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If the operation is defined for all integers a and b [#permalink]

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18 Sep 2010, 20:29

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cmugeria wrote:

If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a II. a@0 = a III. (a@b)@c = a@(b@c)

(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II and III

We have that: \(a@b=a+b-ab\)

I. \(a@b = b@a\) --> \(a@b=a+b-ab\) and \(b@a=b+a-ab\) --> \(a+b-ab=b+a-ab\), results match;

II. \(a@0 = a\) --> \(a@0=a+0-a*0=0\) --> \(0=0\), results match;

III. \((a@b)@c = a@(b@c)\) --> \((a@b)@c=a@b+c-(a@b)*c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc\) and \(a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab-ac+abc\), results match.

Re: If the operation @ is defined for all integers a and b [#permalink]

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15 Jan 2012, 17:17

I was careless in writing the equation for option 3, else it was easy. I got it wrong but once you write it you clearly see that C also satifies the equation.

Re: If the operation @ is defined for all integers a and b [#permalink]

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18 Dec 2012, 08:45

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?)

Re: If the operation @ is defined for all integers a and b [#permalink]

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19 Dec 2012, 03:24

Expert's post

fguardini1 wrote:

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?)

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.

Re: If the operation @ is defined for all integers a and b [#permalink]

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28 Jan 2013, 22:58

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subhashghosh wrote:

(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

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?

Re: If the operation @ is defined for all integers a and b [#permalink]

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23 Oct 2013, 13:34

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

Re: If the operation @ is defined for all integers a and b [#permalink]

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27 Oct 2013, 14:13

Would it be smart to pick numbers for each of the variables? I selected 1&3 only since I got mixed up with the letter variables. Is picking number the most efficient way to approach this problem?

Re: If the operation @ is defined for all integers a and b [#permalink]

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28 Oct 2013, 12:40

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jlgdr wrote:

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.

Re: If the operation @ is defined for all integers a and b [#permalink]

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31 Oct 2013, 05:39

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

Re: If the operation @ is defined for all integers a and b [#permalink]

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31 Oct 2013, 05:55

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jlgdr wrote:

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

You can always test values that's an alternative

Let a=1,b=2 and c=3

Definition => a@b=a+b-ab

Option 3

III. (a@b)@c = a@(b@c)

a@b = 1 + 2 -2 = 1 1@3 = 1+3 -3 = 1

B@c = 2@3 = 5-6 = -1 1@-1 = 1 - 1 - ( -1 * 1) =1

LHS = RHS _________________

Click +1 Kudos if my post helped...

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Re: If the operation @ is defined for all integers a and b [#permalink]

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01 Mar 2014, 16:08

Just took this question today, and I was also wondering if there were a way to solve it more quickly than performing the heavy manipulations that are in III or guessing numbers.

Re: If the operation @ is defined for all integers a and b [#permalink]

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13 Jul 2014, 22:50

Bunuel wrote:

cmugeria wrote:

If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a II. a@0 = a III. (a@b)@c = a@(b@c)

(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II and III

We have that: \(a@b=a+b-ab\)

I. \(a@b = b@a\) --> \(a@b=a+b-ab\) and \(b@a=b+a-ab\) --> \(a+b-ab=b+a-ab\), results match;

II. \(a@0 = a\) --> \(a@0=a+0-a*0=0\) --> \(0=0\), results match;

III. \((a@b)@c = a@(b@c)\) --> \((a@b)@c=a@b+c-(a@b)*c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc\) and \(a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab+abc\), results match.

Re: If the operation @ is defined for all integers a and b [#permalink]

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14 Jul 2014, 03:07

Expert's post

PathFinder007 wrote:

Bunuel wrote:

cmugeria wrote:

If the operation @ is defined for all integers a and b by a@b=a+b-ab, which of the following statements must be true for all integers a, b and c?

I. a@b = b@a II. a@0 = a III. (a@b)@c = a@(b@c)

(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II and III

We have that: \(a@b=a+b-ab\)

I. \(a@b = b@a\) --> \(a@b=a+b-ab\) and \(b@a=b+a-ab\) --> \(a+b-ab=b+a-ab\), results match;

II. \(a@0 = a\) --> \(a@0=a+0-a*0=0\) --> \(0=0\), results match;

III. \((a@b)@c = a@(b@c)\) --> \((a@b)@c=a@b+c-(a@b)*c=(a+b-ab)+c-(a+b-ab)c=a+b+c-ab-ac-bc+abc\) and \(a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab+abc\), results match.

Re: If the operation is defined for all integers a and b [#permalink]

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20 Jan 2015, 23:41

I have a question regarding this problem. Don't we have to test for negative numbers while expanding the equation? This is where I got stuck and it became time consuming for me at which point I guessed and moved on. Please advise why that's not a factor.

Re: If the operation is defined for all integers a and b [#permalink]

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26 Jul 2015, 07:16

Yes, tested for negative values and got stuck. For a@0 = a. If we put a is negative then it doesn't hold. Any thoughts on this? Where am I going wrong?

Re: If the operation is defined for all integers a and b [#permalink]

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26 Jul 2015, 07:23

Expert's post

Yoshit wrote:

Yes, tested for negative values and got stuck. For a@0 = a. If we put a is negative then it doesn't hold. Any thoughts on this? Where am I going wrong?

It does: lets say a=-4 ---> a@0=-4@0 = -4+0-0*-4 = -4 =a. This HAS to be true even by the underlying algebra!

Thus satisfies the given equation.

Can you show your steps for calculations? _________________

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