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Open Letter to Harper's

Harper's Magazine published a long essay by Nicholson Baker in the current (September 2013) issue, entitled: "Wrong Answer: the Case against Algebra II". Baker is taking up the same anti-algebra banner that was waved by Andrew Hacker in last year's NY Times. Baker rightly points out many problems with the current math curriculum in the schools, but I think he makes many errors of logic and interpretation.

Baker starts off by establishing the difficulty of algebra by quoting a solution of Cardano to the partial cubic equation from 1545, stating that "the algebraic rules that Cardano described and codified are variants of the techniques that students are taught, with varying degrees of success, today". It is as if I argued that  the reading of English is too difficult for students to learn by giving a quotation from Chaucer, whose language is a "variant" of what students are taught today. In point of fact, (1) the notation that we use today has been refined over the last 5 centuries or so, precisely to make the techniques of algebra as easy as possible, (2) Cardano did not "codify" algebra in any reasonable meaning of the term, and (3) the solution of the cubic equation in radicals was no longer being taught in Algebra II when I took it 50 years ago and is, practically speaking, a fairly useless algorithm today.

It is no secret that large numbers of students dislike high school mathematics. Baker seems to feel that this proves that high school mathematics is too difficult for most children to master. It never seems to have crossed his mind that perhaps it is the way that algebra is taught is the problem, not the fact that it is taught at all. In fact, advances in understanding how people learn make it possible to teach high school mathematics to anyone except the most learning disabled. The problem is that these methods are not being used because teaching mathematics well is not an easy task. It is one that we entrust to teachers who are overworked, underpaid, and inadequately trained. In the elementary grades, where the foundations need to be set that will allow the child to master algebra, teachers often themselves have a misunderstanding and dislike of mathematics, which is all too easily passed on to children.

All this might not be so bad if we subscribe to Baker's theory that most students only need a one-year survey course of mathematics in high school: they need to learn "courtesy and kindness, the times tables, fractions, percentages, ratios, reading, writing, some history -- the rest is gravy, really". He cites the fact that American technology ruled the world in the 1950s when only a fourth of students took high school algebra, and seems to think this proves that, as a country, it is fine if most students don't learn algebra. I find it impossible to follow this logic. Haven't a few things changed since the 1950s that might make American technological supremacy less sure and teaching our children mathematics a tad more important?

One of the things that has changed is that we now at least pay lip service to the ideal of equity in education. The idea that it is fine that 3/4 of students (mostly female and people of color) will never be able to enter technical fields that require college mathematics is no longer acceptable to most of us. When algebra, geometry, and trigonometry were not required subjects, the lower classes went to the lower classes where they learned a little shop math or "general math". I don't believe we want to go back to those days. I don't think that students should be allowed to have their future position in society cemented in ninth grade because they could opt out of a subject they might not like. As a nation, we need to teach mathematics better and make it more relevant, not make it optional. Anything less is a cop-out.

A7. Comparing areas

In the diagram below, ABCD is a square, DCE is an equilateral triangle, F is the intersection of AE with CD, G is the intersection of BE with CD, IFJ and HGK are perpendicular to CD, and L is on CB so CF = CL. Prove that the area of triangle AFL is equal to the area of rectangle FGKJ.