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Square roots to 1 d.p.

Hi I'm a trainee teacher with a background in engineering. As part of the UK GCSE mathematics syllabus, students are expected to calculate, without a calculator the square root of a number to 1 d.p. The method expected is shown below, with the example of $$ Find\;\sqrt{32}\;to\;1\;d.p. $$

$$ \sqrt{25}<\sqrt{32}<\sqrt{36} \\ 5<\sqrt{32}<6 \\ \text {Try}\;5.6^2 = 31.36 \\ \therefore\;5.6<\sqrt{32}<6 \\ \text {Try}\;5.7^2 = 32.49\\ \therefore 5.6<\sqrt{32}<5.7\\ \text {Consider midpoint};5.65^2=31.9225\\ \therefore\;5.65<\sqrt{32}<5.7\\ \therefore\sqrt{32}=5.7\;\text {to 1 d.p.} $$

The last step is the sections that is puzzling me, if we can see that 32.49 is closer to 32 than 31.36, can we not say our answer is 5.7 without having to create an inequality. My mentor has said that its because x^2 is not linear so although 5 is half way between 1 and 9, the square root is 5 is not halfway between the square roots of 1 and 9, which I do understand. But this is not the same as the problems, because you would say $$ \sqrt{4}<\sqrt{5}<\sqrt{9}\\ 2<\sqrt{5}<3\\ \text {as 4 is closer to 5 then } \sqrt{5} \text { is closer to }\sqrt{4}=2\\ \sqrt{5}=2.2360...\\ \text {so the above is true.} $$

I'm struggling to see this in context as to why it's not true but cannot figure it out.

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4 Answers 4

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Try this problem: Find $\sqrt{31.924}$ to one decimal place.

$$ \sqrt{25} < \sqrt{31.924} < \sqrt{36} \\ 5 < \sqrt{31.924} < 6 \\ \text{Try } 5.6^2 = 31.36 \\ 5.6 < \sqrt{31.924} < 6 \\ \text{Try } 5.7^2 = 32.49\\ 5.6 < \sqrt{31.924} < 5.7 $$

And now we observe that $31.924 - 31.36 = 0.564$ while $32.49 - 31.924 = 0.566$. Therefore $31.924$ is closer to $5.6^2$ than to $5.7^2$. But $\sqrt{31.924} \approx 5.65013$ rounded to one decimal place is $5.7.$

Here's an interesting twist, however: working with decimal numbers, the input number $x$ (that you are taking a square root of) must have at least three decimal places in order to set up an example in which $x$ is closer to $a^2$ than to $b^2$ (where $a$ and $b$ are two consecutive one-decimal-digit numbers) and yet $\sqrt{x}$ is closer to $b$ than to $a.$

Let $a = n/10,$ where $n$ is an integer, and let $b = a + 0.1 = (n+1)/10.$ Then $a^2 = n^2/100$ and $$b^2 = \frac{(n+1)^2}{100} = a^2 + \frac{2n+1}{100}.$$ Consider all the two-place decimal numbers between $a^2$ and $b^2$; all such numbers up to $a^2 + \frac{n}{100}$ are closer to $a^2,$ and all such numbers from $a^2 + \frac{n+1}{100}$ upwards are closer to $b^2.$ But also $$ (a + 0.05)^2 = \frac{\left(n+ \frac12\right)^2}{100} = \frac{n^2 + n + \frac14}{100} = a^2 + \frac{n + \frac14}{100}.$$ So the square roots of all numbers from $a^2$ to $a^2 + \frac{n}{100}$ are closer to $a$ than to $b$ and the square roots of all numbers from $a^2 + \frac{n+1}{100}$ to $b^2$ are closer to $b$ than to $a.$ The only case in which it is incorrect to use the "closest square" method to decide which way to round $\sqrt{x}$ is when $$ a^2 + \frac{n + \frac14}{100} \leq x < a^2 + \frac{n + \frac12}{100}.$$ And it takes at least three decimal places to write such a number $x$ in decimal notation.

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  • $\begingroup$ This is great David the big problem I had was finding a counter example with an integer and you've shown that it must be a number to 3d.p. for it to occur. Thanks! $\endgroup$
    – george
    Nov 20, 2021 at 21:31
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The mathematical question seems to be: if $a^2 < t < b^2$ and $t$ is closer to $a^2$ than it is to $b^2$, is it automatically true that $\sqrt t$ is closer to $a$ than it is to $b$? The answer is no, because taking square roots does not preserve distances—more precisely, because $\sqrt x$ is a concave function, it shrinks distances to the right more than it does to the left (consider $16-9.61=6.39>5.61=9.61-4$ yet $\sqrt{16}-\sqrt{9.61}=0.9<1.1=\sqrt{9.61}-\sqrt4$).

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  • $\begingroup$ Thanks for the counter example. I had originally been looking for a counter example with a < sqrt(x) < a+0.1 but @David K has shown that this can not occur $\endgroup$
    – george
    Nov 20, 2021 at 21:33
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As an aid to the various theoretical arguments, an example.

Consider $\sqrt{2.403}$. We quickly find $1.5^2 = 2.25$ and $1.6^2 = 2.56$. We compute \begin{align*} 2.403 - 2.25 &= 0.153 \\ 2.56 - 2.403 &= 0.157 \text{,} \end{align*} so $2.25$ is closer. If these differences were linear in the differences of the roots, we would expect $1.55^2 > 2.403$ and therefore we would report $1.5$ is the square root to one decimal place.

However, $1.55^2 = 2.4015 < 2.403$, so to one decimal place, we get $1.6$ for the square root.

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  • $\begingroup$ Thanks for the counter example. I had originally been looking for a counter example with a < sqrt(x) < a+0.1 but @David K has shown that this can not occur $\endgroup$
    – george
    Nov 20, 2021 at 21:32
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You can use this approximation for $a,b$ integers.

$$\sqrt{a^2+b}=\left(a+\frac b{2a}\right)-\frac{\left(\frac b{2a}\right)^2}{2\left(a+\frac b{2a}\right)}$$

See here for details: https://math.stackexchange.com/a/2866233/399263.

This approximation is a Taylor expansion of order $1$ followed by one step of Newton's method.

Note that for $\sqrt{x}$ for $x$ real you can multiply by a big suitable square $n^2$ and calculate the square root of the integer closest to $n^2x$ so that the result is still good at 1 decimal place.

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  • $\begingroup$ Thank you but I'm not looking for alternative methods, as this method is part of the syllabus $\endgroup$
    – george
    Nov 20, 2021 at 21:28

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