 valtricotinevh

2022-07-15

Discrete math sets help?
How would I do this question?
Suppose $U=\left\{1,2...,9\right\}$, $A=$ all multiples of 2, $B=$ all multiples of 3, and $C=\left\{3,4,5,6,7\right\}$. Find $C-\left(B-A\right)$. Rihanna Robles

Expert

Step 1
The multiples of 3 in U are 3,6, and 9, so $B=\left\{3,6,9\right\}$. The multiples of 2 in U are 2,4,6, and 8, so $A=\left\{2,4,6,8\right\}$. For any sets X and Y, the set difference X∖Y is the set of objects that are in X but not Y. Thus, B∖A is the set of digits that are in B but not in A. Look at the members of B: 3 and 9 are in B but not A, so they’re in B∖A. 6, on the other hand, is in B and A, so it’s not in B∖A. And nothing else is in B, never mind whether it’s in A or not, so nothing else is in B∖A. Thus, $B\setminus A=\left\{3,9\right\}$.
Step 2
Now see if you can apply the same sort of reasoning to find C∖{3,9}. Lorelei Patterson

Expert

Step 1
If B is the set of multiples of 3 in the universe, then the elements of B are just 3,6,9; that is,
$B=\left\{3,6,9\right\}$
Step 2
Likewise, A is the set of even numbers inside U, so $A=\left\{2,4,6,8\right\}$. Then $B-A$ is the set of things in B that aren't in A. We see that 3 and 9 are in B not in A, so $B-A=\left\{3,9\right\}$

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