Quarter Wit, Quarter Wisdom: Some Inequalities, Mods and Sets

Quarter Wit, Quarter WisdomToday, let’s look at a question that involves inequalities and modulus and is best understood using the concept of sets. It is not a difficult question but it is still very tricky. You could easily get it right the first time around but if you get it wrong, it could take someone many trials before he/she is able to convince you of the right answer. Even after I write a whole post on it, I wouldn’t be surprised if I see “but I still don’t get it” in the comments below!

Anyway, enough of introduction! Let’s get to the question now.

Question: If x/|x| < x, which of the following must be true about x?

(A)   x > 1

(B)   x > -1

(C)   |x|< 1

(D)   |x| = 1

(E)    |x|^2 > 1

Solution:

First thing we do is tackle the mod. We know that |x| is just the absolute value of x.

So, x/|x| can take only 2 values: 1 or -1
If x is positive, x/|x| = 1 e.g. if x = 4, then 4/|4| = 1
If x is negative, x/|x| = -1 e.g. if x = -4, then -4/|-4| = -4/4 = -1
x cannot be 0 because we cannot have 0 in the denominator of an expression.

Now let’s work on the inequality.

x > x/|x| implies x > 1 if x is positive or x > -1 if x is negative.
Hence, for this inequality to hold, either x > 1 (when x is positive) or -1 < x< 0 (when x is negative)

x can take many values e.g. -1/3, -4/5, 2, 5, 10 etc.

Now think – which of the following MUST BE TRUE about every value that x can take?

(A)   x > 1

or

(B)   x > -1

I hope that you agree that x > 1 doesn’t hold for every possible value of x whereas x > -1 holds for every possible value of x. Mind you, every value greater than -1 need not be a possible value of x.

This concept might need some more work. Let me explain with another example.

Forget this question for a minute. Say instead you have this question:

Example 1: x > 2 and x < 7. What integral values can x take?
I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line.

The top arrow shows x > 2 and the bottom arrow shows x < 7. You see that the overlapping area includes 3, 4, 5 and 6. That is the region that satisfies both the inequalities.

Now consider this:

Example 2: x > 2 or x > 5. What integral values can x take?
Let’s draw that number line again.

So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. the overlapping part) as was the case in example 1. OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 …

Now go back to this question. The solution is a one liner.

x/|x| is either 1 or -1.
So x > 1 or x > -1
So which values can x take? All values included in at least one of the sets. Therefore, x > -1.

Note that the confusion lies only between the first two options. The other three options are rejected outright.

(C)   |x|< 1 implies -1 < x < 1. Definitely doesn’t hold.

(D)   |x| = 1 implies x = 1 or -1. Definitely doesn’t hold.

(E)    |x|^2 > 1 implies either x < -1 or x > 1. Definitely doesn’t hold.

So what do you say? Are you convinced that the answer is (B)?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

25 Responses

  1. Amit says:

    Hi Karishma,
    I have just started following your blogs and I really find them useful. You are doing a great job.

    Question on above article
    How and Why did you convert below two conditions to OR
    1. When X is negative then 0 < x 1

    Also, if we are converting above two conditions to OR; can I state that modulus values divided between positive and negative are always an OR

    • Karishma says:

      When x takes a value, it will be either positive OR negative (or 0 but that’s irrelevant here).
      When you say |x| = 5, you take 2 cases:
      Case 1: x >= 0
      x = 5
      Case 2: x < 0
      -x = 5
      x = -5

      The OR is implicit here, isn't it? x cannot be both positive and negative at the same time.

  2. Oh I loved this one…truly! Not really sure about it’s testing potential but it was a quick hit nevertheless…

  3. Moni says:

    Very useful Article Karishma!

    Wanted to understand the number line part better.

    If we take another number line scenario , then how the following would be derived
    x<-1 or 0<x<2

    from
    x<2 or, x<0 or, x<-1

    • Karishma says:

      x<-1 or 0<x<2
      There is no conflict here. x is either less than -1 or between 0 and 2. It can lie in only one range at a time so no problem.

      x<2 or, x<0 or, x<-1
      Check out the regions represented by each of these inequalities. x can take any value which lies in one or more of these regions.
      The range you get is x < 2.

      • Moni says:

        Karishma,
        I understand it. But, how about the values from -1 to -0.1 , as we are given x<0 or x<2 ?

        Where does the values -0.1, -0.2 ,…………………….,-0.9, -1 come in the range of
        0< x <2 ?

        Please explain.

        • Karishma says:

          Are you talking about the first part or the second?

          In the range x<-1 or 0<x<2, x cannot take the values from -1 to 0.

          In the range x<2 or, x<0 or, x<-1, x can take any value as long as x < 2.

          • Moni says:

            Hey Karishma,
            Looks like you didn’t get my question.
            I will repeat my question in a lucid manner:

            Actually, this is only one expression &
            the expression x<-1 or, x<0 , or x<2 is minimized to another expression that x<-1 or, 0<x<2

            I am not getting how does this derive?

