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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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2 Answers

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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I'm confused, why are there $ signs everywhere? –  Deza Nov 2 '11 at 7:30
    
@Deza: If you’re seeing dollar signs, try refreshing the page; the $\LaTeX$ didn’t load properly. –  Brian M. Scott Nov 2 '11 at 8:51
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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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Second day, second downvote on the old answer without justifying the reason. Who are you, mysterious brave guy? –  Ilya Apr 19 '12 at 22:10
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