## Significant Figures

Your calculator probably gives you answers to twelve or fourteen digits. But those are usually not all reliable.

Every measurment has some

uncertaintyassociated with it.How long is this block? With the equipment (the ruler) and your skill, you may decide that it is

L _{1}= 1.31 mlong. With better equipment, you might be able to measure it more accurately. But you know it is between 1.3 m and 1.4 m in length and you decide it is 1.31 m long. To indicate that digits beyond this are uncertain, we might write

L _{1}= 1.31????? mNow measure the length of another block,

This time we have better equipment--perhaps just a ruler with better markings. Now we decide that the length of this block is

L _{2}= 0.548 mThat means we know its length is between 0.54 m and 0.55 m. To indicate that digits beyond this are uncertain, we might write

L _{2}= 0.548???? mWhat is the length of these two block placed together?

Of course, we "know" the total length is just the sum of the two lengths,

L _{tot}= L_{1}+ L_{2}But how do we handle the fact that the two blocks' lengths are know with different uncertainty or with different accuracies? Write the addition with the trailing question marks to understand this better,

While the final digit 8 or .008 in L

_{2}= 0.548 m may be quite well known, the corresponding part of L_{1}'s measurement isnotwell known at all. Only the final 1 or 0.01 in L_{1}= 1.31 m is known. That means the third digit beyond the decimal in L_{tot}is uncertain and, therefore, meaningless. So we write it only as L_{tot}= 1.85 m.This prototype addition problem provides the generalization we need for significant figures in adding and subtracting.

In addition and subtraction,the number of decimal places in the answer is equal to the smallest number of decimal places in any of the terms being added or subtracted.What about multiplication and division?

Again, let us start with block number one,

With the equipment (the ruler) and your skill, you decide, as before, that it is

L = 1.31 m long. You know it is between 1.3 m and 1.4 m in length and you decide it is 1.31 m long. To indicate that digits beyond this are uncertain, we might write

L = 1.31????? m We are going to find the

areaof this block, so measure its width.Adn we might determine that its width is

W = 0.2345 m with our better equipment, skill, and care.

Now what is the area?

We know that area is length multiplied by width,

A = L W So, if we just enter that into a calculater we find,

A = L W = (1.31 m)(0.2345 m)

A = 0.307 195 m ^{2}But is that reasonable? Do we really know the area that accurately? Carry out the multiplication by hand,

and we find that the calculator is, indeed, correct. Now, as before, let's add question marks to these number to indicate which digits are uncertain and again carry out the mathematical operation by hand,

You might think that we should have A = 0.307 1?? m

^{2}. But the mathematics is even "worse" than that. Because we don't know what 4 + 5 + ? is in the long-hand multiplication above, we do not know that the next digit, from 3 + 3 + 5 is really going to be the 1 from 11. So that digit, too, is uncertain, as we have written above.Again, we will generalize from this prototype for multiplication (and division).

When

multiplying(ordividing) quantities, thenumberof significant figures in the answer is equal to thesmallestnumberof significant figures in the quantities.In our example, the length L has

threesignificant figures while the width W hasfoursignificant figures. The answer, the area A, therefore, is known only tothreesignificant figures.Calculators provide a vast array of digits. Do not be misled. Remember,

GIGO(Garbage In, Garbage Out).Youmust determine which of those figures from your calculator is significant.

Significant Figures Return to Table of Contents, Ch 1 Introduction(c) Doug Davis, 2001; all rights reserved