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Jvcll
 
 

math

by Jvcll Thu Sep 06, 2007 10:15 pm

Question Bank, Inequalities, # 10

If r is not equal to 0, is r^2/|r|<1?


(1) r > -1

(2) r < 1

(1) AND (2) SUFFICIENT: Together, the statements tell us that r is between -1 and 1. The square of a proper fraction (positive or negative) will always be smaller than the absolute value of that proper fraction.

The correct answer is C.
why in this question is not possible the following when c is the answer if r=1/2 then r^2=1/2^2=1/4 then lrl = to -1/2 (since from an absolute value we have positive and negative ) then result = 1/4>-1/2 . I understand when r=-1/2 then r^2=-1/2^2 = 1/4 and lrl=1/2 then this gives us 1/4<1/2
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by StaceyKoprince Sun Sep 09, 2007 4:57 pm

why in this question is not possible the following when c is the answer if r=1/2 then r^2=1/2^2=1/4 then lrl = to -1/2 (since from an absolute value we have positive and negative ) then result = 1/4>-1/2 . I understand when r=-1/2 then r^2=-1/2^2 = 1/4 and lrl=1/2 then this gives us 1/4<1/2


A variable inside an absolute value does mean that the variable could be positive or negative, but you start your question by saying "if r = 1/2" - so you've just set that variable to be positive 1/2, not negative 1/2. You have to keep using that value for r throughout the problem once you set it.
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raheel11
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Re: math

by raheel11 Thu May 26, 2011 7:47 am

Hi,

I am having a hard time understanding how from

r^2 < |r|

we arrived at the conclusion that ---------------> -1<r<1

What approach is to be used in arriving at the conclusion above? An algebraic approach or an approach of plugging numbers will be more useful?

Thanks
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Re: math

by jnelson0612 Sat May 28, 2011 3:19 pm

raheel11 Wrote:Hi,

I am having a hard time understanding how from

r^2 < |r|

we arrived at the conclusion that ---------------> -1<r<1

What approach is to be used in arriving at the conclusion above? An algebraic approach or an approach of plugging numbers will be more useful?

Thanks


Think about it this way . . . most numbers when squared become larger. For example, 5 squared is 25. So in this case, we need to think about which numbers become smaller when squared. The first thought that should occur to you is fractions between 0 and 1. For example, 1/2 squared is 1/4.

Now, we may have one other possibility since the denominator has the absolute value of r. What if I square a number between -1 and 0? Well, the squaring will make that number positive, but it will still be smaller than the absolute value of that number (since that particular value will be positive). For example, -1/2 squared is 1/4, and the absolute value of -1/2 is 1/2. So these numbers fit too.

Personally, I think the knowledge that fractions between 0 and 1 become smaller when squared is all the information you need to solve, and then you can just test numbers from there to help you include -1 to 0. Much better than algebra!
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Re: math

by raheel11 Sun Jun 19, 2011 10:00 am

jnelson0612 Wrote:
raheel11 Wrote:Hi,

I am having a hard time understanding how from

r^2 < |r|

we arrived at the conclusion that ---------------> -1<r<1

What approach is to be used in arriving at the conclusion above? An algebraic approach or an approach of plugging numbers will be more useful?

Thanks


Think about it this way . . . most numbers when squared become larger. For example, 5 squared is 25. So in this case, we need to think about which numbers become smaller when squared. The first thought that should occur to you is fractions between 0 and 1. For example, 1/2 squared is 1/4.

Now, we may have one other possibility since the denominator has the absolute value of r. What if I square a number between -1 and 0? Well, the squaring will make that number positive, but it will still be smaller than the absolute value of that number (since that particular value will be positive). For example, -1/2 squared is 1/4, and the absolute value of -1/2 is 1/2. So these numbers fit too.

Personally, I think the knowledge that fractions between 0 and 1 become smaller when squared is all the information you need to solve, and then you can just test numbers from there to help you include -1 to 0. Much better than algebra!



Thank you! That was helpful.
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Re: math

by jnelson0612 Sun Jun 19, 2011 11:12 pm

Good to hear! :-)
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