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

Question

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

Answer 1

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... :-)

Answer 2

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)

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

those could do it: A = {a million,3,5} B = {a million,3,9} C = {a million,3,7} A ? B = {a million,3} - meaning A and B share aspects a million and 3 A ? C = {a million,3} - meaning A and C share aspects a million and 3 yet needless to say B and C at the instant are not the comparable.

Answer 7

No, they are not equal. It is an order of operations problem.

Answer 8

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.
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Answer 9

A-B = A&B'
(A-B)-C = (A&B')-C=(A&B')&C' = A&B'&C'

A-(B-C) = A&(B-C)' = A&(B&C')' = A&(B'||C) = (A&B'||A&C)

SO THEY ARE NOT EQUAL