What is the meaning of $1$ in a relative error?

If we measure a length and is measured as $$12.5$$ meters long, accurate to $$0.1$$ of a meter this means the absolute error is $$0.05$$m.
The relative error is: $$\frac{0.05}{12.5} = 0.004$$. This means that the measurement is accurate to $$\frac{4}{1000}$$ or $$0.4$$%. or equivalently that for each unit we measure we introduce an error of $$0.004$$.

Now if we compare the $$10^{-50}$$ with $$10^{-6}$$ then the absolute error is: $$|10^{-50} - 10^{-6}| = 10^{-6}$$.
The relative error then is $$\frac{absolute\space error}{measured\space value} = \frac{10^{-6}}{10^{-6}} = 1$$

What is the meaning of $$1$$ here? How is it interpreted and how from a small absolute difference we get something that seems to indicate $$100$$% error?

• What are the $10^{-50}$ and $10^{-6}$? If the measured value is $10^{-6}$ but the lowest possible value is $10^{-50}$, then the relative error may really be that large. Commented Nov 7, 2021 at 0:16
• @peterwhy: the $10^{-50}$ should be the actual value and $10^{-6}$ the measured/estimated value
– Jim
Commented Nov 7, 2021 at 0:17
• Then the absolute error, or that absolute difference between the actual and measured values, are almost $100\%$ of the measured value and not small relative to the measured value. Commented Nov 7, 2021 at 0:21
• @peterwhy: that helps thank you
– Jim
Commented Nov 8, 2021 at 17:38

Briefly, when the absolute value of the relative error exceeds unity, you can no longer trust the sign. If $$T$$ is the target value and $$A$$ is the approximation, then the absolute error is $$E = T-A$$ and the relative error is $$R=\frac{E}{T}.$$ It follows that $$A = T - (T-A) = T - \frac{T-A}{T}T = T-RT= T(1-R).$$ Typically, we do not know the relative error, but we have an upper bound for the absolute value. Now if $$|R|<1,$$ then $$A$$ and $$T$$ have the same sign. If $$|R|\geq1$$, then it is entirely possible that $$A$$ and $$T$$ have different sign.

Computing the correct sign is critical in root finding applications. If $$f : \mathbb{R} \rightarrow\mathbb{R}$$ is continuous and $$f(x_1)$$ and $$f(x_2)$$ have different signs, then $$f$$ has a zero in the interval between $$x_1$$ and $$x_2$$. If we cannot trust the computed value of the sign of $$f(x_i)$$, then we cannot make this determination with certainty.

• In this case though $R= 0$ right?
– Jim
Commented Nov 7, 2021 at 10:27
• @Jim: To what case are you referring? Commented Nov 7, 2021 at 13:12
• In my post, the relative error is $1$. So $A = T$ in your formula right?
– Jim
Commented Nov 7, 2021 at 15:59
• @Jim. No, in your post you do not have $R=0$ because $10^{-50} \not =0$ and in general $T \not =A$. Commented Nov 7, 2021 at 20:23
• The relative error in the post is $1$. So based on your formula $A = T(1 - 1) = T\cdot 0 = A$. What does this mean?
– Jim
Commented Nov 7, 2021 at 22:04