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    Can You Do Square Roots By Heart ?
    By Tommaso Dorigo | June 26th 2014 02:26 PM | 6 comments | Print | E-mail | Track Comments
    About Tommaso

    I am an experimental particle physicist working with the CMS experiment at CERN. In my spare time I play chess, abuse the piano, and aim my dobson...

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    When I was five years old I used to be sort of an attraction to relatives. One of my mother's brothers is an engineer, and he was amazed by my ability to do complex calculations by heart. But to me it was only amusing to observe their amazement for what I considered a triviality.

    On one occasion - I remember it as it was yesterday - my uncle picked me up and while he kept me with his arms he asked me "Ok, let's see this. Tommaso, what is the square root of 5968?". Mind you, I do not remember the exact number; I only recall it was between 5000 and 7000. I watched up into the void for two seconds, and I replied "77.3". Uncle Ciccio put me down and ran for the pocket calculator - he did have one, although they were a real novelty those years.

    He came back, poked on the machine, and read the result on the display: "77.253! How the hell did you do it ??" Everybody in the room looked at me as if I had just landed from Mars. He must be a genius ! Yes, how the hell can a five-years-old kid compute the square root of a four-digit number ? As old Groucho used to say, "A child of five could understand this. Go fetch a child of five!"

    Really, there is nothing magical about learning how to extract reasonable estimates for the square root of large numbers. By the time we are five we have learned much more complex tasks, in fact. Still, given the widespread mathematical illiteracy of the world around us, a 5yo kid knowing how to perform a calculation beyond the powers of most grown-ups seems like a prodigy. But it really isn't.

    What you need, in order to learn how to do the same thing yourself, is just the will to observe some basic facts about integer numbers. Do you know how much is 5 by 5 ? And 6 by 6 ? You know this, right ? Then you also know what is the square root of 30 - it has to be something between 5 and 6, as 30 is almost exactly halfway between 25 and 36. Yes, your guess - 5.5 - is right on the money, or almost so.

    Let us make it harder. What is the square root of 28 ? If you knew nothing else but the fact that 5x5=25 and 6x6=36, you could be tempted to interpolate between those two numbers. 28 is 3 above 25, and there's 11 between 25 and 36. So, since 3 is a bit more than a fourth of 11, you could guess that the answer must be 5+0.28: 0.25 is a fourth, 0.33 is a third, and we said 3 is more than a fourth of 11, but less than a third of it. So your guess is 5.28; the real answer is instead 5.29. Quite close, isn't it ? Enough to surprise your friends.

    At the root of the above calculation is the fact that within restricted domains, the quadratic function that describes how the square of numbers grow is very well approximated by a linear function. A linear function is what we implicitly assumed in our calculation, as we broke down the difference between 25 and 36 into a interval, 11, and we took a bit more than a fourth of it to compute the square root of 28, knowing that 28-25 was a bit more than a fourth of 11.

    Once one grasps the simple concept of how to interpolate between two numbers, a whole new world opens up. What is the square root of 5968, then ? Well, you need to remember that the square of 7 is 49, that the square of 8 is 64, and that every time you add a zero to the number that gets squared, you add two to the product - hence 70x70 is 4900, duh. But now things get a bit more complicated.

    If all you knew was that 70x70 is 4900 and 80x80 is 6400, getting a good estimate for the square root of 5968 would not be too straightforward. You would need to reason as follows: "there's 1500 between 4900 and 6400, and there's 432 between 5968 and 6400. 432 is a bit more than a fourth of 1500, so I'd say we miss 2.9 (a fourth of 80-70=10 is 2.5) from 80 - the answer is about 77.1". The answer is pretty good (as noted above, the correct result is 77.253), highlighting that the linear interpolation method works even if your bases for extrapolation are far away from the point you want to evaluate. However, the calculation can be above the head of many laymen. How can a child of five do that ?

    A child of five may have less neurons than you, but he has time and memory in excess. To a motivated child, learning by heart the powers of all numbers from 1 to 100 is a child's game, indeed. 1,4,9,16,25,36,49,64,81,100,121,144,169... It is not too hard. But even only knowing that 75x75 is 5625 is a huge help in the calculation above, as we are much closer to the target 5968...

    And then there is the final trick, which, I'll admit, I had indeed grasped at 5 years of age, but not everybody know otherwise. The difference between two consecutive powers, say x and x+1, is x+x+1. That is to say: if the 2nd power of 75 is 5625, then the power of 76 is 5625+75+76, or 5776. That of course comes from the formula of the square of (x+1)^2=x^2+2*x+1, but a child does not need to know the formula to observe it, if given a succession of squares.

