- #1

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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...

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- Thread starter ZeroPivot
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- #1

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xy + compliment(xy) = 1

is it true? because x+compliment(x) = 1

maybe its not true...

- #2

CompuChip

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Now set ##a = xy##.

Note, however, that ##\overline{xy} \neq \overline{x} \overline{y}##.

- #3

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Now set ##a = xy##.

Note, however, that ##\overline{xy} \neq \overline{x} \overline{y}##.

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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(guys, thanks for the compl

what if x=1 and y=0 the xy=0 and compliment(xy)=0 …

No, compl

- #5

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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

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xy + compliment(xy) = 1

…

is it true? because x+compliment(x) = 1

maybe its not true...

i meant compliment(x)compliment(y) = 0

but thanks.

So you meant, is ##xy + \bar{x}\bar{y} = 1## ?

No.

- #7

CompuChip

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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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