Why Division by Zero Can Lead to Absurd and Disastrous Results

In an earlier post, Why is Division by Zero is Forbidden, I discussed why division by zero is a mathematically meaningless operation. The purpose of this post is to show how attempted or actual division by zero can lead to absurd and disastrous results. I will also take a look at two rational polynomial functions to show how division by zero at point x = k can be interpreted. It’s important to understand why division by zero results in an undefined value, but it’s more important to be aware of actual or potential division by zero, and how to avoid, interpret, and deal with division by zero.

The first example is based on an article from Wikipedia that was brought to my attention by one of my students. On 21 September 1997, the U.S. Navy cruiser USS Yorktown (CG-48) was rendered dead in the water due to a failure in the ship’s propulsion system. A crew member entered a zero in a data base field causing attempted division by zero which in turn caused a buffer overflow in a computer network. As a result, all machines in the network were shut down until the problem was corrected. This undetected data entry error caused the crew to frantically work several hours to make the ship seaworthy. In light of standard programming practices, it’s difficult to understand why the possibility of a division by zero error was not considered by the network control software programming team. It’s not reasonable to assume that a human will always enter the correct value in a data base field; humans make mistakes.

The next example gives a flawed 9-step mathematical proof that 1 = 2. A class discussion to determine the flaw in the proof can be very interesting and enlightening. One only needs to understand the basic properties of real numbers and simple factoring to follow the logic of the proof. The fatal error in the proof occurs in step 8 because p – q = 0 which in turn leads to division by zero. It’s fun to see the initial reaction of a class when you announce that you will give a proof that 1 = 2. When presenting the proof, I have students give a justification for each step.

In terms of a practical application, let’s see how to avoid the error message #DIV/0! in an Excel spreadsheet cell. It’s important that spreadsheet users are aware of the possibility of division by zero, and they know how to handle this type of error. Suppose the formula =C2/D2 is in some cell of an Excel spreadsheet such as cell E10. If cell D2 is blank or has a value equal to zero, the value of cell E10 will be set to #DIV/0!. This division by zero problem can be avoided by setting the value of cell E10 to blank on division by zero with the conditional statement =IF(D2=0,” “,C2/D2) in cell E10. Another way to avoid this problem is to set the contents of cell E10 to “Undefined” on division by zero with the conditional statement =IF(D2=0,”Undefined”, C2/D2) in cell E10.

Next consider the expression (x² + x – 6) / (x + 3) = (x + 3)(x – 2) / (x + 3) and the graph of the function y = (x² + x – 6) / (x + 3) shown below. Depending on the class and mathematical level of the students, there are a variety of interesting questions and points of discussion that can be raised. Remind students that the hole in the graph at (-3, -5) has an infinitely small diameter, and therefore can only be seen by a mathematically aware person or God.

• Why is the expression (x² + x – 6) / (x + 3) mathematically equivalent to x – 2 except when x = -3?
• Why is the graph of the function y = (x² + x – 6) / (x + 3) appear to be equal to the graph of y = x – 2?
• What is the value of f(-3)?
• What is the limit of f(x) as x approaches -3 or when x is arbitrarily close to -3?
• Why is the function y = (x² + x – 6) / (x + 3) continuous at every real number x except when x = -3?
• Why does the graph of the function y = (x² + x – 6) / (x + 3) contain a single hole at the point (-3, -5)?
• Graph the function y = (x² + x – 6) / (x + 3) on your favorite graphing utility. Most likely the graph is just the graph of the line y = x – 2 without any indication that the graph contains a hole? Why is this so?

Lastly, consider the rational polynomial function f(x) = y = 3(x – 2)(x + 4) / ((x + 2)(x – 5)). The graph of this function and a table of key function values are shown below. Depending on the class and mathematical level of the students, there are a variety of interesting and important concepts that can be discussed. Teachers can use rational polynomial functions to give students a gentle and concrete introduction to important math concepts. In this example, students are confronted with division by zero when x = -2 or 5. Notice how the graph of f(x) behaves near x = -2 or x = 5. Warning! Initially, many students will not get the same results shown below because they failed to enclose (x + 2)(x – 5) in a pair of matching parentheses when they entered the expression in their graphing calculator. I often have to remind students that calculators obey the order of operation rules.

• By only looking at the equation for y = f(x), how can you tell that f(2) = 0 and f(-4) = 0?
• From a geometric point of view, what does f(2) = 0 tell us? What does f(-4) = 0 tell us?
• Verify that f(0) = 2.4. From geometric point of view, what does this tell us?
• By only looking at the equation for y = f(x), how can you tell that both f(-2) and f(5) are undefined?
• As shown in the table below, pick values of x that are very close to -2 or 5, and then use a calculator to verify that the absolute value f(x) is a relatively large number.
• When x is close to -2 or 5, the absolute value of (x + 2)(x – 5) is relatively small compared to the absolute value of 3(x – 2)(x + 4). Therefore |f(x)| is relatively large because dividing by smaller numbers gives us larger numbers.
• Since f(-2) and f(5) are undefined, the lines x = -2 and x = 5 are vertical asymptotes or poles. This tells us that the graph of y = f(x) is literally ripped apart along the vertical asymptotes. In other words, the function y = f(x) is not continuous at x = -2 or x = 5. Continuous functions can be drawn without lifting the pencil from the sheet of paper.
• As illustrated in the table below, pick values of x, negative or positive, that have very large absolute values, and then use a calculator to verify that f(x) is very close to 3. This tells us that the line y = 3 is a horizontal asymptote. In other words, as x approaches ±∞, f(x) approaches 3. You can verify that f(-1,000,000) = 2.999985 and f(1,000,000) = 3.000015.

I will close this post by mentioning that the free handout, Rapid Curve Sketching, explains how one can quickly sketch the graph of a polynomial or rational polynomial function when the equation of the function is completely factored. You can also download this handout by visiting our free instructional content page in the algebra and pre-calculus tab.