Understanding Powers – The Theoretical Advantage

by on July 21st, 2013

By Roy Nissan-Cohen, GMAT instructor

I’ve recently stumbled upon a difficult Data Sufficiency question in one of the BTG forums. Take a couple of minutes to solve it before reading on.

Is a > 0 ?

(1)  a^3a < 0
(2) 1a^2 > 0

Judging by the replies on the thread, most GMAT aspirants would set out to solve it by plugging in values, which is indeed a viable, albeit lengthy, approach. Others would conjure some manner of algebraic factoring solution; again, a valid, if time consuming, option.

However, there is also a third and better option. Instead of diving into a complicated set of plug ins, or working out long calculations, this question is solvable using purely theoretical terms – something many GMAT takers rarely try. In this case, simply trying to understand the quantitative situation outlined in the problem can reward one with a less-than-a-minute solution – assuming one has the required theoretical knowledge.

The four regions of the number line

Before setting out on our logical solution, let’s lay down some groundwork. Different numbers behave differently when raised to a power. The most simple example of this is the difference between a smaller-than-one fraction (a ‘proper fraction’) such as ½ and a greater-than-one number such as 2. When raised to the second power, 2 becomes 4; it gets bigger. On the other hand,  ½ raised to the second power becomes ¼; it gets smaller.

These concepts are true for all numbers within these two regions:

Region 1: Numbers greater than 1. For these numbers, the greater the exponent, the greater the result.

2^1 is 2,
2^2 is 4,
2^3 is 8.

Region 2: Positive ‘proper’ fractions (between zero and 1). These actually go the opposite way: the greater the exponent, the smaller the result.

(1/2)^1 is 1/2,
(1/2)^2 is 1/4 – smaller than half,
(1/2)^3 is 1/8 – smaller than the previous powers.

Negative numbers are a bit more tricky, but they still follow a certain pattern. To keep things simple, let’s put aside even exponents for the time being. Since an even exponent on a negative base will always result in a positive value, they transcend the general effect a power has on a negative base.

Region 3: Negative proper fractions, i.e., -1 < x < 0 (for example, -1/2), become greater when raised to an odd power – they”move” more to the right on the number line.

(-1/2)^1 = -1/2,
(-1/2)^3 = -1/8, which is greater than -1/2,
(-1/2)^5 = -1/32, which is greater than both.

Region 4: Negative numbers smaller than -1, such as -2 or -5. These become smaller when raised to an odd power.

(-2)^1 = -2
(-2)^3 = -8, which  is smaller than -2,
(-2)^5= -32, which is smaller than both.

The important concept to remember here is that numbers can’t ‘change groups’ when raised to a (positive) power. Raising 3 (member of region 1) to any positive exponent will not turn it into a fraction or a negative. Likewise, raising ⅓ (member of region 2) to any positive exponent will not turn it into a number greater than 1.

The Solution

See how the concepts above are applied to the question:

Statement (1): a^3a < 0

First, simplify the inequality by adding a to both sides:

a^3 < a

The result is much more than your day-to-day inequality. This particular inequality is actually a signpost which presents us with significant information about our variable. Ask yourself – when is this true? What kind of number is greater than its own third power?

As outlined above, there are exactly 2 types of numbers that satisfy this weird path in the given inequality: positive fractions smaller than 1 in region 2 (such as ½, ⅓ and so on), and negative numbers smaller than -1 in region 4 (such as -2, -5 etc.). So statement (1) indicates that a‘s value can be either 0 < a < 1 OR a < -1. Therefore, statement (1) is insufficient to determine whether a > 0.

Statement (2): 1a^2> 0

Similarly to how we treated the first statement, let’s begin by simplifying the inequality:

1 > a^2

Once again, the resulting inequality actually tells us much more than it initially shows. What sort of numbers result in less than 1 when squared? Again, exactly two types of numbers comply: positive fractions smaller than 1, and negative fractions greater than -1. In either case, a^2 would have been a positive proper fraction, and therefore smaller than one. In algebraic terms, this statement teaches us that -1< a <1, which is still insufficient to determine whether a is positive.

