# Math Help - sets and relations

1. ## sets and relations

Let A = {x belongs to Z absolute value -1 less than or equal to x less than or equal to 2}, B={2x - 3 absolute value x element of A} C={x element of R absolute value x = a over b, a element A, b element of B}

2. Originally Posted by tygracen
Let A = {x belongs to Z absolute value -1 less than or equal to x less than or equal to 2}, B={2x - 3 absolute value x element of A} C={x element of R absolute value x = a over b, a element A, b element of B}
first, i think you mean "such that" or | or : as opposed to "absolute value"

secondly, what's your question? you just defined 3 sets, what are we supposed to do with them? ...

3. ## sets and relations

Originally Posted by Jhevon
first, i think you mean "such that" or | or : as opposed to "absolute value"

secondly, what's your question? you just defined 3 sets, what are we supposed to do with them? ...
Sorry, I am not good with this stuff at all!!!!!!!!
It is "such that"
list the elements of A, B, and C
list the elements of (A union B) X B
list the elements of B\C
list the elements of A and the symmetric difference of C

4. lets work through these. remember the notations used, what they mean and how sets are defined
Originally Posted by tygracen
Sorry, I am not good with this stuff at all!!!!!!!!
It is "such that"
list the elements of A, B, and C
ok, so $A = \{ x \in \mathbb{Z} \mid -1 \le x \le 2 \} = \{ \text{The set of all integers between -1 and 2 inclusive} \}$

once you are able to interpret what the symbols mean, it is not so hard, right? now, lets write out the set

$A = \{ -1, 0, 1, 2 \}$

and $B = \{ 2x - 3 \mid x \in A \}$

that means we evaluate the expression 2x - 3 for all the elements of A

so, not to confuse you, but say $f(x) = 2x - 3$

then, $B = \{ f(-1), f(0), f(1), f(2) \}$

so, $B = \{ -5, -3, -1, 1 \}$

now, finally, $C = \left \{ x \in \mathbb{R} ~ \bigg| ~x = \frac ab, ~a \in A, ~b \in B \right \}$

so, you take each element of A and divide it by all the elements of B, one by one until you exhaust the elements of A. thus,

$C = \left \{ \frac {-1}{-5}, \frac {-1}{-3}, \frac {-1}{-1}, \frac {-1}{1}, 0, \frac {1}{-5}, \frac {1}{-3}, \frac {1}{-1}, \frac {1}{1}, \frac {2}{-5}, \frac {2}{-3}, \frac {2}{-1}, \frac {2}{1} \right \}$

$\Rightarrow C = \left \{ \frac 15, \frac 13, -1, 0, 1, - \frac 15, - \frac 13 , - \frac 25, - \frac 23, -2, 2 \right \}$

and you can write those in order if you like

Now, can you do the questions? i will remind you of the definitions

list the elements of (A union B) X B
list the elements of B\C
list the elements of A and the symmetric difference of C
Let $X$ and $Y$ be sets. Then,

$X \cup Y = \{ x \mid x \in X \mbox{ or } x \in Y \}$

$X \times Y = \{ (x,y) \mid x \in X,~ y \in Y \}$

$X \backslash Y = \{x \mid x \in X \mbox{ and } x \notin Y \}$

$X \Delta Y = (X \backslash Y) \cup (Y \backslash X)$ (symmetric difference)

5. ## sets and relations

Originally Posted by Jhevon
lets work through these. remember the notations used, what they mean and how sets are defined
ok, so $A = \{ x \in \mathbb{Z} \mid -1 \le x \le 2 \} = \{ \text{The set of all integers between -1 and 2 inclusive} \}$

once you are able to interpret what the symbols mean, it is not so hard, right? now, lets write out the set

$A = \{ -1, 0, 1, 2 \}$

and $B = \{ 2x - 3 \mid x \in A \}$

that means we evaluate the expression 2x - 3 for all the elements of A

so, not to confuse you, but say $f(x) = 2x - 3$

then, $B = \{ f(-1), f(0), f(1), f(2) \}$

so, $B = \{ -5, -3, -1, 1 \}$

now, finally, $C = \left \{ x \in \mathbb{R} ~ \bigg| ~x = \frac ab, ~a \in A, ~b \in B \right \}$

so, you take each element of A and divide it by all the elements of B, one by one until you exhaust the elements of A. thus,

$C = \left \{ \frac {-1}{-5}, \frac {-1}{-3}, \frac {-1}{-1}, \frac {-1}{1}, 0, \frac {1}{-5}, \frac {1}{-3}, \frac {1}{-1}, \frac {1}{1}, \frac {2}{-5}, \frac {2}{-3}, \frac {2}{-1}, \frac {2}{1} \right \}$

$\Rightarrow C = \left \{ \frac 15, \frac 13, -1, 0, 1, - \frac 15, - \frac 13 , - \frac 25, - \frac 23, -2, 2 \right \}$

and you can write those in order if you like

Now, can you do the questions? i will remind you of the definitions

Let $X$ and $Y$ be sets. Then,

$X \cup Y = \{ x \mid x \in X \mbox{ or } x \in Y \}$

$X \times Y = \{ (x,y) \mid x \in X,~ y \in Y \}$

$X \backslash Y = \{x \mid x \in X \mbox{ and } x \notin Y \}$

$X \Delta Y = (X \backslash Y) \cup (Y \backslash X)$ (symmetric difference)
OK, that makes more sense, now let's see if I am on the right track:
(A U B) = {-1,0,1,2,-3,-5} (A U B) X B = {(-5,-1), (-3,0), (-1,1), (1,2)}
B\C = (0,2)
A sym. dif C = {1/5, 1/3, -1/5, 1/3, 2/5, 2/3, -2}
Is this anywhere close to right?
Thanks so much for your help!

6. Originally Posted by tygracen
OK, that makes more sense, now let's see if I am on the right track:
(A U B) = {-1,0,1,2,-3,-5}
yes

(A U B) X B = {(-5,-1), (-3,0), (-1,1), (1,2)}
nope

the first elements in the pairs should be the elements of the first set in the product, while the elements of the second set are the second elements in the pair.

example: X = {1,2,3} and Y = {a,b,c}

then X x Y = {(1,a), (1,b), (1,3), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c)}

B\C = (0,2)
no, B\C is a set, not an ordered pair. it is the set containing all elements of B that are not in C. so you take the set B and if there are any elements in there that are also in C, you throw them out.

A sym. dif C = {1/5, 1/3, -1/5, 1/3, 2/5, 2/3, -2}
Is this anywhere close to right?
nope. try this one again, using the new insight you gained from the last two

you may want to take a look here. they have definitions and examples. Note that "\" is the sybol for set difference. so A\B is the same as A - B. different notations for the same thing