What a pleasure it was to meet with the Colorado State University (CSU) Math Club last Thursday. Under president Joshua de Jong, the Math Club has grown in both membership and the variety of activities they participate in, such as sending nine members to the 2017 Joint Mathematics Meetings.
Prof. Patrick Shipman invited me to speak to the Math Club, as he shares my passion for mathematical pedagogy, and he kindly attended the presentation, entitled Algebra Without All the Symbols. The talk was based on the premise that the way we solve algebra problems relies far too heavily on a bulky, cumbersome notation that gets in the way of learning and actual problem solving for many students. Modern algebra notation, which dates back some three centuries, uses a lot of symbols, such as:
Arithmetic operators: +, , ×, /
Relational operators: >, <, =, ≈, ≤, ≥, ≠
Grouping symbols: ( ), [ ], { }
Variables: x, y, z
Unfortunately, these symbols and their associated rules lead to many unfortunate mistakes, and we have all “enjoyed” finding a wrong answer in line 17 of some problem, and then spending several minutes tracing the error back to a missed negative sign in line 6!
Furthermore, the usual way algebra is taught focuses on solving for “unknowns” in a way that often misses the key point. For example, consider the following problem:
If 1/7 of the Cindy’s money equals 1/4 of Donna’s money, and Cindy has $93 more than Donna, how much does each woman have?
Nearly every college graduate (all with technical degrees, sometimes including graduate degrees) that I have asked approaches the problem basically as follows:
C/7 = D/4
C = D+93
(D+93)/7 = D/4
D/7 + 93/7 = D/4
93/7 = D/4 – D/7
93/7 = 7D/28 – 4D/28
93/7 = 3D/28
93/7 × 28/3 = D
124 = D
C = 124+93 = 217
This is the traditional approach, and though it works, it is far too long and convoluted for many students (does anybody forget the parentheses in step 3?), and even top students take a while to solve this problem.
A much simpler approach is to ignore what is unknown (C and D) and focus on what we do know. To be specific, some “miracle quantity” (we’ll call it •) just happens to be 1/4 of what Donna has, and 1/7 of what Cindy has! Once we focus on this “miracle quantity,” the solution becomes obvious.
Cindy  Donna 

•••••••  •••• 93  Because the • is 1/7 of Cindy’s money, her total is seven dots. Similarly, the • is 1/4 of Donna’s money, so her total is four dots. We add $93 to Donna so that her money equals Cindy’s. 
•••  93  We get rid of four dots on both sides. 



•  31  We then divide by 3 to see that the dot is $31. Multiply by 4 and 7 to obtain $124 for Donna and $217 for Cindy. 
The first solution is very notationintensive and errorprone; the second is reasonably clear to fourth and fifthgrade students. Furthermore, even very experienced and talented mathematicians appreciate the simplicity of the “dot method.” As Prof. Shipman commented, “It is a revelation to be able to focus on the quantities that we do know rather than the quantities that we do not know.”
I thank Prof. Shipman and Mr. de Jong for inviting me to CSU, and I look forward to seeing them again sometime soon.