            Also in the new expression we are not covering the values in the range (-1 to 0) e.g. -0.1, -0.2 ,…………………….,-0.9, -1

            It would be great if you could clarify this confusion .

          • Karishma says:

            These two:
            x<-1 or, x<0 , or x<2
            and
            x<-1 or, 0<x<2
            are NOT equivalent.

            They represent two different ranges.
            x<-1 or, x<0 , or x<2 represents x < 2
            x<-1 or, 0<x<2 represents x<2 except the values from -1 to 0.

  4. vasanth kailash says:

    By comparing with ur eg,
    |__|____|____|_____|
    -1 0 1

    |______________|
    x>-1(when x is negative)

    or |_____|
    x>1(when x is positive)

    so we take x>-1, as it has a larger set covering all possible values…. But wont x>-1 be wrong in the range -1<X<1,……As per this solution, for (x / |x|) -1….. but when x= -0.5..then we get

    -0.5/|-0.5|= -0.5.. -0.5 is not less than -0.5… please clarify this…

  5. vasanth kailash says:

    **please ignore my last comment.. here is my question without mistakes**

    By comparing with ur eg

    x>-1(when x is negative)

    or
    x>1(when x is positive)

    so we take x>-1, as it has a larger set covering all possible values…. But wont x>-1 be wrong in the range -1<X<1,……As per this solution, for (x / |x|) -1….. but when

    x= -0.5..then we get

    -0.5/|-0.5|= -0.5.. -0.5 is not less than -0.5… please clarify this…

    • Karishma says:

      When x = -0.5,

      -0.5/|-0.5|= -1 (not -0.5)
      -1 is less than -0.5

    • Karishma says:

      Also, let me repeat what I wrote in the post above:
      “Mind you, every value greater than -1 need not be a possible value of x but every value of x for which the inequality holds will be greater than -1.”

  6. Moni says:

    Thanks for clarifying ,Karishma !

  7. vasanth kailash says:

    ////When x = -0.5,
    -0.5/|-0.5|= -1 (not -0.5)
    -1 is less than -0.5////

    silly me!! need to reduce my careless mistakes….

    ////“Mind you, every value greater than -1 need not be a possible value of x but every value of x for which the inequality holds will be greater than -1.”////

    i understand now….. by the way, i recently read your graphical method and number line method… it was very helpful… i love doing in graphical method now…thank you sooo much karishma!!

  8. Ali Asad says:

    Hi Karishma,

    Thanks for such wonderful articles. Your blog is a real treasure trove of intellectual stimulation.

    Just wanted to clarify one point. When you say

    “Hence, for this inequality to hold, either x > 1 (when x is positive) or -1 < x1 immediately
    by plugging it back into the equation. Once I get the solutions by opening up the modulus, my modus operandi is to plug in the values back into the inequality to check whether such values satisfy the equation or not. In this case , say plug in x=2

    => 2/2<2 False

    So the only valid solution to the equation is -1<x<0 and we can pick the choices.

    Is my reasoning correct?

    Thanks in advance.

    • Karishma says:

      I am not sure I understand what you have done here. Why do you say that x = 2 does not satisfy the inequality?

      2/2 < 2
      1 < 2
      This is true so x = 2 satisfies.
      Every value that satisfies the inequality is greater than -1.

  9. Rishabh says:

    So x > -1 is correct because:

    x > -1 can include both the cases i.e. x being positive or x being negative whereas x > 1 can only include the case when x is positive.

    Am I correct?

  10. Rishabh says:

    Thank You, the concept was totally new to me & it took me sometime to get hold of it.
    It was pretty helpful!

  11. H says:

    ‘ for this inequality to hold, either x > 1 (when x is positive) or -1 < x 0 , 2 cases follow:

    1. x(x-1) > 0 —–> x1

    2. x(x+1) -1<x1 or -1<x<0

    or is there any better way to check the ranges ?

    • Karishma says:

      I am not sure I understand your question properly. Do you mean to ask how we get the range x > 1 or -1 < x < 0?

      If yes, then the easier method is the one I have used. Consider the given expression:

      x/|x| x/|x| i.e. x > 1
      If x is negative, x > x/|x| i.e. x > -1. This means x can take values in the range -1 < x < 0

  12. Felix says:

    i dont understand why B is the answer… if x > -1, then X could be “0″. this would not hold.

  13. Arun says:

    Excellent post. Thanks a lot.

    My doubt on the example problem is as follows.

    Since, the range of x between 0 < x < 1 do not satisfy the inequality x/|x| -1 is the answer.

    I am missing something.

    Kindly explain.

    • Karishma says:

      Think on the following:
      This is a MUST BE TRUE question. They are not asking you the range of x.
      It is NOT necessary that every value in the range should be a value of x.
      It is, however, necessary that every value that x can take should lie in the range given.

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