# By considering bounds, work out V to a suitable degree of accuracy

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 $1.0795$.
I'm confused, why are there $signs everywhere? – Derek Nov 2 '11 at 7:30 @Deza: If you’re seeing dollar signs, try refreshing the page; the$\LaTeX$didn’t load properly. – Brian M. Scott Nov 2 '11 at 8:51 Let us denote$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.