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Assume that $a = a_m \times 10^m + \dots > 0$ is an approximation of $A$ with $n$ correct significant digits. Prove that: $$ \delta(a) \leq \frac{5 \times 10^{-n}}{a_m} $$ I know that if $a$ has $n$ correct significant digits then $$ \delta(a) < 5 \times 10^{-n} $$ Since $1 \leq a_m \leq 9$ then $$ \frac{5 \times 10^{-n}}{a_m} \leq 5 \times 10^{-n} $$ So my guess is that when $a_m > 1$ since we can rewrite the above equation as $$ \frac{5 \times 10^{-n}}{a_m} < 5 \times 10^{-n} $$ I have to show that $$\delta(a) = \frac{5 \times 10^{-n}}{a_m}$$ could be true but I have no idea how to show that

Mathemagician
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If $n$ digits are correct without rounding ​ \begin{align} \delta(a)=\frac{|a-A|}{|A|}\le\frac{10^{m-n+1}}{a_m10^m+a_{m-1}10^{m-1}+\dots}\le\frac{10^{m-n+1}}{a_m10^m}=\frac{10^{-n+1}}{a_m} \end{align} If $n$ digits are correct with rounding ​you get a half of it \begin{align} \delta(a)=\frac{0.5\times10^{-n+1}}{a_m}=\frac{5\times10^{-n}}{a_m} \end{align} Example: $A=0.114,a=0.11,n=2,m=-1,a_m=1,\delta(a)=\frac{0.004}{0.114}\approx0.035\le\frac{5\times10^{-n}}{a_m}=0.05$.

Vítězslav Štembera
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  • If $n$ digits are correct with rounding ​you get a half of it. I don't get why it happens. – Mathemagician Jan 14 '22 at 11:25
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    If you have two digits correct with rounding, for example $A=0.114, a=0.11$, then difference is maximally half of the digit, i.e. $0.005$. If you have two digits correct without rounding, for example $A=0.119, a=0.11$, then difference is maximally the full digit, i.e. $0.01$. – Vítězslav Štembera Jan 14 '22 at 11:47