Combine both statements.

Statement (1): 0 < a < 1 OR a < -1

Statement (2): -1 < a < 1

Each of the two statements limits a to two possible regions, but only one of those regions is common to both: positive less-than-one fractions, or 0 < a < 1. a must be a positive fraction, and is therefore always positive. We have a definite answer to our question, so both statements together are sufficient and the answer is C.

The entire solution can easily take less than 45 seconds – provided that you hold the 4 regions of the number line at your finger tips. Thus, memorize these regions and their respective behaviors, and practice recognizing them. Needless to say, not all GMAT problems allow for such purely logical solutions. However, it would be a shame to miss out on those that do. 

12 comments

  • Nice explanation!

    Ans. D for the OG problem.

  • I wrote the statement 2 as (1-a)(1+2)>0
    which means that either
    1-a>0 and 1+a>0 (1)
    or
    1-a<0 and 1+a<0 (2)

    I figured out that it should b -1<a<1 from (1)
    but I am not able to verify that it applies as well from (2)
    What's the mistake in my reasoning?

    • For the OG problem- an easier approach is to know that a decimal number when is squared, it's getting smaller. Taking the root from a decimal number, you will get an result that is bigger, but less than the number itself. Let me know if I am wrong.

    • You have to be careful when you deal with negative numbers --- remember -- you ALWAYS have to flip the inequality sign.

      Think about this example:

      (x^2) > 25

      We know x can be positive or negative..

      So the two solutions will be:
      x>5 or x 0;
      (1-a)(1+a) > 0;
      Case 1 -- both terms positive: (1-a)>0 and (1+a) > 0;
      Case 2 -- both negative: - (1-a) < 0 and -(1+a) a -1 --- same outcome as case 1 -- there is symmetry between both cases as expected.

      The takeaway is -- when dealing with negative numbers on inequalities -- do not forget the *sign!

    • Sorry there was a typo in my previous post

      -- for my example --

      x^2 > 25

      The two solutions are x > 5 and x < - 5

  • Why not take the second given condition out from the first one:
    (1-a^2)>0=>(a^2-1) 0 => a(a^2-1)<0 since we already know (a^2-1) 0

  • I see thanks Srini !

  • statement 1)
     a^3 - a
    a(a^2 -1)  
    a<0 and
    (a^2-1) a^2 -1<a<1 
    Because 2 answers a<0 and -1<a not suff

    statement 2)
    1-a^2>0 =>
    a^2 -1<a not suff because we can not answer that a is positive or negative

    statement 1 and statement 2

    1)  -1<a<1 
    2)  -1<a0 or <0

    What's the mistake in my reasoning?

    • C is not suff as well because we can say "a" is >0 or "a",<0

      therefore d is correct .

      What I miss here ?

  • Hello People!

    Please help me explain this problem. I am still trying to clear my understanding on inequalities. This is how I approached this problem:

    Stat 1:
    (a^3-a)<0
    a(a^2-1)<0
    a<0 or (a^2-1)<0
    a<0; -1<a0
    multiply both sides with -1.
    (a^2-1)<0
    a^2<1
    -1<a<1

    Since a can be both -ve and +ve : Not sufficient.

    Combining 1 and 2
    -1<a<1
    Since a can still be both -ve and +ve : Not sufficient.

    Answer E!

    Please help and let me know where I am wrong in my concept?

  • Hello People!
    Please help me explain this problem. I am still trying to clear my understanding on inequalities. This is how I approached this problem:
    Stat 1:
    (a^3-a)<0
    a(a^2-1)<0
    a<0 or (a^2-1)<0
    a<0; -1<a0
    multiply both sides with -1.
    (a^2-1)<0
    a^2<1
    -1<a<1
    Since a can be both -ve and +ve : Not sufficient.
    Combining 1 and 2
    -1<a<1
    Since a can still be both -ve and +ve : Not sufficient.
    Answer E!
    Please help and let me know where I am wrong in my concept?

    Sorry for the repeat post. I forgot to check notification box before submitting.

  • Excellent write up !!

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