<s>[INST] Calculating square roots to 1 decimal place 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.
 [/INST] 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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