Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

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What is the Question to This Answer?

Posted by mark schwartz on May 11, 2016


In Greenleaf’s 1862 New Elementary Algebra, he devotes considerable time to fractions, including exercises which require considerable mastery of the operations with fractions. He introduces the section on fractions with a discussion of unit fractions, which are fractions with a 1 in the numerator. He establishes a very handy algorithm for adding them, and then posits the question “Doesn’t it seem that if given the answer one could identify the fractions that added together gave that answer?” He doesn’t answer his own question, but I found it intriguing enough to play with it and it led to an interesting relationship with another standard Algebraic operation.

The Story:

Given the answer to the addition of two unit fractions, can one find those two fractions?

For example, given the answer 5/6, what two fractions added together give this answer? For those familiar with adding fractions, the answer of 1/2 and 1/3 might be found after a little trial and error. On the other hand, finding the “question” to the answer 14/24 might take a little longer.

But there is one caveat of caution to be considered throughout this discussion, particularly if you are the one creating the answers to be used. In all cases, if the answer is one that can be reduced, use the non-reduced form first. If not, it is possible that the reduced form will give no “question” or a “question” that is correct but different from the one used to generate the answer. For example, 14/24 can be reduced to 7/12, and the answer 14/24 gives the question 1/2 + 1/12, whereas the answer 7/12 gives the question 1/3 + 1/4. So, in presenting this kind of problem, students may be asked to find both sets of answers. But continuing this caveat of caution, 20/96 can be reduced to 5/24 but 5/24 gives no answer. So, when generating the answer, check out the possible questions.

Let’s start by identifying some useful algebraic strategies that already exist and can be used to find the question to the answer.

There is a method presented in Algebra books for finding the sum of two fractions without finding a lowest common denominator. Yes, there is such a procedure and it is:

m/a + n/b = (mb + an)/ab

For purposes of this discussion, remember the caveat provided earlier. Further, since we are discussing this premised on unit fractions, the value of ‘m’ and ‘n’ in this expression would both be equal to 1, and the expression is then

Expression 1.               (a + b)/ab

This expression is linked to the quadratic equation! Take a look.

Using ax2 + bx + c = 0, where a = 1, the solution is to find two factors of the constant ‘c’ which when added together are equal to ‘b’, the coefficient of the x-term, which is exactly what is seen in Expression 1. In the quadratic, the value “b” is the numerator of Expression 1, while the value “c” is the denominator of Expression 1. Here’s how to use this.

We were given the answer of 5/6 and asked to find the two fractions that added together gave this answer. If the numerator 5 is seen as the coefficient of the x-term ‘b’ in the quadratic (the sum of two factors) and the denominator 6 is seen as the constant ‘c’ in the quadratic (the product of two factors), the quadratic equation can be stated as

x2  – 5x + 6 = 0

The x-term is negative because the values in the two fractions are all positive, and when solving the quadratic equation for the values of x, it becomes apparent that the factors of the constant term both have to be negative and therefore when added together give a negative value for the coefficient of the x-term.

Factoring this equation gives (x – 2)(x – 3) = 0 and setting both to zero and solving for x gives 2 and 3, which are the denominators of the unit fractions 1/2  and 1/3, which when added together give the answer 5/6.

So, try another one. Given the answer 8/15, what two fractions added together give this answer? Using the above concept, we have

x2  – 8x + 15 = 0

Factoring, we get (x – 3)(x – 5) = 0 and setting both to zero and solving for x gives 3 and 5, which are the denominators of the unit fractions 1/3  and 1/5, which when added together give the answer 8/15.

This type of problem may never appear in today’s texts but if it ever does, thank Mr. Greenleaf for providing an algorithm for solving it.


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