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I keep getting this question in my GCSE papers, but I have no idea how to solve it, and everywhere I look there doesn't seem to be a simple answer. The general question goes like this:

$$v=\sqrt{\frac{a}{b}}$$

$a = 6.43$ correct to 2 decimal places.

$b = 5.514$ correct to 3 decimal places.

By considering bounds, work out the value to $v$ to a suitable degree of accuracy.

(Sorry about the tagging, not sure what this fitted into)

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Since $a=6.43$ to two decimal places, $a$ lies between $6.425$ and $6.435$. Similarly, $b$ lies between $5.5135$ and $5.5145$. The smallest possible value of $a/b$ occurs when $a$ is as small as possible and $b$ as large as possible; the largest possible value of $a/b$ occurs when $a$ is as large as possible and $b$ as small as possible. Thus, $$\frac{6.425}{5.5145}\le \frac{a}b\le \frac{6.435}{5.5135},$$ and $$\sqrt{\frac{6.425}{5.5145}}\le v\le \sqrt{\frac{6.435}{5.5135}}\;.$$ These bounds on $v$ are approximately $1.07940$ and $1.08034$, so we know that $v$ is between $1.075$ and $1.085$ and hence that $v=1.08$ is correct to two decimal places. Can we go one place further? The best approximation of $v$ to three decimal places is clearly $1.080$, but it isn’t correct to three places, because $v$ isn’t guaranteed to be between $1.0795$ and $1.0805$: $v$ could be just a hair under $1.0795$.

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  • $\begingroup$ I'm confused, why are there $ signs everywhere? $\endgroup$ – Derek Nov 2 '11 at 7:30
  • $\begingroup$ @Deza: If you’re seeing dollar signs, try refreshing the page; the $\LaTeX$ didn’t load properly. $\endgroup$ – Brian M. Scott Nov 2 '11 at 8:51
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The first thing to do is work out the error interval (bounds) for a and b:

6.425 ≤ a < 6.435

5.5135 ≤ b < 5.5145

As we want to work out v by considering these bounds, we have to find both the upper and lower limit of the value of v. To find the upper value, we must take the highest possible value of a and divide it by the lowest possible value of b. To find the lower value, we must take the lowest possible value of a and divide it by the highest possible value of b.

Upper bound of v = √(6.435 ÷ 5.5135) = 1.080340323

Lower bound of v = √(6.425 ÷ 5.5145) = 1.079402689

Now to find out what the suitable degree of accuracy is, you find the most accurate value that both bounds round to. It sounds complicated, but it’s easy once you understand it.

To one digit, they both round to 1. To one decimal place, they both round to 1.1. To two decimal places, they both round to 1.08. To three decimal places, the upper value rounds to 1.080, however the lower value rounds to 1.079. These values are now different, therefore the most accurate value that you can take is 1.08.

So v = 1.08 because all values round to 1.08 ( 2dp).

I hope this helps anyone reading this! I am also studying for GCSEs and I have a feeling the other responses on this question are looking at it from too advanced a level. My answer will get you full marks on this question at GCSE. Good luck to everyone! X

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Let us denote $a' = 6.43,b' = 5.514$ and $a = a'+\delta_1$ and $b = b'+\delta_2$ with $\delta_1\leq 0.01$ and $\delta_2\leq 0.001$. We know that $$ v = \sqrt{\frac ab}. $$

Let us denote $v' = \sqrt{\frac{a'}{b'}}$ and put $\delta = v-v'$. Note that $v,v'\geq 1$ then $$ |\delta| = |v-v'| = \frac{|v^2-v'^2|}{v+v'} \leq\frac12|v^2-v'^2| = \frac 12\left|\frac{a'+\delta_1}{b'+\delta_2}-\frac{a'}{b'}\right| $$ $$ =\frac 12\left|\frac{(a'+\delta_1)b' - a'(b'+\delta_2)}{b'(b'+\delta_2)}\right|\leq \frac12\cdot\frac{\delta_1}{5}\leq 0.001 $$ so you have $v\approx \sqrt{\frac{a'}{b'}}= 1.07987...$ with an error smaller than $0.001$ or up to 3 decimal places.

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  • $\begingroup$ Second day, second downvote on the old answer without justifying the reason. Who are you, mysterious brave guy? $\endgroup$ – Ilya Apr 19 '12 at 22:10

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