In the expression a $ b, the $ symbol represents one of the

Question

In the expression a $ b, the $ symbol represents one of the following arithmetic operations on a and b (in the order the variables are shown): addition, subtraction, multiplication, and division. Given that it is not true that a $ b = b $ a for all possible values of a and b, a pair of nonzero, non-identical values for a and b is chosen such that a $ b produces the same result, no matter which of the operations (under the given constraints) that $ represents. The nonzero value of b that cannot be chosen, no matter the value of a, is

A. -2
B. -1
C. -1/2
D. 1
E. 1/2

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I would like to have some discussion before posting the correct answer

A. -2
B. -1
C. -1/2
D. 1
E. 1/2

I would like to have some discussion before posting the correct answer

Bunuel · Accepted Answer

Since it's NOT true that  a$b=b$a  for all possible values of  a  and  b , then $ is neither addition not multiplication (because  a+b=b+a  and  ab=ba  for all possible values of  a  and  b ).

So, we have that $ is either subtraction or division.

Next, we are told that  a$b produces the same result, no matter which of the operations (under the given constraints) that $ represents  so no matter whether $ is subtraction or division  a$b  will produce the same result, so  a-b=\frac{a}{b}  -->  ab-b^2=a  -->  a=\frac{b^2}{b-1}  -->  b  cannot be 1, because in this case  b-1=0  and we cannot divide by zero.

Answer: D.

Hope it's clear.

Zarrolou · Answer

"Given that it is not true that a $ b = b $ a for all possible values of a and b"

so $ cannot be + or *
because
 a * b = b * a

a + b = b + a

for every a,b

we know that
 a/b = a - b

a = ab - b^2

a - ab = -b^2

a(1 - b) = -b^2

a = -b^2/(1 - b)

B cannot be 1

DoItRight · Answer

Great explainations. I have 1 question. In the equation, why do you solve for a instead of solving for b? Is there something in the wording that tells you to solve "a=" instead of "b=?"

a/b=a-b
ab-b^2=a After this step, how do I know whether I try to simplify a or b?

Zarrolou · Answer

I tried to get all "a" and "b" terms on different sides, that's the main point.

However if you try to simplify b you get:

b(a-b)=a

b=\frac{a}{(a-b)}

and you go nowhere.

jjack0310 · Answer

Simplifying b does not work. i.e. it wont give you any result:

You will have:

b^2 = ab - a
b = a - a/b

The value of b cannot be 1 because substituting b = 1 give you: 1 = a-a --> 1 = 0?? Makes no sense

gps5441 · Answer

Need some help
let b=-2
a=1/2
a/b=1/2/-2=1/-4
a-b=1/2-(-2)=5/2 
hence b cannot be -2?

Bunuel · Answer

Why should a be 1/2 if b = -2?

b = -2 is possible: a - (-2) = a/(-2) --> a = -4/3.

AkP1992 · Answer

a$b ≠ b$a for any nonzero a,b.
So b cannot be 1.
That's the answer.
Do we need to solve any further?

Posted from my mobile device

Rickooreoisb · Answer

@bunnel - I need your help. I feel it is some misreading but even after reading the question unable to clearly find out what am I missing

Since it's NOT true that a$b=b$a for all possible values of a and b, then $ is neither addition not multiplication (because a+b=b+aa+b=b+a and ab=baab=ba for all possible values of a and b).

So, we have that $ is either subtraction or division.

But can we not use addition and subtraction or multiple and division or addition and division ?

because the answer varies in this scenario

Bunuel · Answer

The question says:  “the $ symbol represents one of the following operations”  and  “it is not true that a $ b = b $ a for all possible values of a and b.”  That condition is enough to eliminate addition and multiplication entirely, because those always give the same result whether you do a $ b or b $ a. So $ can only be subtraction or division. It’s not about picking “addition and subtraction” or “multiplication and division” as sets. Only one operation is represented, and it must be one of the two where order matters: subtraction or division.

In the expression a $ b, the $ symbol represents one of the

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In the expression a $ b, the $ symbol represents one of the

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