Can anyone find (A - B) - C and A - (B - C). Are these sets equal?

Can anyone find (A - B) - C and A - (B - C). Are these sets equal?

Can anyone find (A - B) - C and A - (B - C). Are these sets equal?
These are not congruent.





Consider the distributive property...


(A - B) - C = A - B - C


A - (B - C) = A - B + C





I'm trying to think of an example of sets... but I am wondering what your context is... what are you working on?





In an elementary mathematics situation...


consider A = 10, B = 5 and C = 2


(A - B) - C = (10 - 5) - 2 = 5 - 2 = 3


and


A - B - C = 10 - 5 - 2 = 5 - 2 = 3





A - (B - C) = 10 - (5 - 2) = 10 - 3 = 7


and


A - B + C = 10 - 5 + 2 = 5 + 2 = 7





I'm not sure if this helps, if can elaborate on the context I might be able to shed more light onto the subject... :-)
Reply:No, they are not equal. It is an order of operations problem.
Reply:No , because The associative property of real numbers which means ( Changing the grouping does not affect the result ) is not applied on subtraction


Just addition and Multiplication
Reply:Let m be an element of Sets A and C, but not of set B.





Then m cannot belong to (A-B)-C:


m belongs to A and m does not belong to B, means that m belongs to (A-B). However, because m also belongs to C, it is removed when performing the -C operation.





However, m belongs to A - (B-C):





m belongs to C but not to B, therefore, m does not belong to (B-C).


m belongs to A and it is not removed when performing the - (B-C) operation.





Therefore:


m belongs to A - (B - C),


but


m does not belong to (A - B) - C





-----





Let A be the set of all positive integers from 1 to 9


Let B be the odd subset of A.


Let C be the even subset of A.


Let m=2 (or any other even element of A)





A={1, 2, 3, 4, 5, 6, 7, 8, 9}


B={1, 3, 5, 7, 9}


C={2, 4, 6, 8}





(A-B) = {2, 4, 6, 8}


(A-B)-C = {2, 4, 6, 8} - {2, 4, 6, 8} = empty set


(m does not belong to the empty set)





(B-C) = {1, 3, 5, 7, 9}


because no elements of C are contained in B, subtracting C from B has no effect on B; (B-C) = B


A-(B-C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 3, 5, 7, 9} =


A-(B-C) = {2, 4, 6, 8} = C


C is not equal to the empty set


(m belongs to C)
Reply:A-B = A%26amp;B'


(A-B)-C = (A%26amp;B')-C=(A%26amp;B')%26amp;C' = A%26amp;B'%26amp;C'





A-(B-C) = A%26amp;(B-C)' = A%26amp;(B%26amp;C')' = A%26amp;(B'||C) = (A%26amp;B'||A%26amp;C)





SO THEY ARE NOT EQUAL
Reply:no they arent.


the parentheses changes everything.





see i will use the nubers 7(a) 6(b) and 5(c) as an example.





(a-b)-c


(7-6)-5


you always have to do parentheses first.


soo 7-6 is 1.


you can take the parentheses away.


1-5= -4





the next one...


a-(b-c)


7-(6-5)


once again...parentheses first.


soo 6-5 is 1.


7-1 is 6.





see? perentheses does count.


the first one was -4 and the second one was 6.





hope this helped.


sometimes when i explain math, no one gets me.
Reply:No, they're not equal


Say A is 1, B is 2, C is 3. The first problem would go like this


(1 - 2) - 3. You'd do the problem in the parenthesis first, you'd get -1, then -3 more would be -4.





The second equation, would be


1 - (2 - 3)


Again, parenthesis first.. you'd get -1, then minus the 1 on the outside (A) would be -2. Different answers.
Reply:They are not equal.


when we open (A-B)-C it is A-B-C


but when we open A-(B-C) it is A-B+C


minus sign has to be multiplied with +B AND -C, which are in the bracket, and which will become -B+C.


WANT TUTIONS IN MATHEMATICS, TELL. IT WILL ALL BE FREE FOR TWO HOURS EVERY SATURDAY ON NET.

elephant ear