Many GMAT and Executive Assessment (EA) exponent problems – especially data sufficiency ones – require you to consider fractional bases. By this I mean proper fractions with values between -1 and 1, not improper fractions whose numerators exceed their denominators.

There are four “kinds” of bases separated by the three “boundary points” of -1, 0, and 1. Numbers in each of the four “zones” separated by these values behave similarly as bases of exponents. On many DS problems, we need to consider numbers less than -1, negative fractions, positive fractions, and numbers greater than 1, as well as the boundary points of -1, 0, and 1. This sounds like a lot of work, but practice will build your “spidey sense” for when a certain kind of base leads to an exception and an insufficient statement.

Let’s handle the “boundary points” first. As mentioned in a prior article, the boundary point of 1 is simple because it is “immune” to exponents: 1^{anything} = 1. Likewise, 0^{positive} = 0. For -1, (-1)^{even} = 1 and (-1)^{odd} = -1. Since (-1)^{anything} equals either 1 or -1, it is worth noting that the *absolute value* of -1^{anything} is 1. If you are unfamiliar with absolute value, don’t worry, it’s a simple concept: absolute value is a number’s *distance from 0* on a number line. The absolute value of a positive number is . . . itself. The absolute value of a negative number is simply the positive version of the number. For positive numbers, “normal” value and absolute value are the same and trend together. For negative numbers, absolute value increases as “normal” value decreases. Absolute value is notated with vertical lines on either side of a value, variable, or expression. |(-1)^{anything}| = 1.

Now for the four ranges of numbers. Numbers greater than 1 are the simplest. For these numbers, the higher the exponent, the higher the overall value. *For numbers greater than 1, higher powers have higher values.* Numbers less than -1 are only slightly more complex. *For numbers less than -1, higher powers have higher absolute values, but odd powers are negative and even powers are positive.*

Now for the positive and negative fractions: the more times you multiply a fraction by itself, the closer the resulting value gets to 0. (¾)^{2} or ¾ * ¾, which we can read as “¾ *of* ¾,” is less than ¾. *For positive fractions, higher powers have lower values*. *For negative fractions, higher powers have lower absolute values, but odd powers are negative and even powers are positive.*

Let’s demonstrate all of our rules with the examples of 2, -2, ½, and -½.

2 < 2^{2} < 2^{3} . . .

|2| < |2^{2}| < |2^{3}| . . .

(-2)^{5} < (-2)^{3} < (-2) < 0 < 2^{2} < 2^{4} < 2^{6} . . .

|-2| < |(-2)^{2}| < |(-2)^{3}| . . .

½ > (½)^{2} > (½)^{3} . . .

|½| > |(½)^{2}| > |(½)^{3}| . . .

-½ < (-½)^{3} < (-½)^{5} < 0 < (-½)^{6} < (-½)^{4} < (-½)^{2}

|-½| > |(-½)^{2}| > |(-½)^{3}| . . .

The patterns for the negatives can take some getting used to, so study these rules frequently and, more importantly, build your fluency with practice problems! Here’s a straightforward one:

Is x^{2} greater than?

- x
^{2}is greater than 1. - x is greater than -1.

For statement 1, it helps to remember that only numbers greater than 1 or less than -1 can have powers greater than 1. Powers of fractions are always fractions. So if x^{2} is greater than 1, x is either greater than 1 or less than -1. If x is less than -1, then x^{2}, which according to the statement is greater than 1, is greater than x. And if x is greater than 1, it still gets larger when it is squared, so x^{2} is always greater than x, and **statement 1 is sufficient.**

Statement 2 tells us that x is greater than -1. If we remember our boundary points, we can solve this one without having to think about fractions. X could be 0 or 1, and in either case, x2 is equal to, not greater than, x. But for any number greater than 1, x^{2} *is* greater than x. **So statement 2 on its own is insufficient, and the correct answer is A.**

Let’s try another DS problem:

Is xy > x^{2}y^{2}?

- 0 < x
^{2}< 1/4 - 0 < y
^{2}< 1/9

To verbalize the question: is the absolute value of the product xy greater than the square of the product xy?

Statement 1 tells us that x^{2} is a positive fraction, which means that x itself is a fraction with a greater absolute value, but we don’t know whether it is positive or negative. Without knowing anything about y, this isn’t enough. Statement 1 alone is insufficient. Statement 2 is similar and also insufficient. Taking the statements together, we know that the absolute value of x is greater than x^{2}, and the absolute value of y is greater than y^{2}. It follows that the absolute value of the product xy is greater than x^{2}y^{2}. We don’t know whether x and y are positive or negative, but we’re talking absolute value so it doesn’t matter. The statements together are sufficient, and **the correct answer is C.**

Finally, let’s see what happens when fractional bases meet negative exponents:

(1/2)^{-3}(1/4)^{-2}(1/16)^{-1}=

A. (1/2)^{(-48)}

B. (1/2)^{(-11)}

C. (1/2)^{(-6)}

D. (1/8)^{(-11)}

E. (1/8)^{(-6)}

This problem benefits from “translating” out of the negative-exponent form. (½)^{-3} = 2^{3}, (¼)^{-2} = 4^{2}, and (1/16)^{-1} = 16. Ideally, you recognize that everything in the expression can be converted to powers of 2. 4^{2} and 16 both equal 2^{4}, so the full expression is 2^{3} * 2^{4} * 2^{4}, or, remembering your exponent rules, 2^{11}. If you noticed how the fractional base and the negative exponent “canceled” each other, you should recognize 2^{11} as (½)^{-11}, **answer choice B.**

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**Contributor: ***Elijah Mize (Apex GMAT Instructor)*