# algebraic laws of sets

• May 25th 2010, 11:45 PM
may
algebraic laws of sets
Excuse me,how to do this type of question?

Using the laws of the algebra of sets show that

(a-b)∩(a∩b)'=a∩b'
• May 26th 2010, 05:00 AM
$(A \backslash B) \cap (A \cap B)'$

$= (A \cap B') \cap (A' \cup B')$

$= A \cap B' \cap (B' \cup A')$

$= A \cap B'$

Where we used the following rules...

$(A \backslash B) = (A \cap B')$

$(A \cap B)' = (A' \cup B')$

$B' \cap (B' \cup A') = B'$
• May 26th 2010, 05:01 AM
Wilmer
• May 26th 2010, 06:43 AM
Hello everyone

All that Deadstar says is correct, but I think the statement
$B' \cap(B'\cup A') = B'$
(although obvious) requires further proof - it's not one of the usual Set Laws (although some people do include it as an Absorption Law).

You prove it like this:
$B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$
$=B'\cup(\oslash \cap A')$

$=B'\cup\oslash$

$=B'$

• May 26th 2010, 06:45 AM
may
$(A \backslash B) = (A \cap B')$

$(A \cap B)' = (A' \cup B')$

$B' \cap (B' \cup A') = B'$[/quote]

how to know when to apply this rule?
• May 26th 2010, 06:48 AM
Hello may
Quote:

Originally Posted by may
Where we used the following rules...

$(A \backslash B) = (A \cap B')$

$(A \cap B)' = (A' \cup B')$

$B' \cap (B' \cup A') = B'$

how to know when to apply this rule?

Sorry - there's no quick answer to that - just lots of practice!

• May 26th 2010, 06:56 AM
may
Quote:

Hello everyone

All that Deadstar says is correct, but I think the statement
$B' \cap(B'\cup A') = B'$
(although obvious) requires further proof - it's not one of the usual Set Laws (although some people do include it as an Absorption Law).

You prove it like this:
$B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$
$=B'\cup(\oslash \cap A')$

$=B'\cup\oslash$

$=B'$

which formula Grandad are using to change fr

[/SIZE][INDENT] $B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$
[INDENT] $=B'\cup(\oslash \cap A')$

$=B'\cup\oslash$
[SIZE=3]
• May 26th 2010, 06:59 AM
may
Quote:

Hello maySorry - there's no quick answer to that - just lots of practice!

HOW to do this type of question?Must we memorise All the formula?
• May 26th 2010, 07:05 AM
Hello may
Quote:

Originally Posted by may
which formula grandad are using to change fr
$B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')$

$=B'\cup(\oslash \cap A')$

This uses the Distributive Law:
$(P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)$
where I have replaced $P$ by $B'$, $Q$ by $\oslash$ and $R$ by $A'$.
Quote:

Originally Posted by may
HOW to do this type of question?Must we memorise All the formula?

Yes, I'm rather afraid that you do. As I said - you'll need lots of practice!

$\oslash \cap A' = \oslash$ uses what I have always known as an Identity Law, but others sometimes refer to as a Domination Law.

See, for example, here.
• May 27th 2010, 08:11 AM
may
Quote:

Hello mayThis uses the Distributive Law:
$(P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)$
where I have replaced $P$ by $B'$, $Q$ by $\oslash$ and $R$ by $A'$.

Yes, I'm rather afraid that you do. As I said - you'll need lots of practice!

$\oslash \cap A' = \oslash$ uses what I have always known as an Identity Law, but others sometimes refer to as a Domination Law.

See, for example, here.

Oh,thanks Grandad at last i start to figure out how to do this type of question.thank you very much(Wink)
• May 27th 2010, 08:22 AM
may
Quote:

Hello mayThis uses the Distributive Law:
$(P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)$
where I have replaced $P$ by $B'$, $Q$ by $\oslash$ and $R$ by $A'$.

Yes, I'm rather afraid that you do. As I said - you'll need lots of practice!

$\oslash \cap A' = \oslash$ uses what I have always known as an Identity Law, but others sometimes refer to as a Domination Law.

See, for example, here.

Oh,thanks Grandad at last i start to figure out how to do this type of question.thank you very much(Wink)

erm i want to ask another question..same topic do i need to put the question in new thread or at here?
• May 27th 2010, 01:31 PM