For example, something measured as 3 inches isn't 3 inches. It may be close to 3 inches but never exactly 3 inches. You could say it was more like 3 and 1/16 inches. But even that is not exact. It is only accurate to the nearest 1/16 inch. 
Example 2: A cake recipe calls for 1 quart of milk. You know that 4 cups equals 1 quart, so you measure out 4 cups. Do you have a quart? In theory you should, but measurements are not exact, so you will not have exactly a quart. You are close enough for baking a cake, but for more critical recipes, the inaccuracies can cause problems.
Even if you measured a quart directly without measuring four cups, you still could not get exactly one quart. You can only get as close as the measuring tools and your ability to read them allow. 
The Hubble telescope is a good example. Engineering had the correct theoretical sizes needed for the curvature of the mirror that collects and focuses the light. Measuring the curvature of the mirror was another matter. The builders of the mirror made large measurement mistakes that caused the mirror to have a blurred image. This required a second mirror to correct the previous measurement mistakes. The picture below illustrates before and after images from the telescope. 
Measurement is important to everyone: Athletes, artists, aviators, architects, accountants, archeologists,... and that's just the "A's" 
A measurement, however, is only good if you know its accuracy. Unfortunately, the accuracy of a measurement isn't always made clear. The purpose of the following exercises is to teach you how to make the accuracy of your measurements clear. 
Measurements reported as 1 decimeter, 1.0 decimeters, and 1.00 decimeters may seem the same, but their accuracies are very different. [Note: a decimeter is one tenth of a meter. Sizes shown are close to actual sizes.]  
1 decimeter implies that it could actually range from a half (0.5) decimeter to just under 1.5 decimeters. It's been rounded to the nearest whole decimeter.  
The ".0" in 1.0 decimeter implies accuracy to the nearest tenth. The actual length could range from 0.95 decimeters (which got rounded up to 1.0) to just under 1.05 decimeters (which got rounded down to 1.0)  
The ".00" in 1.00 decimeter implies accuracy to the nearest hundredth. The actual length could range from 0.995 decimeters (which got rounded up to 1.00) to just under 1.005 decimeters (which got rounded down to 1.0)  
As you can see, measurements that include more decimal places increase the accuracy tremendously. 
Which of the following measurements is more accurate? 
If you were a carpenter and a customer called and said he wanted a floortoceiling bookshelf and the height of the wall was 8 feet, what would you do?  
A. Build a bookshelve to as close to 8 feet as you could.  
B. Build a resizable bookcase that could stretch from 6 feet to 10 feet.  
C. Ask the customer to clarify his accuracy, by measuring to the nearest 1/4".  
D. Go measure the height yourself.  
An atlas says the distance from Phoenix to Honolulu is 3,300 miles. You also learned in your geography class that continental drift causes the US to separate from the Hawaii at 2 centimeters per year. How far will Phoenix be from Honolulu in 10 years?  
A. 3,300 miles plus 20 centimeters.  
B. Still 3,300 miles.  
C. Convert 3,300 miles to centimeters, then add 20.  
A newspaper reports that a certain basketball player will make a total of 12 million dollars next year from playing and from endorsement of products. The player said he will donate exactly $50,000 to charity. How much will he have left?  
A. $11,950,000.  
B. Still 12 million dollars.  
C. Between $12,450,000 and $11,450,000.  
D. Around 12 million.  


A celebrity's accountant reports that his client made $17,245,820.28 last year and gave $15.25 to his least favorite charity. How much did he have left after this specific charity donation?  
A. 17 million dollars.  
B. $17,245.805.03  
C. $15.25 is too small to bother.  

Bill Gates, who owned Microsoft, was reported in the newspaper to be worth 15 billion dollars. Let's say you buy Windows Vista software for $90 and Bill's share is two dollars. How much is he worth now?  
A. $15,000,000,002.  
B. Still 15 billion  
C. $2 is insignificant compared to billions.  

You tell your friend you can both park in your one car garage if the total length of your two cars is less than 35' 6 3/4". You measured your Jaguar at 18' 3 1/4". Your friend said his car was 17 feet in length. Should you close the garage door?  
A. Yes, their total length is only 35' and 3 1/4". That's 3 1/2" to spare.  
B. No.  

You just tested your pool and the results indicated the chlorine concentration at 50 ppm (parts per million, which means 50 chlorine atoms for every 1 million H20 molecules.) However, your friend spills his 12 ounces of beer into your 2,000 gallon pool. What is the concentration of chlorine now?  
A. Less than 50 ppm.  
B. Still 50 ppm. The extra 12 ounces is too small to affect reading.  
C. 50 ppm chlorine, 1 ppm alcohol.  
When adding or subtracting two amounts, does the following statement make sense?  
If the inaccuracy of one amount is larger than the other amount, don't bother in doing the addition or subtraction.  
For example, in the picture a measurement of '9 cm.' is only accurate to the nearest whole centimeter, which means it could vary from 8.5 cm. to just under 9.5 cm. (a range of 1 cm) The other number is .4 cm., which is smaller than the 1 centimeter variation. In other words, if we tried to add the two together to get 9.4, we are indicating our answer to be accurate the nearest tenth of a centimeter. However, we can't because our measurement of 9 cm. was not measured that accurately. 
Most textbooks teach significant figures with a set of rules. I don't like that approach because rules don't teach you why. You are just following the rules blindly. Not much learning happens there. Let's take a closer look at how you get significant numbers. 

