## Dhaka Higher Secondary 2011/3

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**don't post problems (by starting a topic)**in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.

### Dhaka Higher Secondary 2011/3

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

### Re: Dhaka Higher Secondary 2011/3

Hint:

Every logical solution to a problem has its own beauty.

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**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)### Re: Dhaka Higher Secondary 2011/3

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

### Re: Dhaka Higher Secondary 2011/3

$B=\{\phi\}$ isn't it?

**A man is not finished when he's defeated, he's finished when he quits.**

### Re: Dhaka Higher Secondary 2011/3

Remember set of empty set can be inside a set.

"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please

Please

**install LaTeX fonts**in your PC for better looking equations,**learn****how to write equations**, and**don't forget**to read Forum Guide and Rules.### Re: Dhaka Higher Secondary 2011/3

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

You should explain your answers ...

You should explain your answers ...

Every logical solution to a problem has its own beauty.

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**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)### Re: Dhaka Higher Secondary 2011/3

Oh shit! I haven't saw the braccet around $\phi$ i saw $A=\phi$ ok ok okz! Now, trying again.

**A man is not finished when he's defeated, he's finished when he quits.**

- leonardo shawon
**Posts:**169**Joined:**Sat Jan 01, 2011 4:59 pm**Location:**Dhaka

### Re: Dhaka Higher Secondary 2011/3

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

$P(A)=\{\phi,empty set\}$

$A \cup B=\{\phi,empty set\}$ ,

$A \cap B=\phi$

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.

BRAC University.

*long way to go .....*### Re: Dhaka Higher Secondary 2011/3

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.Every logical solution to a problem has its own beauty.

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**Important**: Please make sure that you have read about the**Rules, Posting Permissions and Forum Language**)### Re: Dhaka Higher Secondary 2011/3

$A=\{\emptyset\}$

$A \cup B=P(A)$

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?

$A \cup B=P(A)$

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?

**A man is not finished when he's defeated, he's finished when he quits.**