- #1
ZeroPivot
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Homework Statement
xy + compliment(xy) = 1
Homework Equations
The Attempt at a Solution
is it true? because x+compliment(x) = 1
maybe its not true...
Boolean Algebra Homework: Solving xy + compliment(xy) = 1
- Thread starter ZeroPivot
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Discussion Overview
The discussion revolves around the Boolean algebra expression xy + compliment(xy) = 1, exploring its validity and implications. Participants engage in a homework-related inquiry, examining the properties of Boolean operations and their relationships.
Discussion Character
- Homework-related
- Debate/contested
- Mathematical reasoning
Main Points Raised
- One participant questions the truth of the expression, suggesting that it may not hold true based on the identity x + compliment(x) = 1.
- Another participant agrees with the identity but notes that compliment(xy) does not equal compliment(x) compliment(y).
- A participant provides a counterexample using x=1 and y=0, arguing that this leads to 0 + 0, which does not equal 1.
- Another participant corrects the previous claim, stating that compliment(xy) actually equals 1 in that scenario.
- Further clarification is offered regarding the expression, with a participant suggesting that the intended question might be whether xy + compliment(x) compliment(y) equals 1, to which another participant responds negatively.
- It is noted that if xy + compliment(x) compliment(y) were true, then compliment(xy) would equal compliment(x) + compliment(y), which is not the case.
Areas of Agreement / Disagreement
Participants express differing views on the validity of the original expression and the implications of Boolean identities. There is no consensus reached regarding the correctness of the claims made.
Contextual Notes
Participants rely on specific Boolean identities and properties, but there are unresolved assumptions about the definitions and interpretations of the terms used in the expressions.
xy + compliment(xy) = 1
is it true? because x+compliment(x) = 1
maybe its not true...
- #2
CompuChip
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Yes. Take ##x + \overline{x} = 1## and rename x to a: ##a + \overline{a} = 1##.
Now set ##a = xy##.
Note, however, that ##\overline{xy} \neq \overline{x} \overline{y}##.
Now set ##a = xy##.
Note, however, that ##\overline{xy} \neq \overline{x} \overline{y}##.
- #3
ZeroPivot
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CompuChip said:
what if x=1 and y=0 the xy=0 and compliment(xy)=0 then 0+0 != 1
- #4
tiny-tim
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Hi ZeroPivot! 
(guys, thanks for the compliments, but it's complements!
)
No, complement(xy) = 1
(guys, thanks for the compliments, but it's complements!
)ZeroPivot said:
No, complement(xy) = 1
- #5
ZeroPivot
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tiny-tim said:Hi ZeroPivot!
(guys, thanks for the compliments, but it's complements!
No, complement(xy) = 1
i meant compliment(x)compliment(y) = 0
but thanks.
- #6
tiny-tim
Science Advisor
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Hi ZeroPivot!
So you meant, is ##xy + \bar{x}\bar{y} = 1## ?
No.
ZeroPivot said:
ZeroPivot said:
So you meant, is ##xy + \bar{x}\bar{y} = 1## ?
No.
- #7
CompuChip
Science Advisor
Homework Helper
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tiny-tim said:Hi ZeroPivot!
So you meant, is ##xy + \bar{x}\bar{y} = 1## ?
No.
Because if it were, ##\overline{xy}## would equal ##\bar x \bar y##. In fact it equals ##\bar x + \bar y##.
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