algebraic laws of sets

may

May 2010
13
0
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'
 
Oct 2007
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\(\displaystyle (A \backslash B) \cap (A \cap B)'\)

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

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

\(\displaystyle = A \cap B'\)

Where we used the following rules...

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

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

\(\displaystyle B' \cap (B' \cup A') = B'\)
 
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Grandad

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Dec 2008
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Hello everyone

All that Deadstar says is correct, but I think the statement
\(\displaystyle B' \cap(B'\cup A') = B'\)
(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:
\(\displaystyle B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')\)
\(\displaystyle =B'\cup(\oslash \cap A')\)

\(\displaystyle =B'\cup\oslash\)


\(\displaystyle =B'\)

Grandad
 
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may

May 2010
13
0
\(\displaystyle (A \backslash B) = (A \cap B')\)

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

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

how to know when to apply this rule?
 

Grandad

MHF Hall of Honor
Dec 2008
2,570
1,416
South Coast of England
Hello may
Where we used the following rules...

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

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

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

how to know when to apply this rule?
Sorry - there's no quick answer to that - just lots of practice!

Grandad
 
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may

May 2010
13
0
Hello everyone

All that Deadstar says is correct, but I think the statement
\(\displaystyle B' \cap(B'\cup A') = B'\)
(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:
\(\displaystyle B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')\)
\(\displaystyle =B'\cup(\oslash \cap A')\)

\(\displaystyle =B'\cup\oslash\)


\(\displaystyle =B'\)

Grandad
which formula Grandad are using to change fr

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

\(\displaystyle =B'\cup\oslash\)
 
Last edited:

may

May 2010
13
0
Hello maySorry - there's no quick answer to that - just lots of practice!

Grandad
HOW to do this type of question?Must we memorise All the formula?
 

Grandad

MHF Hall of Honor
Dec 2008
2,570
1,416
South Coast of England
Hello may
which formula grandad are using to change fr
\(\displaystyle B' \cap(B'\cup A') = (B' \cup \oslash)\cap(B'\cup A')\)

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

Grandad

PS ... and to answer your extra question, following your edit:

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

See, for example, here.
 
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may

May 2010
13
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Hello mayThis uses the Distributive Law:
\(\displaystyle (P \cup Q)\cap (P \cup R) = P \cup (Q \cap R)\)
where I have replaced \(\displaystyle P\) by \(\displaystyle B'\), \(\displaystyle Q\) by \(\displaystyle \oslash\) and \(\displaystyle R\) by \(\displaystyle A'\).

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

Grandad

PS ... and to answer your extra question, following your edit:

\(\displaystyle \oslash \cap A' = \oslash\) 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)