16
Generalization
In Part 1 you learned how to read and do proofs, which is one example of a mathematical thinking process. In Part 2, you will learn a number of other thinking processes related to creating and using proofs.
16.1 WHAT IS GENERALIZATION?
The first process to be described is generalization, in which you create from an original mathematical concept—such as a problem, a formula, a definition, a proposition, or a mathematical object—a broader concept that includes not only the original concept but also something new and different. The original concept that you start with is referred to as the special case and the new broader concept you create is called the generalization. To draw an analogy, think of the special case as a set S and the generalization as a set T that contains all of S and something more (see Figure 16.1). The advantage of generalization is that any results and insights you obtain for the generalization apply not only to the special case that gave rise to the generalization but also to any other special case you might encounter in the future, thus saving you the time and effort of having to derive those results again. Therefore, after creating a generalization, it is important to verify the special case, that is, to make sure that the generalized concept includes the special case. How you do so depends on the type of mathematical concept. This process is now illustrated with a variety of different mathematical concepts, starting with the problem of solving one linear equation in one unknown.
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Fig. 16.1 A Generalization from a Special Case.
Generalizing a Linear Equation to a Quadratic Equation
You know that a mathematical problem involves using given data to obtain a desired quantity of interest, as illustrated in following problem:
The Problem of Solving a Linear Equation
Given real numbers a and b, find a value for the real number x so that
ax− b = 0. (16.1)
In the foregoing problem, a and b are the data and the solution is:
x = b
a (provided that a6= 0).
One generalization of the problem of solving a linear equation is the following:
The Problem of Solving a Quadratic Equation
Given real numbers p, q, and r, find a value for the real number x so that
px2+ qx + r = 0. (16.2)
You can verify the special case by observing that the quadratic equation in-cludes the linear equation, and more. By “include” in this case is meant that there are specific values for the data p, q, and r of the quadratic equation
16.1 WHAT IS GENERALIZATION? 181
that, when substituted in (16.2), reduce the quadratic equation to the linear equation ax− b = 0 in (16.1). Specifically, on setting
p = 0, q = a, and r =−b
in the quadratic equation px2+ qx + r = 0 in (16.2) you obtain precisely the linear equation ax− b = 0 in (16.1). Furthermore, note that when p 6= 0, the quadratic equation includes new problems that are not linear. When verifying the special case, be careful of overlapping notation. For instance, if the quadratic equation is written as ax2+ bx + c = 0, then the symbols a and b overlap with those in the linear equation ax− b = 0 but the symbols a and b have different meanings in each case. When this happens, it is best to rewrite the general problem using symbols for the data that do not overlap with those of the original problem, just as you learned to do in Chapter 3.
After using generalization to create a new problem, the next step is to develop a solution procedure for that problem. In some cases, this is relatively easy to do using the solution to the original problem; however, if the new problem is significantly different from the original problem, then you might have to develop a completely new solution, as is necessary for the foregoing quadratic equation whose solution is:
Solution to the Quadratic Equation
Given values for the real numbers p, q, and r with p6= 0 and q2− 4pr ≥ 0, the solutions to the quadratic equation in (16.2) are:
x =−q ±p
q2− 4pr
2p . (16.3)
Unfortunately, you cannot use the solution in (16.3) to solve the linear equation in (16.1) because, to do so, you need to substitute p = 0 in (16.3), which you cannot do because you would be dividing by 0. When the solution to the general problem does not lead to the solution of the special case, you should check if:
• A mistake was made in the generalized problem or its solution.
• A different type of generalized problem or solution is more appropriate.
In the foregoing example of the quadratic equation, it is possible to cre-ate a different solution that does provide a solution to the special case of the linear equation. To see how this is done, divide the quadratic equation px2+ qx + r = 0 in (16.2) through by x2 (assuming that x6= 0) to obtain:
On letting y = 1/x, the foregoing equation becomes
p + qy + ry2= 0. (16.4)
The quadratic formula applied to (16.4) leads to the following solutions for y:
y = 1
x =−q ±p
q2− 4pr
2r (provided that r6= 0). (16.5) Taking the reciprocal of both sides of (16.5) leads to the following solution to the original quadratic equation in (16.2):
Alternative Solution to the Quadratic Equation
Given values for the real numbers p, q, and r with q2− 4pr ≥ 0 and also
−q ±p
q2− 4pr 6= 0, the solutions to the quadratic equation in (16.2) are:
x = 2r
−q ±p
q2− 4pr. (16.6)
The solution in (16.6) provides the solution to the special case of the linear equation in (16.1). Indeed, on substituting p = 0, q = a, and r =−b in (16.6) you obtain the following solution to the linear equation, in which the negative square root is used to avoid a division by 0:
x = 2r
Note that using (16.6) to solve the linear equation requires more computa-tional effort than solving the linear equation directly. This disadvantage of generalization is caused by the fact that the solution to the generalized prob-lem uses more data than does the solution to the special case.
Generalizing a Linear Equation to Two Equations in Two Unknowns It is often possible to generalize a problem in more than one way. For example, you can generalize the problem of solving one linear equation in one unknown to the following problem:
The Problem of Solving Two Linear Equations in Two Unknowns Given real numbers p, q, r, s, t, and u, find values for the real numbers x and y so that
px + qy = t, (16.7)
rx + sy = u. (16.8)
The next step is to verify that the linear equation ax− b = 0 in (16.1) is a special case of solving two equations in two unknowns by an appropriate substitution of values for the data in (16.7) and (16.8). There are several different ways to do so in this case. For instance, substituting
p = a, q = 0, t = b, (16.9)
r = 0, s = 0, u = 0, (16.10)