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S98-04

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For any operation \(?\) that acts on two numbers \(x\) and \(y\), the commutator is defined as \(x?y - y?x\). For which of the following operations is the commutator not equal to zero for some values of \(x\) and \(y\)?

I. \(x?y = xy\)

II. \(x?y = (x - y)^2\)

III. \(x?y = x^3 - 3x^2y + 3xy^2 - y^3\)


A. I only
B. II only
C. III only
D. II and III only
E. None of the above
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 01:52
Official Solution:


For any operation \(?\) that acts on two numbers \(x\) and \(y\), the commutator is defined as \(x?y - y?x\). For which of the following operations is the commutator not equal to zero for some values of \(x\) and \(y\)?

I. \(x?y = xy\)

II. \(x?y = (x - y)^2\)

III. \(x?y = x^3 - 3x^2y + 3xy^2 - y^3\)


A. I only
B. II only
C. III only
D. II and III only
E. None of the above


To solve this problem, you can't let the strangeness of the jargon intimidate you. Rather, just accept the definition of the commutator without trying to understand it too much right away in the abstract; then plug in specific operations one at a time.

For instance, in the first statement, the operation \(x?y\) is defined as \(xy\) (the product of \(x\) and \(y\)). Since the commutator is generally defined as \(x?y - y?x\), we can write an equation:

Commutator \(= x?y - y?x\)

Substitute the first statement's definition. Note that \(y?x = yx\) (we switch the positions of the variables).

Commutator \(= xy - yx\)

Since \(xy\) equals \(yx\), we get the following:

Commutator \(= xy - xy = 0\)

This means that for the operation of multiplication, the commutator always equals zero, for all values of \(x\) and \(y\). As a result, the first statement does not fit the question, which asks for cases in which the commutator does not equal zero for some values of \(x\) and \(y\). We can now rule out answer choice (A).

Next, we consider statement II. We write the definition of the commutator again, but we substitute the second definition of \(?\).

Commutator \(= x?y - y?x\)

We substitute the second statement's definition, switching the \(x\) and \(y\) to replace \(y?x\).

Commutator \(= (x - y)^2 - (y - x)^2\)

Commutator \(= x^2 - 2xy + y^2 - (y^2 - 2yx + x^2)\)

Commutator \(= x^2 - 2xy + y^2 - y^2 + 2yx - x^2\)

Commutator \(= 0\)

By now, we might notice that the commutator is always zero when \(x?y = y?x\), or in other words, when switching \(x\) and \(y\) in the operation gives you the same result. For instance, with multiplication (in the first statement), you can write \(xy\) or \(yx\) to represent the same thing. In the second case, \((x - y)^2\) always equals \((y - x)^2\), since both expressions expand to equal \(x^2 - 2xy + y^2\). Asking whether the commutator is zero is a fancy way of asking whether switching the variables gives you the same result.

We can now eliminate any answer choice that includes statement II. Eliminate (B) and (D). We are left with answer choices (C) and (E).

Now we can examine the third statement's definition:
\(x?y = x^3 - 3x^2y + 3xy^2 - y^3\)

Although this expression is complicated, we can see that if we switch x and y, we do NOT get the same expression, after rearrangement:
\(y?x = y^3 - 3y^2x + 3yx^2 - x^3\)

Notice, for instance, that in \(x?y\), the \(x^3\) term is positive, but in \(y?x\), that same term is negative. The overall expressions will only be equal when \(x = y\).

Thus, for this operation, the commutator \(x?y - y?x\) will NOT always be zero, and this definition fits the question.


Answer: C
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New post 27 Jun 2017, 11:55
Are there any more questions which involve commutators? Request you to post! A really good question once you understand its about how we interpret the graphic involved..

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New post 27 Jun 2017, 23:10
Madhavi1990 wrote:
Are there any more questions which involve commutators? Request you to post! A really good question once you understand its about how we interpret the graphic involved..


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Re: S98-04 [#permalink]

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New post 29 Aug 2017, 11:17
xy operation (multiply) is always commutative
and the second one as well even consisting subtraction because of the square commutative as well and as we have subtraction of the very same values, both equations are 0s

So we need actually to evaluate only the 3rd one, so it's either C or E
And there is actually no need for substitution as the "-" in this equation is not a commutative operation, so it will not equal to 0


What are the odds of having the questions related to commutators on real gmat?

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Re: S98-04   [#permalink] 29 Aug 2017, 11:17
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