neokeynesian

11-14-2004, 08:20 AM

0 > a > b

Col A

ab

Col B

(ab)^2

Answer: D

Col A

ab

Col B

(ab)^2

Answer: D

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neokeynesian

11-14-2004, 08:20 AM

0 > a > b

Col A

ab

Col B

(ab)^2

Answer: D

Col A

ab

Col B

(ab)^2

Answer: D

thebullfighter

11-14-2004, 11:00 AM

0 > a > b

=> a & b are -ve. pick a=-2 & b=-3

col. A: 6

col. B: 6^2 = 36

point is, as a>b, both a & b can not be -1, so the value of their product (will be positive, both -ve) will be greater than 1 for sure. so, clearly it's square will be greater than th product.

hence, B>A

=> a & b are -ve. pick a=-2 & b=-3

col. A: 6

col. B: 6^2 = 36

point is, as a>b, both a & b can not be -1, so the value of their product (will be positive, both -ve) will be greater than 1 for sure. so, clearly it's square will be greater than th product.

hence, B>A

krakow

11-14-2004, 11:06 AM

let's get a:=-1, b:=-2

ab = 2, (ab)^2 = 4; (A<B)

------------------------------

let's get a:=-1/2, b:=-2

ab= 1, (ab)^2 = 1; (A=B)

:grad:

ab = 2, (ab)^2 = 4; (A<B)

------------------------------

let's get a:=-1/2, b:=-2

ab= 1, (ab)^2 = 1; (A=B)

:grad:

econphd123

11-14-2004, 05:52 PM

krakow's right, D is the answer

thebullfighter

11-14-2004, 07:28 PM

yap. i realised tht.

danielman

11-17-2004, 01:00 PM

In this case both expressions are conmutative and positive, so it is safe to think of the given condition 0>a>b as just a>0 and b>0 (it is not the same, but for these expressions is equivalent).

So the question may be restated as

A) 1

B) ab

and, of course, the answer is D.

So the question may be restated as

A) 1

B) ab

and, of course, the answer is D.

krakow

11-17-2004, 01:37 PM

In this case both expressions are conmutative and positive, so it is safe to think of the given condition 0>a>b as just a>0 and b>0 (it is not the same, but for these expressions is equivalent).

So the question may be restated as

A) 1

B) ab

and, of course, the answer is D.

true,

but how comutative implies this?

So the question may be restated as

A) 1

B) ab

and, of course, the answer is D.

true,

but how comutative implies this?

danielman

11-17-2004, 02:06 PM

true,

but how comutative implies this?

because they give you an order relationship between a and b, but none of the expressions would be altered if you assumed the oposite relationship.

but how comutative implies this?

because they give you an order relationship between a and b, but none of the expressions would be altered if you assumed the oposite relationship.

krakow

11-17-2004, 03:09 PM

because they give you an order relationship between a and b, but none of the expressions would be altered if you assumed the oposite relationship.

could you explain what is order relationship in this case, please?

could you explain what is order relationship in this case, please?

danielman

11-17-2004, 03:54 PM

could you explain what is order relationship in this case, please?

Just to be less or greater than. My quotiation isn't a big deal. Only wanted to point out that even when they say that a>b, you don't have to worry about it because you could assume that b>a and the answer wouldn't change.

Just to be less or greater than. My quotiation isn't a big deal. Only wanted to point out that even when they say that a>b, you don't have to worry about it because you could assume that b>a and the answer wouldn't change.

krakow

11-17-2004, 04:20 PM

sorry for incoveniences; I still don't get the idea..

what is the purpose of assuming opposite relationship, and how 'commutative' relates to this problem..

I was trying to figure it out, but tell me if it is what you mean..

1 {relatioship} ab / * ab (multiply both sides by ab, we can do this, because ab is positive)

1*ab {rel.} ab * ab

ab {rel.} (ab)^2

I didn't need comutativeness (or,how it is in english.. :p "commutation"?) to proceed like this..

(I would have need it if I needed expression a^2*b^2...)

what is the purpose of assuming opposite relationship, and how 'commutative' relates to this problem..

I was trying to figure it out, but tell me if it is what you mean..

1 {relatioship} ab / * ab (multiply both sides by ab, we can do this, because ab is positive)

1*ab {rel.} ab * ab

ab {rel.} (ab)^2

I didn't need comutativeness (or,how it is in english.. :p "commutation"?) to proceed like this..

(I would have need it if I needed expression a^2*b^2...)

danielman

11-17-2004, 04:47 PM

sorry for incoveniences; I still don't get the idea..

what is the purpose of assuming opposite relationship, and how 'commutative' relates to this problem..

I was trying to figure it out, but tell me if it is what you mean..

1 {relatioship} ab / * ab (multiply both sides by ab, we can do this, because ab is positive)

1*ab {rel.} ab * ab

ab {rel.} (ab)^2

I didn't need comutativeness (or,how it is in english.. :p "commutation"?) to proceed like this..

(I would have need it if I needed expression a^2*b^2...)

Don't worry. It actually isn't important. I was only trying to say that you don't have to think further in order to do something with this piece of information (a>b) because it doesn't change the result. Just that. As I said, it isn't a big deal.

what is the purpose of assuming opposite relationship, and how 'commutative' relates to this problem..

I was trying to figure it out, but tell me if it is what you mean..

1 {relatioship} ab / * ab (multiply both sides by ab, we can do this, because ab is positive)

1*ab {rel.} ab * ab

ab {rel.} (ab)^2

I didn't need comutativeness (or,how it is in english.. :p "commutation"?) to proceed like this..

(I would have need it if I needed expression a^2*b^2...)

Don't worry. It actually isn't important. I was only trying to say that you don't have to think further in order to do something with this piece of information (a>b) because it doesn't change the result. Just that. As I said, it isn't a big deal.

swethav12

11-17-2004, 08:01 PM

0 > a > b

Col A

ab

Col B

(ab)^2

we can try it another way too

a and b may not integers they may be decimal numbers less than 1 too like for example 0.5 and 0.4 and as we know square of a decimal number is aways less than the original number.

so we cannot predict correctlywhich one greater from the given data so answer should be D

Col A

ab

Col B

(ab)^2

we can try it another way too

a and b may not integers they may be decimal numbers less than 1 too like for example 0.5 and 0.4 and as we know square of a decimal number is aways less than the original number.

so we cannot predict correctlywhich one greater from the given data so answer should be D