If the operation @ is defined for all integers a and b by a@b = a+b-ab

(I) and (II) are obviously correct.

For (III)

(a+b-ab)@c = (a + b - ab)@c = a + b - ab + c - c(a + b - ab) = a + b - ab + c - ac - ab + abc

a@(b@c) = a@(b + c - bc) = a + b + c - bc - a(b + c - bc) = a + b + c - bc -ab - ac + abc

Answer - E

Sorry for bumping up an old thread, I have a doubt: my approach for solving the question was to assume that the operation in this case was the union between two sets, a and b, and consequently the three points were the properties of the union of sets. Is that a correct approach or might it be too risky in the actual exam? (or is it even a wrong assumption, and I got it right out of luck?)

I. a@b = b@a
\(a+b-ab=b+a-ab\) TRUE!

II. a@0 = a
\(a+0-0 = a\) TRUE!

III. (a@b)@c = a@(b@c)
\(a+b-ab+c-ac-bc+abc = b+c-bc + a - ab-ac+abc\) STRIKE OUT DUPLICATES ON RHS and LHS! TRUE!

Answer: I,II, and III or (E)

That's not correct. Stem defines some function @ for all integers a and b by a@b=a+b-ab. For example if a=1 and b=2, then a@b=1@2=1+2-1*2.

Hope it's clear.

Been looking at this for a while and still can't figure it out.. I thought that we must ALWAYS first do the calculations in the brackets and open them, and then do the remaining calculations. as we have a@(b@c), how come do you straight come up to (a + b - ab)@c, when it's b@c in the brackets, not a@b anymore.. finding this one a bit confusing.. thanks for explaining in advance

EDIT:

OK, I think I get it now.. pls, have a look at my upload and let me know if I am correct.. this is the left side of the equation in (III). With the right one, we do the exact same thing, right?

Hey there folks, sorry to bump on an old thread. Just wondering, is there a way to evaluate statement 3 faster?
I believe this questions takes around 2 minutes and evaluating statement 3 takes a lot of work and is prone to errors.

Just wondering if I'm doing this the correct/most efficient way

Thanks guys
Cheers!

J

In this problem we have been asked to check the commutative and associative property of the given function. These properties are defined as below:

Commutative: In mathematics, a binary operation is commutative if changing the order of the operands does not change the result.

Associative: Within an expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not changed. That is, rearranging the parentheses in such an expression will not change its value.

If you're wondering if commutativity implies associativity in mathematics then the answer is NO. However, for simple addition and multiplication functions commutativity does imply associativity and hence in such cases option 3 need not be tested if option 1 is true. However, the only way to solve such problems which involve functions other than simple addition and multiplication would be to solve the expression completely as stated above.

Well, can't argue that learning the definitions is in fact quite interesting and thank you for that.
Nevertheless, I was really intereted in solving statement 3 quicker/faster/more efficient
So, being able to recognize if operations in the different order given will yield same result without having to go through all the long distribution process.

I will try to come up with a faster way but if anyone else come's up with a nice and elegant approach I'd be happy to give some nice Kudos for the collection

Cheers
J

Given: a#b=a+b-ab

(i) a#b=b#a
a#b = a+b-ab
b#a = b+a-ba --> we can rewrite this as a+b-ab
which is same as a#b. Thus it is true

(ii) a#0=a
a#0 = a+0-(a*0) = a. Thus true

(iii) (a#b)#c=a#(b#c)
(a#b)#c
Consider a#b = x. Thus (a#b)#c ==> x#c => x+c-(x*c)
Now a#b = x => a+b-ab =x. Sub in the above equation
a+b-ab+c-[(a+b-ab)*c) = a+b+c-ab -[ac+bc-abc] = a+b+c-ab-ac-bc+abc--(1)

a#(b#c)
Consider b#c =y. Thus a#(b#c) --> a#y -> a+y-ay
Now b#c=y --> b+c-bc =y. Sub in the above equation
a+b+c-bc-[a*(b+c-bc)] --> a+b+c-bc-[ab+ac-abc] --> a+b+c-bc-ab-ac+abc --(2)
(1) =(2)
Thus (a#b)#c=a#(b#c) is true.

Hence all are true

Check functions related questions in our Special Questions Directory.

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions

Solution:

I and II are true as they prove the commutative and identity property of + and *
a + b - ab + c - c(a + b - ab) = a + b - ab + c - ac - ab + abc

In III-
LHS = (a@b)@c = a + b - ab + c - c(a + b - ab)

= a + b - ab + c - ac - bc + abc

= a + b + c - bc -ab - ac + abc


RHS =a@(b@c)=a@(b + c - bc)

= a + b + c - bc - a(b + c - bc)

= a + b + c - bc -ab - ac + abc = LHS (option e)

Devmitra Sen
GMAT SME

my friend, avigutman

I would love to know how you approach tackling statement 3. I trust that you will show me the way! Unfortunately, I cannot wrap my heard around it despite the expert responses.

woohoo921 The exact placement of parentheses is inconsequential in a sum or in a product, and the operation is just the sum of two terms minus their product. So, intuitively, it shouldn't matter. If you know that algebra isn't a strength for you, then your best move here is to guess based on that intuition.
If you're going to try to prove it algebraically, consider not writing down the steps all the way to the end. Start with the parentheses in each side.
LHS: (a+b-ab)@c
RHS: a@(b+c-bc)
Then perform the @ operation for each side, and try to write down the terms in the same order.
LHS: a+b+c-ab-ac-bc+abc
RHS: a+b+c-ab-ac-bc+abc

is it possible to use the number '0' and accordingly - all these options are possible since its asking ' which is possible'

The question does not ask "which is possible", it asks "which of the following MUST be true".

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