    If you know that to go from x^2 to (x+1)^2 you just need to add 2x+1, everything becomes much, much easier. What is the square of 141 ? Well, it is the square of 140 plus 281. Since the former is (14x14)x100, or 19600, the answer is 19881. Doh!

    Playing by heart with square numbers and square roots will make you a faster calculator. At some point you need not restrict yourself to square roots and squares: you can easily raise by one level and start dealing with other powers. It gets harder, but it can be done. And more often than not, you can connect a difficult question to things you already know.

    I once was challenged by a colleague, as we discussed calculus by heart, to do the seventh square of 11. I stupidly said "oh well, that's too hard" without thinking at it for a second, but I should have known better: it was quite easy to get a really good estimate. In fact, one just has to note that the answer can't be large. If you take 2, you know that 2^7 is 128, so we are way below that; and of course above 1^7=1 too. Then one might notice that 2^3 is 8 and 2^4 is 16, so 2^3.5 must be close to 11 - and 3.5 is half of 7. Since we all know that the square root of 2 is 1.41, from geometry we took in middle school, then it just should hit you: 1.4x1.4 is 2, 2^3 is 8, 8x1.4 is about 11... so the seventh square of 11 is very close to 1.4.

    I guess the bottomline is: if you don't loathe from the start the idea of fiddling with numbers by heart, try doing the exercise of guessing square roots of large numbers - you may get very good at it in the matter of an afternoon!

    Comments

    At one point in elementary school They(tm) got quite concerned that I didn't have multiplication tables memorized. 7's and 8's were particularly problematic for me. To calculate 8x8 I'd just 6x6=36 (that was easy to memorize) and then compute something like 36+6+7+7+8 to get the answer (of course later i learned binary and memorized 2^n's which etched the answer to that into my brain). I'd solve the problem geometrically in my head by picturing a 6x6 square and what I'd need to complete that out to an 8x8 square -- I might have actually used 36+6+6+8+8 since that'd probably made the addition easier.

    I chose physics because it's a career where people are not supposed to know how to calculate anything by hearth.

    When I was at the beginning of my PhD, I got invited by McKinsey Italy to an event whose purpose was to attract potential future employees. (Those were the long forgotten days when an Italian job market still existed, and consultancy companies still hired, and young physicists not only found real jobs, but where even valued for their skills.)
    One of the days of the event was organized this way: groups of 7-8 people seated around a table, and at each table a McK guy presented a hypothetical case (ours was, I think, "you are considering whether to start or not being a taxi driver in Rome"). He was providing all the needed data one by one and asked each of us, by rotation, to calculate some relevant quantity by hearth. The problem was that I had been for so many years ideologically contrary to any calculation done by hearth, that I now was simply unable to do even a simple addition. And anyway, I couldn't see the point of that: couldn't we just deduce and write down the right formula, and then insert all the data in the formula at the end? The answer was that a consultant must impress the clients by delivering numbers fast.
    So we ended that nice event with me thinking that they were dumb and with them thinking that I was dumb and stubborn. I was the only one at the event who did not apply for a job at McK, having decided that if I had to leave academia I would have rather worked as a low-wage clerk than as a way-too-paid consultant, but anyway they proactively communicated that they didn't want me for the job anyway :)

    dorigo
    Quite atypical... I do believe that it is helpful for a physicist to manage quick-and-dirty guesstimates on the fly, but I concur that it is not a mandatory ability. As for the McK guy, he might have a point -in the consulting market it might be quite important. But I'm happy you did not apply there.

    Cheers,
    T.
    Another neat mental-math trick that 's not seen too often in an age of calculators is one that I figured out on my own in late-elementary school. If I wanted to compute the average of similar numbers, like 34, 45, 64, 38 and 56, I began by guessing, for instance 50. I then mentally added the differences from 50: -16-5+14-12+6= -13. I then divided that result by the number of data: -13/5= -2.6 and finally added it to the estimate of 50, which gave the exact average of 47.4.

    Also, for Americans who have to convert Celsius to Fahrenheit, instead of using F = (9/5) C + 32, it's easier to mentally double the Celsius temperature and subtract 10% and then add 32,  since 2C - 0.10*(2C) = 1.8C = (9/5) C .




    dorigo
    Ah, these are both good ones!
    Cheers,
    T.
    Hah. Good. I actually feel validated for clinging to my predilection of performing mental math. I'm a bit surprised that my methodology is similar to yours, which tends to confuse most people but works amazingly well for me, especially for quick and 'dirty' estimates that are nonetheless accurate. But I'm just a writer.

    Thanks for the interesting read.