# An advanced set question that I could not solve.

• Jun 26th 2010, 04:18 AM
PythagorasNeophyte
An advanced set question that I could not solve.
The question confuses me mainly in the $x^2$ part. Here's how it goes:

Given that E(universal set) ={x:x is positive even number},
A={x:0 is equal to and less than x is equal to and less than 10} and
B={x:1 is equal to and less than $x^2$ is less than 82}, list the elements of B, and A n B.

I don't how $x^2$ in group B can be used. Can you help me explain and help me in completing the question with workings? Thank you so much and have a nice day!
• Jun 26th 2010, 04:27 AM
Ackbeet
So your domain of discourse is the positive even integers. And you have

$A=\{x:0\le x\le 10\},$ and

$B=\{x:0\le x^{2}<82\}.$

This is correct? And you're asked to list the elements of $A$ and $A\cap B$, correct?

If all this is correct, what ideas do you have so far?
• Jun 26th 2010, 04:41 AM
PythagorasNeophyte
Quote:

Originally Posted by Ackbeet
So your domain of discourse is the positive even integers. And you have

$A=\{x:0\le x\le 10\},$ and

$B=\{x:0\le x^{2}<82\}.$

This is correct? And you're asked to list the elements of $A$ and $A\cap B$, correct?

If all this is correct, what ideas do you have so far?

Thank you for your response, but I am looking for the elements of $B$ and $A\cap B$. Nonetheless, I am still dubious of how the element $x^2$ in the set $B$ could be substituted, so I'm stuck in listing the elements of set $B$. I also do not know why $A=\{x:0\le x\le 10\},$ and $B=\{x:0\le x^{2}<82\}.$ both have $x:0$ rather than $x:x$.
• Jun 26th 2010, 05:28 AM
Defunkt
Quote:

Originally Posted by PythagorasNeophyte
Thank you for your response, but I am looking for the elements of $B$ and $A\cap B$. Nonetheless, I am still dubious of how the element $x^2$ in the set $B$ could be substituted, so I'm stuck in listing the elements of set $B$. I also do not know why $A=\{x:0\le x\le 10\},$ and $B=\{x:0\le x^{2}<82\}.$ both have $x:0$ rather than $x:x$.

What do you mean by
Quote:

I also do not know why .... both have $x:0$ rather than $x:x$
?

That is just a matter of notation - $\{x: \ 0 \leq x \leq 10 \}$ means the set of all $x$ such that x is greater or equal to 0 and lesser or equal to 10.
Likewise, $\{ x:0\le x^{2}<82\ \}$ is the set of all $x$ such that $x^2$ is greater or equal to 0, and lesser than 82.
• Jun 26th 2010, 10:30 AM
ragnar
Quote:

Originally Posted by PythagorasNeophyte
Thank you for your response, but I am looking for the elements of $B$ and $A\cap B$. Nonetheless, I am still dubious of how the element $x^2$ in the set $B$ could be substituted, so I'm stuck in listing the elements of set $B$. I also do not know why $A=\{x:0\le x\le 10\},$ and $B=\{x:0\le x^{2}<82\}.$ both have $x:0$ rather than $x:x$.

The following sets are the same:

{x| 0 is less than or equal to x, and x is less than or equal to 10} (This is basically what you wrote for the definition of set A.)

{x| x is greater than or equal to 0, and x is less than or equal to 10} (This way of writing it has the form "The set of x, such that x..." which you are used to, but this way of writing it says the same thing as the other way.)

$\{ x| 0 \leq x \leq 10 \}$

Given that A and B are the two sets that you're working with, and you're only considering positive even integers, listing the individual elements of each set should be very easy, so there's no need to substitute $x^{2}$. What you want to find is the set of elements which are in $A$ and which are also in $B$.