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# 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?

### 9 Answers

- 1 decade agoFavorite Answer
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... :-)

Source(s): I am a high school mathematics teacher and a college calculus tutor. - RaymondLv 71 decade ago
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)

- Aidan ConnorLv 41 decade ago
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.

- Anonymous1 decade ago
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

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- 1 decade ago
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.

- hardageLv 44 years ago
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.

- Anonymous1 decade ago
No, they are not equal. It is an order of operations problem.

- 1 decade ago
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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- 1 decade ago
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