Textbook

You will see the terms *defined*, *undefined*, and *not defined* in many kinds of GRE problems. We’ll briefly cover some definitions, but for the purposes of the GRE, it is best to just explore the scenarios where the terms become relevant.

There are four main situations where you will find an undefined result:

- The result after dividing anything by 0 is not defined.

$x/0=undefined$

- The square root of any negative number is not defined.

$x when:x =undefined<0 $

- Zero to the zero power is not defined.

$0_{0}=undefined$

- A vertical line (one that goes straight up, like $x=5$) will have an undefined slope because there is no “run” in the rise/run slope equation (we’ll cover this concept later). The line does not tilt left or right, so it always has the same $x$ value everywhere along the line, meaning the change in the $x$ is $0$. Since you can’t divide by $0$, the slope is undefined.

Knowledge of undefined numbers is relevant for questions with multiple values for a specific variable. Here’s an example that drives the point home:

$x_{2}+7x=0$

$y/x=z$$z$ is defined and $z<x$

Quantity A: $x$

Quantity B: $y$

We can start to solve this question by factoring the first equation.

$x_{2}+7xx(x+7) =0=0 $

The values of $x$ that satisfy this equation are $0$ and $−7$.

However, the second equation is $y/x=z$, which has $x$ as the denominator of a fraction. Dividing by $0$ results in an undefined value, but the question tells us that $z$ is defined. So although $x$ could be either $0$ or $−7$ in the first equation, the value $x=0$ won’t work for the second equation, meaning that the only possible value of $x$ must be $−7$.

Considering that $x$ is negative, and $z$ is less than $x$, we know that $z$ must also be negative. Because $x$ is negative, $y$ must be positive to make $y/x=z$ negative. Know that we’ve determined $y$ is positive and $x$ is negative, we have enough information to know that Quantity B is greater than Quantity A.

There are generally two kinds of situations you may see the term *defined* in the GRE: *functions* and *sequences* (we’ll cover these terms later as well, for now just focus on *defined* and *undefined*).

The phrase *is defined for* or *defined by* can sometimes be replaced by *works with* to help you better understand the question. For example:

The function $f$ is defined for all values other than $1$ for $x$ by $f(x)=(x−2)/(x−1)$

It may make more sense to restate the sentence as:

The function $f$

works withall values other than $1$ for $x$ by $f(x)=(x−2)/(x−1)$

In this case, $x$ cannot be $1$ because the denominator becomes $0$, which would make the result undefined. That is why $x$ *works with* or *is defined for* all numbers other than $1$.

When a sequence is *defined by* a rule, it just means that all the terms in the sequence are determined by that rule. Here is an example question that uses the *defined* term with a sequence.:

A sequence is defined by the following equation: $S_{n}=S_{n−1}_{2}$

If the first term in the sequence is $3$, what is the average of the next $3$ terms in the sequence?

Essentially, the sequence states that each term ($S_{n}$) equals the previous term ($S_{n−1}$) squared.

We can show the values in the sequence using a table like the one below. The $n$ represents the position of the term, so for example, $n=1$ means the first term. The $S_{n}$ represents the value of that term in the sequence. For example, $S_{1}$ is the value of the first term in the sequence, which for this question, is stated to be $3$.

$n$ | $S_{n}$ |
---|---|

$1$ | $S_{1}=3$ |

$2$ | $S_{2}=(S_{1})_{2}=3_{2}=9$ |

$3$ | $S_{3}=(S_{2})_{2}=9_{2}=81$ |

$4$ | $S_{4}=(S_{3})_{2}=81_{2}=6561$ |

As you can see, the next three terms are: $[9,81,6561]$, and their average is $(9+81+6561)/3=2217$.

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