The below image is that of a shoeprint at a crime scene. A ruler is superimposed to do a measurement between a corner of one pattern to the edge of another pattern. This measurement could be used to find the shoe that made the impression, which finds the person who left his or her shoeprint at the crime scene. Let's say the examiner wrote down the distance as 10 centimeters. Unfortunately, 10 cm is ambiguous. Was it 11 cm that got rounded down to 10 cm or was it 8 cm that got rounded up to 10cm? If the examiner wrote "10.", the decimal point means he or she measured it to the nearest centimeter. In other words, the measurement was less than 10.5 centimeters, but greater than 9.5 centimeters. If the examiner wrote 10.3 centimeters, it means it was measured to the nearest tenth of a centimeter. The figures in "10.3" are all significant. The "1" means there were about 10 centimeters of length. The "0" means there were no whole centimeters to add, and the ".3" means there were 3 tenths of a centimeter extra length. That's 3 significant figures. With a good eye an examiner can look between the two millimeters lines (tenths of a centimeter) and estimate that the edge was about half way in between those lines. That means he or she could write down the distance as "10.35" centimeters, which has 4 significant figures. It was very important to have enough accuracy (enough significant figures) to allow police officers to see if they had a shoe that matched this print.  
This is a digital balance like the one students use in my portable lab kit for CHM107. It measures to the nearest tenth of a gram. Let's say I had them weigh out some baking soda and it weighed 40.3 grams. That is 3 significant figures because each figure is contributing to the measurement. In other words, the "4" says there are 40 grams. The "0" says there where no single grams. And the ".3" says there were 3 tenths of a gram extra. They all contribute to the final weight of 40.3 grams. They all are significant figures. Even the zero in between 4 and 3 is giving us information. Now let's say we need to use this weight in a formula that needs the weight in kilograms. Turning 40.3 grams into kilograms gives us 0.0403 kilograms. Now we have two zeros in front of the 403. Does that mean the weight now has 5 significant figures? No, the measurement can't get more accurate by simply changing the units. Those zeros to the left of the 403 are not significant. They are only there to place the decimal point correctly when going from grams to kilograms. So these leading zeros are not counted as being significant.  
The above example showed how converting from grams to kilograms (40.3 g > 0.0403 kg) implied a change of accuracy, but you learned that the leading zeros are not counted. The same thing can happen going the other way. If we converted 40.3 grams to milligrams, the measurement becomes 40,300 milligrams. That looks like 5 significant numbers but, again, we can't get more accuracy by simply changing the unit of measurement. So those zeros after the 403 don't count. In that way we retain the same 3 significant figures.  
Let's do another example with a weight measurement. In a lab I say you need 50 grams of sugar to do the experiment. By saying "50 grams" I am implying that the sample can be between 45 and 55 grams. "50" has only one significant figure (5), which means its accurate only to the nearest 10 grams. Not realizing that, a student worked hard to get the sample to show 50.0 grams on the readout. That's 3 significant figures because it was measured to have 50 grams with zero single grams and zero tenths. So the zeros in "50.0" are giving real information about the weight. If the students simply reported they weighed out 50 grams, I could only assume it was rounded. By reporting "50.0 grams" I will assume they took the effort to measure it down to a tenth of a gram. Note: If a situation asked for 50.00 grams of sugar, this balance could not measure to that accuracy. So the point is 50 g, 50.0 g, and 50.00 g are not the same weights. In math, 50, 50.0, and 50.00 are all the same number. But when they are measurements, they are not.  
Using money is a good lesson in significant figures. Let's say you and a friend were going to buy tickets to a concert. The tickets are $30. You call a friend and ask how much money does he have on him. He glances in his wallet and tells you he has $30. You have $30.25 so you know you're good. When you meet him at the ticket window, you find out that he actually has only $29 and 35 cents. So you don't have the $60 to get the two tickets. When you heard your friend say $30, you can only assume that he rounded the money to the nearest $10. He may have had $26 or $34. $30 has only one significant figure. If he said $29, then you know he's rounding the amount to the nearest dollar. So he's between $28.50 and $29.50. However, if he told you he had $29 and 35 cents, he's giving you 4 significant figures. Also, if you told him you have $30 and 25 cents. He knows the $30 is not rounded because you are including the 25 cents. So $30.25 also has 4 significant figures. That zero that sits between 3 and 2 is an accurate measurement saying there are no single dollars.  
Significant number is calculations: 

