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neokeynesian
11-14-2004, 08:20 AM
0 > a > b



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

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:

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.

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?

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.

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?

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.

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...)

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.

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