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If the operation @ is defined for all integers a and b

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Re: If operation # for all integers is defined as a#b=a+b-ab, which of the [#permalink]

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New post 02 Aug 2015, 03:18
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visram04 wrote:
If operation # for all integers is defined as a#b=a+b-ab, which of the following is must be ture for all integers a,b and c?
i) a#b=b#a
ii) a#0=a
iii) (a#b)#c=a#(b#c)

Answer is all 3. Can someone help me understand i) and iii)?



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

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New post 25 Dec 2015, 14:32
I apologize for the rudimentary questions here, but can you break down the strategy for solving this problem? I always get confused with custom character problems. What are we trying to prove here?
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Re: If the operation is defined for all integers a and b [#permalink]

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New post 28 Mar 2017, 10:20
I just picked random positive integers and started testing right away, all 3 worked for me

Though time consuming, but one can arrive pretty quickly at 1,2. 3 took much more time
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If the operation is defined for all integers a and b [#permalink]

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New post 16 Jul 2017, 10:53
While Brunnel provides terrific answers the issues are that he sometimes assumes we have his level of knowledge of math, which we don't. Here is a more simple version of what he did in answer choice (C) which explains the steps.

Essentially, for operator questions one variable becomes another expression. So (A operator B) Operator C means A operator B becomes A and C becomes B. Hopefully, the logic flows through below.
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Re: If the operation @ is defined for all integers a and b [#permalink]

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New post 07 Feb 2018, 08:51
Bunuel wrote:
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.

Hope it's clear.


Hey Bunuel,

Since I always tend to mix up the variables by the end of the problem, please tell if it would be wrong to test the function with random numbers (positive integers)?
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Re: If the operation @ is defined for all integers a and b [#permalink]

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New post 08 Apr 2018, 18:59
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 \(a@(b@c)=a+b@c-a*(b@c)=a+(b+c-bc)-(b+c-bc)a=a+b+c-bc-ab-ac+abc\), results match.

Answer: E.


Hi,
FOr this particular questions, can we substitute numbers? WIll that apprach always work?
Please advise
Re: If the operation @ is defined for all integers a and b   [#permalink] 08 Apr 2018, 18:59

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