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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?

fsdd
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  • Reposted at http://math.stackexchange.com/q/778009/18398 – JRN May 02 '14 at 06:54
  • What is your definition of accurate digits? We cannot answer your question if we don't know how you define your terms. – JRN May 02 '14 at 07:13
  • Accurate means correct.. – fsdd May 02 '14 at 07:19
  • Okay, for $x=0.2234$, how many digits are correct? – JRN May 02 '14 at 07:21
  • Yes thats the question – fsdd May 02 '14 at 07:25
  • You say "we know that $\Delta x\le \frac12\cdot 10^{-n}$ where $n$ is the number of the accurate digits." What does $\Delta x$ mean? – JRN May 02 '14 at 07:29
  • absolute value of x – fsdd May 02 '14 at 07:32
  • But if $x=0.2234$, then wouldn't its absolute value be $0.2234$? I think you're misunderstanding what $\Delta x$ is and it seems that's why you can't answer the question. I suggest that you review all your definitions and concepts again. – JRN May 02 '14 at 07:36
  • Sorry x is a value close to the value that we want to find. Δx is the absolute error of a value close to the original value x.They would be related this way: Δ(Absolute error)=|X-x|≤Δx. – fsdd May 02 '14 at 07:48
  • X is the original value – fsdd May 02 '14 at 07:48

2 Answers2

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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 $0.2001$

Ross Millikan
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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?

JRN
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  • If so, then $n=3$ and you would have three accurate digits. This is consistent with Ross Millikan's answer. – JRN May 02 '14 at 13:13