Inequality GT and LT Operator Question

[james.jt.wu]

Hi There,

I have found the GT and LT operators in Inqualities from the Strategy Guide to be very helpful for many questions. However, I have a question about its application - specifically, can u substitute a GT or LT operator into an existing inequality???


Here is a sample:

1.) x + y <20
2.) y <20

I know you can add the two sides of the inequality, but I have another idea in mind.

y<20 can be expressed as LT(20)... therefore

x + LT(20) < 20 ...
x < 20 - LT(20) ...

x< GT(0) ??!!

However, the above expression seems a bit odd. How do I reconcile < and GT together?

In summary, are we allowed to substitute GT and LT operators into an inequality??

Thank you for your help!

James

[StaceyKoprince]

you can, yes, but then you have to figure out how to interpret things, as you noted. Depending on the math, that could be relatively easy and it could be quite complicated. This works best, actually, when the LT or GT thing you're plugging in is plugged into an equation, not an inequality.

in your example, you have:

x < 20 - LT(20) ...

20 - LT(20) is complicated. It could be less than 20 OR greater than 20. For example, LT(20) could be, say, 15, so 20-LT(20) is a bit greater than zero (5, to be exact). But what if LT(2) is -100? Then what happens to 20-LT(20)?

What CAN'T 20-LT(20) be? Well, LT(20) can't be 20 or greater, so 20-LT(20) can't be zero or less.

But then you've got to reconcile that with x < whatever 20-LT(20) is... basically, x can be pretty much anything. :)

So you actually do have to think out the implications of what you have, but yes, this can work. It just happens not to tell us much in this example.

[james.jt.wu]

Stacey,

Thank you - I still don't quite get LT and GT... especially on DS problems.

Let me give you an example.

Say

X + 13 = 2 Y
Is X<Y?

Statement 1.) Y>6
Statement 2.) X< 13

The easiest way to solve this would be to plug the stem equation into the inqualities, so

2Y - 13 < Y?
Y < 13?

OR

X< (X+13)/2
2X < X + 13
X < 13?

Therefore rephrased - is X<13 or Y <13?? Clearly, the answer is B.

Now let's try to use GT and LT operators and plug into the equation in the stem. For the sake of illustration, let's just look at Statement 2.

Statement 2

X<13, so x= LT(13)

plug this into the equation from the stem for X

LT(13) + 13 = 2 y

LT(26) = 2y
y= LT(13)

So y<13

Putting this with statement 2 together... X<13 and y<13

However, we no longer know whether X<Y! All this says is X and Y are some integers smaller than 13.

Without using the LT and GT operator, we were able to get sufficiency with Statement 2 alone.

Why is this the case? Am I misunderstanding the purpose of these LT and GT operators? Are they meant to be used with min/max and extreme value questions only, and not meant to be mixed in with "regular" inequality algebra?

Thanks Stacey.

James

you can, yes, but then you have to figure out how to interpret things, as you noted. Depending on the math, that could be relatively easy and it could be quite complicated. This works best, actually, when the LT or GT thing you're plugging in is plugged into an equation, not an inequality.

in your example, you have:

x < 20 - LT(20) ...

20 - LT(20) is complicated. It could be less than 20 OR greater than 20. For example, LT(20) could be, say, 15, so 20-LT(20) is a bit greater than zero (5, to be exact). But what if LT(2) is -100? Then what happens to 20-LT(20)?

What CAN'T 20-LT(20) be? Well, LT(20) can't be 20 or greater, so 20-LT(20) can't be zero or less.

But then you've got to reconcile that with x < whatever 20-LT(20) is... basically, x can be pretty much anything. :)

So you actually do have to think out the implications of what you have, but yes, this can work. It just happens not to tell us much in this example.

[tim]

You've identified a limitation with this notation, and this is why it should be used sparingly. If you want to make this more rigorous, you would need to define an additional operator called LT/2 so that y = LT/2 (13) rather than just LT (13). Of course, then we'd need some additional theory to describe how those two operators relate, and at that point you might as well just use epsilons or better yet solve the inequalities themselves!

The bottom line is that these operators are not perfect, and you should use them only as a last resort if you just can't do the algebra the normal way. You won't be guaranteed to get the correct answer, but at least it has a high probability of working and is better than a random guess..