# Tag Archives: softball

## Softball and the Square Root of 2

Early in my teaching career, I was introduced to the Farey Series by the amazing Henry Pollak at a month-long 1993 Woodrow Wilson Leadership Program in Mathematics on the Mathematics of Change at Princeton University. I’ve inspired many (elementary school to college-level) students over the years with the following simple-to-understand mathematical nugget.  It’s been an especially GREAT HOOK for student athletes and sports fans.

Here’s a simple illustration of how I believe Henry introduced these to us almost 20 years ago.  Let’s say a softball player’s a batting average for the season is 0.387.  How many official at-bats has she had?  A quick-and-dirty response notes $0.387=\frac{387}{1000}$ from which you could conclude that she has 387 hits in 1000 at-bats.  That’s certainly a valid answer, but isn’t 1000 at-bats an unrealistically large number for one season?  It also ignores the reality the 0.387 is quite likely a rounded or truncated number.  So how can you answer this question when you’re not even sure how much you have of the number you’re approximating?  It’s much simpler than you might think.

Start by finding the simplest two fractions which bound your target number.  In this case, $\frac{1}{3}\approx 0.3333<0.387<0.5=\frac{1}{2}$.  Create a new fraction closer to the target by adding the two fractional bounds the way many students first learning fraction addition want to do:  add the numerators and add the denominators–$\frac{1+1}{3+2}=\frac{2}{5}=0.4$.  Notice two things about this fraction:

1. It’s value is between (but not at the midpoint!) the original two bounding fractions, and
2. It remains higher than the target 0.387.

Replacing my original upper bound with my new fraction, I now have $\frac{1}{3}\approx 0.3333<0.387<0.4=\frac{2}{5}$.  Repeat the procedure to get $\frac{1+2}{3+5}=\frac{3}{8}=0.375$, a new lower bound.  Doing this a few more times leads to

$\frac{1}{3}\approx 0.3333<0.387<0.5=\frac{1}{2}$
$\frac{1}{3}\approx 0.3333<0.387<0.4=\frac{2}{5}=\frac{1+1}{3+2}$
$\frac{1+2}{3+5}=\frac{3}{8}=0.375<0.387<0.4=\frac{2}{5}$
$\frac{3+2}{8+5}=\frac{5}{13}\approx 0.3846<0.387<0.4=\frac{2}{5}$
$\frac{5}{13}\approx 0.3846<0.387<0.3888\approx\frac{7}{18}=\frac{5+2}{13+5}$

Notice that each time I added the bounding fractions “the wrong way” to get a new fraction between the original bounds, and this new fraction became either my new upper bound or my new lower bound, and then I repeated the algorithm getting ever closer to my target value of 0.387.  Doing this one more time gives $\frac{5+7}{13+18}=\frac{12}{31}\approx 0.387096$ !  Forget the earlier guess of 1000 at-bats, I now know that a batting average of 0.387 can be accomplished in just 12 hits in 31 at-bats!

• $\frac{12}{31}$ is the smallest-denominator fraction whose decimal approximation to three places rounds to 0.387, meaning 31 is the absolute minimum number of at-bats for the original problem.
• If you continue the procedure, you will get a long sequence of fractions which all round to 0.387, eventually culminating with the fraction $\frac{387}{1000}$, bringing you all the way back to the original guess.
From a theoretical perspective, there’s lots here for older students who really want to know how a mediant works.   Given two fractions where $\frac{a}{b}< \frac{c}{d}$, it’s relatively simple for a student comfortable with algebra and inequalities to show that $\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$.  That explains how the procedure above quickly narrows the bounds for the target decimal number.  The hard part, as I recall, is proving that this procedure leads to the smallest-denominator fraction that rounds to the target decimal number.