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If a caliper takes the measurement to the 2 decimal points then is it not a correct measure to the 2 decimal points or is it still an approximation?

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  • $\begingroup$ It is an approximation to the true value of the measured property, which may have infinite decimal places. $\endgroup$ – John Rennie Oct 12 '17 at 8:50
  • $\begingroup$ @JohnRennie If that is the case then we can never find the true value of any quantity.we can only find the approximated value to the accuracy we want.am i right? $\endgroup$ – Remy Oct 12 '17 at 9:01
  • $\begingroup$ Yes, that's correct $\endgroup$ – John Rennie Oct 12 '17 at 9:04
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Provided what you are measuring does not yield an "integer" result(s), we call the majority of measurements we take in Physics approximations, as the result(s) may be irrational, (i.e. a value with infinitely many decimal spaces) which is, in most cases, what the result(s) is.

In the case of your example, the remaining decimal spaces following the $2$ provided by the calliper are rounded.

Think of it as follows, if I had some huge number, say $4.48291\times10^{42}$ and I add one to this value, will it significantly impact this value? Given that the two values are so closely related (proportionally), hopefully you said no. And this same logic applies to very small numbers. For example, when you add $1\times10^{-84}$ to $4.48291\times10^{-42}$, it makes such an insignificant difference that you can in fact round the result, while maintaining a suitable degree of accuracy, to actually "discard" the $1\times10^{-84}$ that you've added. This is what results in an approximation, as the rounded value is not exact but provides a value with a suitable enough degree of accuracy to be representative of the exact result.

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  • $\begingroup$ Of course, I overlooked this in my initial review. Thank you! $\endgroup$ – joshuaheckroodt Oct 12 '17 at 11:35
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    $\begingroup$ so if the measurement yield an integer then it is a true value not an approximation ??? $\endgroup$ – Remy Oct 12 '17 at 15:23
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    $\begingroup$ Yes, when the measurement is done using appropriate tools, and the result is an integer, then $90$ percent of the time, the result is the true value $\endgroup$ – joshuaheckroodt Oct 12 '17 at 15:25
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    $\begingroup$ we try to meet the accuracy we need in the approximation whether the value is true or approximation dose not matter as long as we have achieved the accuracy we needed ? $\endgroup$ – Remy Oct 12 '17 at 15:35
  • $\begingroup$ Exactly, I couldn't have said it better myself @Remy $\endgroup$ – joshuaheckroodt Oct 12 '17 at 15:59

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