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I was reading Kate Nowak’s blog f(t) yesterday and followed a link to the Phillips Exeter Academy website where they’ve posted a pretty amazing collection of problem sets for all levels of math from Pre-algebra to Calculus (and beyond!).

I’ve only just begun to work on some of these, but I’m having a good time so far.

Can you find a fraction so that the difference between the fraction and its reciprocal is exactly equal to 1?

That is $\frac ab - \frac ba=1$

The problem gave the example that $\frac85 - \frac58 = \frac{39}{40}$

D’oh – off by $\frac{1}{40}$

Then the question asks – Can you find another fraction that gets closer than this?

I approached this from a couple of different ways – first I took the original equation $\frac ab - \frac ba=1$ and created a common denominator to get $\frac{a^2 - b^2}{ab} = 1$ or $a^2 - b^2 = ab$ and $a^2 - ab - b^2 = 0$.

I didn’t pursue this past that point, but did come back to it later.

Next, I broke out the spreadsheet and set it up to take all the numbers from 1-30 and create fractions and their reciprocals from these and subtract them.

Looking at all that, I noticed a few places where the difference between 1 and the $\frac ab - \frac ba$ was smaller than $\frac{1}{40}$.

This happened for $\frac {13}{8} - \frac {8}{13}$ and $\frac {21}{13} - \frac {13}{21}$.  Then it was time for class.

The numbers in the fractions that were getting close to 1 had caught my eye yesterday and this morning when I came in, I sat down with it again and saw that they were all consecutive Fibonacci numbers.  So, I made a new spreadsheet with Fibonacci numbers and the fractions and reciprocals and noticed that the difference $\frac ab - \frac ba$ was approaching 1.

At some point yesterday afternoon I went to Wolfram Alpha and typed in $a^2 - ab - b^2 = 0$ just to see what I would get and it provided a relationship between the two variables that comes from treating one of the variables as a constant so that

$a=\frac b2 (1 \pm \sqrt{5})$

In other words a and b cannot both be integers!  And of course, some of you may have picked up on this sooner than I did – the number that produces the exact value of 1 is phi – The Golden Ratio.

$\phi - \frac {1}{\phi}=1$.

FUN STUFF!

2 Responses

1. I recently (re)discovered the Exeter books. On first and second glance it seems like a great problem set and a potential resource to base a very student centered classroom on.

I have given some of the problems from Mathematics 2 to my 9th grade Geometry class with some success. I have trained my students to try problems and to explore possible solutions, but maybe(?) not as much as these problem sets would allow if used earlier and more frequently.

One of the greatest struggles I have had as a math teacher is to find good/challenging/engaging problem sets. Something to get beyond the grind of just doing problems. This seems like close to what I have been looking for. Any further reflections on the material? Have you used it with your students?

I find that I spend a lot of time reinventing the wheel using some highly selective and maybe irrational filtering criteria for what is good and what is not good. The Exeter problems strike me as good in a way that I can not quite define or explain.

2. on March 7, 2012 at 3:55 PM | Reply richbeveridge

I agree. I often find textbooks have simple skills practice and a few difficult (often very difficult) “challenge” problems, but very little that bridges the gap between the two. When I lived closer to “civilization” I would comb used book stores to find old textbooks, and also would get old textbooks that Professors at the local university were throwing away and so could cobble together some good problem sets from that material.

I haven’t used the Exeter problems much because the prep work involved in getting the students up to speed to do these types of problems throws off our pacing to complete the course!

I have some interesting projects on my school web page at:
http://www.clatsopcc.edu/rich-beveridges-homepage/non-standard-problems