# Significant digits rule when determining radius from diameter

vcsharp2003
Homework Statement:
If ##r## and ##d## are radius and diameter of a given circle, then to determine radius from diameter we use the formula ##r = d \div 2##. Suppose, ##d=5## then ##r=5 \div 2 = 2.5##. I have a question regarding significant digits in this calculation. We know the significant digits rule for dividing two numbers is that the resulting value must have as many significant digits as the minimum number of significant digits of dividend and divisor. In this case the minimum number of significant digits is 1 for dividend or the divisor, and therefore why we don't apply this rule and express the answer as a number up to 1 significant digit?
Relevant Equations:
##r = d \div 2##
Probably, to satisfy the significant digits rule for division, we should consider ##r = 5.0 \div 2.0##. But I'm unable to come up with a reason why significant digits rule should not apply to ##r= d \div 2##. Also, if we apply significant digits rule to this calculation then we loose accuracy and so we should ignore the significant digits rule.

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Homework Helper
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the significant digits rule for dividing two numbers is that the resulting value must have as many significant digits as the minimum number of significant digits of dividend and divisor.
It is not a precise science. Common sense must be applied.
if we apply significant digits rule to this calculation then we loose accuracy and so we should ignore the significant digits rule.
Quite.

• vcsharp2003
Homework Helper
Hi,

The golden rule for error propagation is $$\Bigl (\Delta f(x,y)\Bigr )^2 = \left({\partial f\over\partial x}\right )^2(\Delta x)^2 + \left({\partial f\over\partial y}\right )^2(\Delta y)^2\ \ .$$
In the case ##\ r = d/2\ ## there is no uncertainty in the factor 2, so we have $$\Delta r = {\Delta d\over 2}$$the first example formula here.

##\ ##

• vcsharp2003
vcsharp2003
In the case r=d/2 there is no uncertainty in the factor 2
How does this fact imply that significant digits rule should not be applied to given scenario?

vcsharp2003
It is not a precise science. Common sense must be applied.
We can say that accuracy of r is same as accuracy of d since 2 is an exact number. Now, if we assume that ##d## is accurate then ##d \div 2## must also be accurate and we don't need to apply significant digits rule to accurate numbers. Only if there is some uncertainty in values of dividend and divisor, do we need to apply significant digits rule. Does this reasoning sound correct?

Homework Helper
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We can say that accuracy of r is same as accuracy of d since 2 is an exact number. Now, if we assume that ##d## is accurate then ##d \div 2## must also be accurate and we don't need to apply significant digits rule to accurate numbers. Only if there is some uncertainty in values of dividend and divisor, do we need to apply significant digits rule. Does this reasoning sound correct?
It depends on how the value of d is known. If it is given as 5cm in a made-up question you can take it to be exact. If someone measured it and stated the measurement as 5cm then you have a problem. Were they really not able to measure it more precisely than that, or are they just being lazy in not specifying it as 5.0cm, or whatever.

Taking it at face value, we are being told it is between 4.5cm and 5.5cm, so r is between 2.25cm and 2.75cm. If we try to express that by using an appropriate number of digits we have the choice of 2.5cm, implying it is between 2.45cm and 2.55cm, or 3cm, implying it is between 2.5cm and 3.5cm. Neither is satisfactory.
The only solution here is to state it as ##2.50\pm 0.25cm##.

Implying the precision from the number of digits works better when there are more of them to play with. If we start with d=1.005cm then it might not matter if we take r to be 0.503cm.

• vcsharp2003
vcsharp2003
I think the rules of significant digits should only be applied when dealing with measurements, otherwise they have no relevance and should not be applied.

For example, if I ask someone to tell the result of ##1 \div 3## then there is absolutely no need to apply significant digits since its just a Math calculation and not related to measurements. As a student reading the chapter on significant digits, I was always curious when these rules should be used and when they should not.

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