputs are same ,it gives output 0 when both inputs are same x*x so it will give as output 0
    0*y the output of this expression depends on y because
    y=0 output is0
    y=0 output is1
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