# Round Off Error Of One Digit Integer

Find the relative error of the function $$f=\frac{a+2b}{c}$$ given the numbers $$a=1,b=1,c=2$$ that were round up to a one digit integer

So intuitively if we take for example $$4.2$$ and round it to $$4$$ we change the $$10^{-1}$$ digit and if it was between $$0.0$$ to $$0.5$$ we will round it to $$4$$ else if it was between $$0.5$$ to $$0.9$$ we will round it to $$5$$ so it is "half the way" or $$\frac{1}{2}\cdot 10^{-1}$$

1. I do not know if this intuitively explanation is correct.

2. There is a formula $$\frac{1}{2}\beta^{1-p}$$ where $$\beta$$ is the base and $$p$$ is the precision or the significant digits, why in this case it is $$2$$?

• Is this a part of the context where, for example, $c=a+b$? Please state so, if that is the case. (I suspect there is more context to this, because there isn't much to learn/say about three numbers that wouldn't already be valid for one number.) Dec 31, 2019 at 10:33
• @StinkingBishop I have written the full question Dec 31, 2019 at 10:36
• But writing like that $f$ is not a function but a number.. $f = 3/2$ Dec 31, 2019 at 10:38

$$\varepsilon_f \approx \frac{a}{a+2b} \varepsilon_a + \frac{2b}{a+2b}\varepsilon_b - \varepsilon_c$$
So, for $$a=b=1$$ and $$c=2$$, the relative error is approximately given by
$$\frac 13 \varepsilon_a + \frac 23 \varepsilon_b - \varepsilon_c.$$