## Mental Math – Square Roots

As you can probably tell from a previous post, “Completing the Square: Two Forms – Two Methods,”  I like taking square roots.  For instance I got several years of pleasure from a mental mathematics method for calculating square roots that I found in Dead Reckoning: Calculating without Instruments by Ronald W. Doerfler.     Once I got the hang of the technique, I started mentally calculating square roots  as I rode my mountain bike up into the hills behind my house .  This distraction caused me to pedal slower due to the concentration involved but the time passed more quickly.  I used EXCEL to make a list of the square roots from 1 to 200 and had them printed upside down, front and back  on a T-shirt.  I could check my answers as I pedaled by just looking down at the shirt.  This idea really didn’t work well since cotton tees are really only okay for vigorous activity in very hot weather. (They retain moisture making for a cold ride back down the hill).  I ended up checking the boxes on the shirt when I got home.

T-Shirt with Upside Down Square Roots

Closeup of Upside Down Square Roots

The square root method requires you to remember just one four digit number besides the digits of the evolving answer.  I found that I could remember up to 13 or 14 digits on a bike ride.  Each calculation took me around 20 minutes if I didn’t get sidetracked by trying different ways to do intermediate calculations or by checking a calculation several different ways.  These excursions were part of the fun anyway.  The Doerfler method works as follows.

Set up the square root problem so you are calculating the square root of a 3-digit  or 4-digit number and your initial square root estimate is the closest 2-digit number.  So if you have $\sqrt{61}$ use $\sqrt{6100}$ with root estimate $78$, if $\sqrt{2}$ use $\sqrt{200}$ with estimate $14$, if $\sqrt{617}$ use $\sqrt{617}$ with estimate $25$, if $\sqrt{6172}$ use $\sqrt{6172}$ with estimate $78$ and if $\sqrt{61728}$ use $\sqrt{617}$ with estimate $25$ and add the leftover $28$ to the remainder as you calculate. You want to be sure that your initial 2-digit estimate of the root is the correct one.  You can always adjust the decimal at the end of the calculation.

We think of $\sqrt{N}$ as equal to $a+b\cdot10^{-2}+c\cdot10^{-4}+d\cdot10^{-6}+e\cdot10^{-8}+f\cdot10^{-10}+...$, where $a,b,c,d,e,f....$ are two digit numbers possibly negative as close to zero as possible.  Then $N = (a+b\cdot10^{-2}+c\cdot10^{-4}+d\cdot10^{-6}+e\cdot10^{-8}+f\cdot10^{-10}+...)^2$ and using my favorite left-to-right multiplication method ( see this post), $N = a^2+2ab\cdot10^{-2}+(2ac+b^2)\cdot10^{-4}+(2ad+2bc)\cdot10^{-6}+(2ae+2bd+c^2)\cdot10^{-8}+(2af+2be+2cd)\cdot10^{-10}+...$

We pick the 2-digit $a$ to make the remainder $N-a^2$ as small as possible.  Divide the remainder by 2 to get $ab\cdot10^{-2}+(ac+\frac{b^2}{2})\cdot10^{-4}+(ad+bc)\cdot10^{-6}+(ae+bd+\frac{c^2}{2})\cdot10^{-8}+(af+be+cd)\cdot10^{-10}+...$

Now choose a 2-digit $b$ to make this modified remainder as small as possible.  The assumption here is that most(two orders of magnitude) of the modified remainder is composed of the $ab$ term.  Thus $b$ can be found by dividing the remainder by $a$ and choosing the result that gives the smallest new remainder.  We now know $a$ and $b$.  Our remainder is $(ac+\frac{b^2}{2})\cdot10^{-4}+(ad+bc)\cdot10^{-6}+(ae+bd+\frac{c^2}{2})\cdot10^{-8}+(af+be+cd)\cdot10^{-10}+...$.  Subtract $\frac{b^2}{2}$ from the remainder and find $c$ as we found  $b$.  Our new remainder is $(ad+bc)\cdot10^{-6}+(ae+bd+\frac{c^2}{2})\cdot10^{-8}+(af+be+cd)\cdot10^{-10}+...$. Subtract $bc$ and find $d$ like we found $b$ and $c$.  Our new remainder is $(ae+bd+\frac{c^2}{2})\cdot10^{-8}+(af+be+cd)\cdot10^{-10}+...$.  Subtract $bd+\frac{c^2}{2}$ and find $e$.  Continue in this way as long as you can keep $a,b,c,d,e,f....$ and the remainder in your head as you are calculating.  Here is a worked out example.

Calculating the Square Roots of Sixty-One

Notice that besides the digits of the square root, the only other number to keep in your mind is the remainder.   As you can see the calculations are simple 4-digit subtraction and 2-digit multiplication and squaring.  I used standard mental math techniques.  For instance, to square $78$, I would calculate $76\cdot 80+2^2$ since $a^2=(a-b)(a+b)+b^2$.  Multiplying by $80$ is easy since you just double three times and append a zero.

I could never get Doerfler’s method to work for $\sqrt{6}$.  I think this was because the entire remainder was in the $e,f,g....$ terms when I attempted to calculate the $d$ term.  Here is my attempt.

An Attempt at Calculating the Square Root of 6

If you look at the closeup of the T-shirt, you will see that I checked the square root of $6$ box.  What I did was calculate $\sqrt{24}$ and divide by $2$ – one of the few times I have had a practical use for this property: $\sqrt{24} = \sqrt{4}\cdot \sqrt{6}$.

Practicing the techniques of mental math is a real joy – a joy denied to most children these days of the “one algorithm fits all” mentality.  I can get pretty emotional about this.