# Math Help - Write in the simplest way possible using the laws of set algebra

1. ## Write in the simplest way possible using the laws of set algebra

Hi, got an exercise I can't seem too solve, just started with Discrete Mathematics.
I'm supposed to write this: (A - B) (intersection) (A U B) in the simplest possible way.

Really appreciate any help

2. ## Re: Write in the simplest way possible using the laws of set algebra

Originally Posted by Remi
Hi, got an exercise I can't seem too solve, just started with Discrete Mathematics.
I'm supposed to write this: (A - B) (intersection) (A U B) in the simplest possible way.
$A\setminus B=(A\cap B^c)$, I used the 'setminus' notation and $B^c$ is the complement of $B$

Thus $(A\cap B^c)\cap (A\cup B)=[(A\cap B^c)\cap A]\cup[(A\cap B^c)\cap B]$.

You should see at once how to finish.

3. ## Re: Write in the simplest way possible using the laws of set algebra

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

But I don't quite get what you did in the second part there: $(A\cap B^c)\cap (A\cup B)=[(A\cap B^c)\cap A]\cup[(A\cap B^c)\cap B]$
Which law did you use ?

4. ## Re: Write in the simplest way possible using the laws of set algebra

Originally Posted by Remi
$= A \cup (A \cap B) = A$ ?

But I don't quite get what you did in the second part there: $(A\cap B^c)\cap (A\cup B)=[(A\cap B^c)\cap A]\cup[(A\cap B^c)\cap B]$
Which law did you use ?
$C\cap(A\cup B)=(C\cap A)\cup(C\cap B)$ that is distribution.
Let $C=(A\cap B^c)$

5. ## Re: Write in the simplest way possible using the laws of set algebra

Thanks, think I got it now Was my solution correct ?
And could it be done like this:
$[(A\cap B^c)\cap A]\cup[(A\cap B^c)\cap B] = A \cup [(A\cap B^c)\cap B] = A \cup \O = A$

Given that $(A \cap B^c) \cap B = A \cap (B^c \cap B) = A \cap \O = \O$

6. ## Re: Write in the simplest way possible using the laws of set algebra

Originally Posted by Remi
Thanks, think I got it now Was my solution correct ?
And could it be done like this:
$[(A\cap B^c)\cap A]\cup[(A\cap B^c)\cap B] = A \cup [(A\cap B^c)\cap B] = A \cup \O = A$

Given that $(A \cap B^c) \cap B = A \cap (B^c \cap B) = A \cap \O = \O$
You missed one part.
$(A\cap B^c)\cap A=A\cap B^c$