We have x=0.2234 and Δx =0.2*10^-3.Find how many accurate digits does the number have?

Question

So we know that $Δx≤(1/2)*10^{-n}$ where n is the number of the accurate digits. Now I just have to replace $Δx =0.2*10^-3$=$(1/2)*10^{-n}$ and find n. But why have I been given x=0,2234 since I dont need it?

Answer 1

You might be worried about carries. If we had $x=0.1999, \Delta x=0.0002$, even the first digit might change. Given your $x$ it is not a problem.

Added: We are given a measurement of $0.2234 \pm 0.0002$, as that is what $\Delta x$ usually means, the possible error in measuring $x$. We know the range is $0.2232$ to $0.2236$. That shows the first three digits are accurate, as they do not change. In my example, the range would be from $0.1997$ to $0.2001$

Answer 2

You say that $\Delta x\le \frac12\cdot 10^{-n}$ where $n$ is the number of accurate digits. You are given $\Delta x=0.2\cdot 10^{-3}$. Thus, $0.2\cdot 10^{-3}\le 0.5\cdot 10^{-n}$. Assuming that $n$ is an integer, then your $n$ could be $3$, or $2$, or $1$, or even $0$. You need to give a more precise definition of accurate if you are looking for a specific value of $n$. Are you saying that $n$ is the largest integer that satisfies the inequality?