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

Name: Discrete math sets help?How would I do this question?Suppose U = { 1 , 2... , 9 }, A = all multiples of 2, B = all multiples of 3, and C = { 3 , 4 , 5 , 6 , 7 }. Find C − ( B − A ).
Uploaded: 2022-02-23T17:22:57+00:00
Description: Discrete math sets help?How would I do this question?Suppose U = { 1 , 2... , 9 }, A = all multiples of 2, B = all multiples of 3, and C = { 3 , 4 , 5 , 6 , 7 }. Find C − ( B − A ).

Question

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

Rihanna Robles · Accepted Answer

Step 1The multiples of 3 in U are 3,6, and 9, so   B  =  {  3  ,  6  ,  9  }. The multiples of 2 in U are 2,4,6, and 8, so   A  =  {  2  ,  4  ,  6  ,  8  }. 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  ∖  A  =  {  3  ,  9  }.Step 2Now see if you can apply the same sort of reasoning to find C∖{3,9}.

Lorelei Patterson · Accepted Answer

Step 1If B is the set of multiples of 3 in the universe, then the elements of B are just 3,6,9; that is,  B  =  {  3  ,  6  ,  9  }Step 2Likewise, A is the set of even numbers inside U, so   A  =  {  2  ,  4  ,  6  ,  8  }. Then   B  −  A is the set of things in B that aren&#039;t in A. We see that 3 and 9 are in B not in A, so   B  −  A  =  {  3  ,  9  }