Inequality - (|x| < 1)

[vivek.rastogi]

This question is similar to a question that appeared in MGMAT CAT.

If x != 0 , is |x| < 1 ?
1) x|x| < x
2) |x| > x

I understand what is explained in the 5/27/2010 Ron's session, but my only confusion is that when Ron said that this question cannot be completely solved with algebra. Please see my description below and let me know whether we can use the below mentioned approach.

_____________________
Question stem |x| < 1 can be re-written as : -1 < x < 1 (exception x != 0)

Statement 1 :
x|x| < x
x|x| - x < 0
x(|x| -1) < 0
a) x > 0 and |x| - 1 < 0 or |x| < 1 (since in this case x >0)
Since x > o then this inequality gives us 0 < x < 1

b) x < 0 and |x| - 1 > 0 or |x| > 1 (since in this case x <0)
Since x < 0 then this inequality gives us x < -1
Thus Statement 1 is INSUFFICIENT

Statement 2:
|x| > x
As this statement is true it tells us that "˜x’ is always negative , which can be re-written as x < 0
This statement is also INSUFFICIENT as it does not tell us about whether x lies between -1 and 1.

Statement 1 and Statement 2 together:
Statement 1 gives us:
----- 0 < x <1 & x < -1 or we can say x lies between the set [-infinity, -1) (0, 1)
Statement 2 gives us:
----- x < 0 or we can say x lies in the set [-infinity, 0)
Therefore, the only set that satisfies these 2 equations is x < -1 or we can say [-infinity, -1)
Thus, both statement together are SUFFICIENT to confirm answer as NO.

[RonPurewal]

yes, that works.
were you actually able to execute the whole thing within two minutes? for most students, this problem would take well over two minutes.

this is a nice solution, but there definitely will be other problems on which an algebraic solution is either inadvisable or impossible -- make sure that you don't discount the plug-in methods completely.

[norizam]

I'm not sure if it is ok to do this:
1) x|x| < x
I divided each side with x,keeping in mind that x could be + or -
if x >0 then |x| < 1
if x < 0 then |x| > 1
therefore not sufficient
2) |x| > x
x < 0 , x could be a "small" negative number or a "large" negative number, therefore not sufficient.
(1) and (2)
x <0, therefore |x| > 1
sufficient

I'm not sure whether the division in (1) is permissible or not. Help!
I really need to strengthen my algebraic approach since plugging always gives me errors.

[mschwrtz]

Your work is OK if you checked a couple of things.

"1) x|x| < x
I divided each side with x,keeping in mind that x could be + or -
if x >0 then |x| < 1
if x < 0 then |x| > 1
therefore not sufficient"

First: If you have an absolute value question with the variable on both sides of the equation or inequality, you need to check your solution against the original equation/inequality.

Notice that the OP avoided this requirement by transposing and factoring, rather than dividing. Did you check? If so, then you're fine. If not, then you got a bit lucky.

Second: When you divide by a variable, you have to check that 0 is not a possible value of the variable. Did you check that x was not 0 before doing this? If so, then you're fine. If not, then you got lucky. X=/=0 because 0|0| is not less than 0.