Dhaka Higher Secondary 2011/3

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Higher Secondary 2011/3

Unread post by BdMO » Fri Jan 28, 2011 10:25 pm

If $A = \{\Phi\}$, $A \cup B = P(A)$ and $A \cap B = \Phi$, find $B$. $\Phi$ represents empty set.

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Re: Dhaka Higher Secondary 2011/3

Unread post by Zzzz » Sat Jan 29, 2011 7:51 am

Hint:
A set can be element of a set ;)
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Re: Dhaka Higher Secondary 2011/3

Unread post by TTowsif » Sat Jan 29, 2011 9:00 am

I think the ans is empty set, isn't it?

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Re: Dhaka Higher Secondary 2011/3

Unread post by Hasib » Sat Jan 29, 2011 9:29 am

$B=\{\phi\}$ isn't it?
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Re: Dhaka Higher Secondary 2011/3

Unread post by Moon » Sat Jan 29, 2011 9:37 am

Remember set of empty set can be inside a set. ;)
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Re: Dhaka Higher Secondary 2011/3

Unread post by Zzzz » Sat Jan 29, 2011 9:44 am

Ah.. guys are getting closer. One said $B=\Phi$, another said $B=\{\Phi\}$. What's next?

You should explain your answers ...
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Re: Dhaka Higher Secondary 2011/3

Unread post by Hasib » Sat Jan 29, 2011 9:15 pm

Oh shit! I haven't saw the braccet around $\phi$ i saw $A=\phi$ :cry: :? ok ok okz! Now, trying again.
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Re: Dhaka Higher Secondary 2011/3

Unread post by leonardo shawon » Sat Jan 29, 2011 10:31 pm

here $A= \{\phi\}$
$P(A)=\{\phi,empty set\}$
$A \cup B=\{\phi,empty set\}$ ,
$A \cap B=\phi$
if we put the value of set "A" then we can see that set "B" is an empty set . That means. . .
$\{\phi,empty set\} \cap empty set = \phi$
so,,$ B= \phi $


Did i do any mistake....? :(

*** EDITED
Last edited by leonardo shawon on Sun Jan 30, 2011 11:06 am, edited 1 time in total.
Ibtehaz Shawon
BRAC University.

long way to go .....

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Re: Dhaka Higher Secondary 2011/3

Unread post by Zzzz » Sun Jan 30, 2011 7:31 am

Hey . $\Phi $ is a set. $A=\{\Phi\}$ is not an empty set. It has an element which is $\Phi$. If they said $A=\Phi$ then we could say $A$ is an empty set.
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Re: Dhaka Higher Secondary 2011/3

Unread post by Hasib » Sun Jan 30, 2011 8:46 pm

$A=\{\emptyset\}$
$A \cup B=P(A)$
now, $P(A)=\{\{\emptyset\},\emptyset\}$
so,
$\underline{\{\emptyset\}} \cup B=\{\{\emptyset\},\underline{\emptyset\}}$

the underlined are same. And $A \cap B=\emptyset$
So, $B=\{\{\emptyset\}\}$

ri8? :?
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