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vikram789
Joined: 01 Aug 2008
Posts: 14
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 22 Aug 2008 5:23 concept clearing question |
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a= 0110 b =1100
r=a.b
r = 0100
now the question is that we have to find b from r and a . its very simple
r = a.b
r(bar) = (a.b)(whole bar) // taking complement on both sides
r(bar) = a(bar) + b(bar) // using De-morgans law
r(bar) - a(bar) = b(bar)
OR b(bar) = r(bar) - a(bar) // solution
but this was my 2nd attempt i first tried the following way
r=a.b
r=a.b+ a(bar).a //a(bar).a = 0
r= a [ b+a(bar) ]
r.a = a.a [ b+a(bar) ]
r.a = a [ b+a(bar) ]
r =[ b+a(bar) ] // this is WRONG !!!!!!!!! u can chk br taking the values ..... explain ?????
i found the answere to this fallacy but i want to share it with my friends so they they dont commit same mistake. its easy try it
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anssprasad
Joined: 29 Jun 2007
Posts: 80
Helped: 7
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| 22 Aug 2008 9:27 Re: concept clearing question |
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| vikram789 wrote: |
a= 0110 b =1100
r=a.b
r = 0100
now the question is that we have to find b from r and a . its very simple
r = a.b
r(bar) = (a.b)(whole bar) // taking complement on both sides
r(bar) = a(bar) + b(bar) // using De-morgans law
r(bar) - a(bar) = b(bar)
OR b(bar) = r(bar) - a(bar) // solution
but this was my 2nd attempt i first tried the following way
r=a.b
r=a.b+ a(bar).a //a(bar).a = 0
r= a [ b+a(bar) ]
r.a = a.a [ b+a(bar) ]
r.a = a [ b+a(bar) ]
r =[ b+a(bar) ] // this is WRONG !!!!!!!!! u can chk br taking the values ..... explain ?????
i found the answere to this fallacy but i want to share it with my friends so they they dont commit same mistake. its easy try it |
you cant remove a from both sides since division is not defined for boolean algebra.
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vikram789
Joined: 01 Aug 2008
Posts: 14
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| 23 Aug 2008 6:55 Re: concept clearing question |
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yes it is right that subtraction & division and not allowed in Booleam algebra but there is one more fallacy. in the secord last step a.a = a is used but this is WRONG as this identity is for 1 bit only whereas here it is a nibble...
Now i more concept clearer...
1 + 1 = 1 using OR gate
while using half adder we use
1 + 1 =0 and carry =1
where is the fallacy , try it is easy ...
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laseetha
Joined: 09 May 2007
Posts: 76
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| 23 Aug 2008 14:34 concept clearing question |
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what you are asking is not boolean.
see, its binary arithmetic like the decimal or hexadecimal or octal
here the rules as you know
0+0 = 0
0+1 = 1
1+1 = 10 ( this is equivalent to 2 in decimal)
1-1 = 0
1.1 = 1
and 1+1+1 = 11 (this is equivalent to 3 )
1+1+1+1 = 100 ( 4 in decimal)
i think now you might have got some idea
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ytliang
Joined: 04 Aug 2008
Posts: 61
Helped: 5
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| 27 Aug 2008 9:50 Re: concept clearing question |
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| vikram789 wrote: |
yes it is right that subtraction & division and not allowed in Booleam algebra but there is one more fallacy. in the secord last step a.a = a is used but this is WRONG as this identity is for 1 bit only whereas here it is a nibble...
Now i more concept clearer...
1 + 1 = 1 using OR gate
while using half adder we use
1 + 1 =0 and carry =1
where is the fallacy , try it is easy ... |
OR gate is used for logic computation
Half Adder is used for arithmatic computation
They just happened to share the same operating symbol.
Also, a.a (a AND a) is still 'a' regardless of how many bits...
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