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Inequality Basics: You cannot multiply/divide an inequality

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Inequality Basics: You cannot multiply/divide an inequality [#permalink] New post 10 Aug 2007, 12:32
Inequality Basics: You cannot multiply/divide an inequality by a variable unless you know the sign of that variable because you need to know whether or not to switch the direction of the inequality symbol. The inequality p/q > 1 does not necessarily imply that p > q (unless q is positive).


My question: if p/q = r/s. Is ps = rq?
Or here is another one:
if (x^2+y^2)/xy < 2 could it be optimised as (x-y)^2 < 2?

If YES to both my questions, then why is the rule not applicable here? :roll: Any :idea:

Last edited by AugiTh on 10 Aug 2007, 13:19, edited 1 time in total.
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 [#permalink] New post 10 Aug 2007, 12:43
Quote:
if p/q = r/s. Is ps = rq?


just set it up as a fraction = fraction and cross multiply. as long as p/q = r/s than they will multiply out to the same thing as well. since both sides each have to be positive or negative all the signs will multiply out to equal the same thing.

Quote:
(x^2-y^2)/xy < 2 could it be optimised as (x-y)^2 < 2?


nope. plug in 4 for x and 2 for y.

4^2 - 2^2 / 4(2) < 2
12/8 = 1.5 < 2

but (4-2)^2 = 2^2 = 4 which is NOT < 2
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 [#permalink] New post 10 Aug 2007, 12:46
Quote:
Inequality Basics: You cannot multiply/divide an inequality by a variable unless you know the sign of that variable because you need to know whether or not to switch the direction of the inequality symbol. The inequality p/q > 1 does not necessarily imply that p > q (unless q is positive).


My question: if p/q = r/s. Is ps = rq?
Or here is another one:
if (x^2-y^2)/xy < 2 could it be optimised as (x-y)^2 < 2?

If YES to both my questions, then why is the rule not applicable here? Any


With inequalities, you are right; you cannot divide by a variable if you don't know the sign (b/c signs change with negative numbers).

For the first statement, it's not an inequality, so the rule doesn't apply. if p/q = r/s, then by definition, ps = rq. not an inequality equation.

Can't say much for the second statement though. Looks like something one of the DS Gurus could explain.
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Re: Iequalities fundamental question [#permalink] New post 10 Aug 2007, 12:50
AugiTh wrote:
Inequality Basics: You cannot multiply/divide an inequality by a variable unless you know the sign of that variable because you need to know whether or not to switch the direction of the inequality symbol. The inequality p/q > 1 does not necessarily imply that p > q (unless q is positive).


My question: if p/q = r/s. Is ps = rq?
Or here is another one:
if (x^2-y^2)/xy < 2 could it be optimised as (x-y)^2 < 2?

If YES to both my questions, then why is the rule not applicable here? :roll: Any :idea:


augi,

i've been known to mess up inequalities... but i would say you should be able to divide out a variable as long as it isn't zero. if you divide a variable out just take into account that you have two possibilities now; one where it is
<and> because you have to flip it if it is negative. it probably isn't useful in PS cuz maybe it leads to untrue statements, but maybe this is used in DS??


for question 1: yes.

question 2:
x^2-y^2/xy = ((x+y)(x-y))/x+y <2
x-y < 2
sure looks good to me.
(did you mean x-y or (x-y)^2? if you meant (x-y)^2 show me how you derived it so i can see your thought processes)
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Re: Iequalities fundamental question [#permalink] New post 10 Aug 2007, 12:58
"KNOCK KNOCK"

Theres been a mistake in my second equation:
It should be

if (x^2+y^2)/xy < 2 could it be optimised as (x-y)^2 < 2?

Another very very fundamental question:

If an equation equates to something then it is not an inequality. Sounds duh, but for clarifications sake....

i.e if p/q < r/s then the rule applies, but if p/q = r/s then pq=rs.

Am i Correct here????
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Re: Iequalities fundamental question [#permalink] New post 10 Aug 2007, 17:43
AugiTh wrote:
"KNOCK KNOCK"

Theres been a mistake in my second equation:
It should be

if (x^2+y^2)/xy < 2 could it be optimised as (x-y)^2 < 2?

Another very very fundamental question:

If an equation equates to something then it is not an inequality. Sounds duh, but for clarifications sake....

i.e if p/q < r/s then the rule applies, but if p/q = r/s then pq=rs.

Am i Correct here????


for I
i must have misread what you typed

x^2+y^2/xy <2
(x+y)(x-y)/xy <2

i can't go any further than that. how are you arriving at (x-y)^2?

(x-y)^2 = x^2-2xy+y^2

if we put it over /xy

x^2-2xy+y^2 / xy <2

x^2-2xy+y^2 <or> 2xy

x^2-4xy+y^2 <or> 0


nothing is really useful coming out of it... i dunno??

forII

you have either ps<qr>qr. which isn't useful in any situation that i can think of. this is because we do not know if s,q are - or +

when they use an equal sign it is an equality; so what is on the left will always = whats on the right.
Re: Iequalities fundamental question   [#permalink] 10 Aug 2007, 17:43
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