Speaking of area, let's say you need carpet and you need the square feet. You measure the length to be 12.5 feet and your helper measured the width to be 11 feet. In other words, your helper only measured to the nearest foot. So your measurement had 3 significant figures because you measured to the nearest tenth of a foot. Your helper's measurement was only 2 significant figures. (note: I will call significant figures "sig figs" for short). To find area of the room, you multiply width times length. So that's 11ft x 12.5 ft =137.5 square feet. "137.5 sq. ft." implies that you know the area to the nearest tenth of a square foot. Also, "137.5 sq. ft." has 4 sig figs. The accuracy of your answer is only as good as the least accurate measurement. That's because the "11 ft" is between 10.5ft and 11.5 ft, meaning your multiplication could have been 10.5 ft x 12.5 ft = 131.25 sq. ft or 11.5 ft x 12.5 ft = 143.75 sq. ft. So you see there's a spread of about 12 sq. ft. difference between the two. So you can't just multiply 11ft x 12.5 ft and expect the 137.5 sq. ft. is accurate. So the way to give an answer that reflects the true accuracy, is to realize that "11 ft" only has 2 sig figs, so we round off the 137.5 sq. ft. to also have only 2 sig figs. That becomes 140 sq. ft. Someone seeing "140 sq. ft." would realize that means the accuracy is only to the nearest 10 sq. ft., which is a true reflection of what we know from these measurements. In other words, the area is between 135 sq. ft. and 145 sq. ft.  
Just out of high school I worked for a construction company that had a fuel storage tank that needed to be filled. The supervisor asked me to go get the measurements. I measured the height as 4.457 meters (that's to the nearest millimeter). It was hard to measure the diameter, so I called it 2 meters. Volume of a cylinder is V=height x pi x r^{2}. If diameter is 2 meters the radius is 1 meter. So the calculation was 4.457 meters x pi x (1 meter)^{2}. For pi I used 3.14. So the answer came out to be 13.99498 cubic meters. As you might guess, this shows an accuracy much more than can be justified. Even though the 4.457 meter measurement had 4 sig figs, the measurement of 2 meters was only 1 sig fig, so the answer would have to rounded to 1 sig fig. Therefore, "13.99498" is rounded to 10 cubic meters. That wasn't good enough for the supervisor. So I went out to measure the diameter again. This time I measured the circumference, which was easy to do. I got 6.753 meters as the circumference. The formula for circumference is C=pi x D. So to solve for diameter, I divided my circumference by pi. You can look up pi to any number of sig figs. Since my measurement of 6.753 meters had 4 sig figs, I used a value of pi that had at least 4 sig figs otherwise, it would cause the answer to be less accurate. So I used 3.1416 with 5 sig figs. So 6.753 meters divided by 3.1416 gave me a diameter of 2.14954 meters, which is a radius of 1.07477 meters. Using the volume formula again, I got V=(height) 4.457 meters x pi (3.1416) x radius^{2} (1.07477 meters)^{2}. V=16.17427 cubic meters. That should be rounded to 4 sig figs; however, I first want to convert that to gallons. So I look up the conversion of cubic meters to gallons and find 1 cubic meter = 264 gallons, but then found another source that had 1 cubic meter = 264.172052 gallons. The first only was accurate to 3 sig figs, but I wanted 4 sig figs or better, so I used 1 cubic meter = 264.17 gallons. So multiplying 16.17427 cubic meters times the 264.17 gallons per cubic meter gave me 4,272.76 gallons. That has 6 sig figs, which isn't justified. Our two measurements are good to 4 sig figs, so we can round our answer to 4 sig figs. So 4,272.76 gallons becomes 4,273 gallons, which we are confident is accurate to the nearest gallon. A fuel truck was ordered to bring 4,270 gallons, and we were confident it would fill the tank up without spilling over. Of course, then depends on them having the same degree of accuracy with their measurements.  
1. Zeros between numbers are actual measurements and are significant. (103, 0.0403, 2001, 0.002001, etc.) 2. Zeros to the right of whole numbers where there is no decimal point are NOT significant because those can result from unit conversion. (300, 20, 4000, etc.) 3. Zeros to the left of decimal fractions are NOT significant also because they can be from unit conversion (0.03, 0.0045, 0.678, 0.000201, etc.) 4. Any zero in a number written in scientific notation is significant except for the zero in the "x 10" (3.0x10^{2}, 4.100x10^{3}, 6.0780x10^{4}, 7.000x10^{20}, etc.) 
To summarize, we see that zeros between figures are significant. For example, 3004 has 4 sig figs because all figures including the zeros were actually measured. But 3000 has only one sig fig because the zeros could be there simply from a conversion of one unit to another. In 0.00450 the zeros on the left are not significant because they could be there from a conversion to another unit; however, the zero on the right is indicating that a measurement was taken to one more level below the "5". So 0.00450 has 3 significant figures. 
I found a website that has a calculator that gives the answer to the correct number of signficant figures. It also has some interactive tutorials. Below is what the calculator looks like. The blue "interactive tutorials" is the link to the tutorials. Check out this URL: http://sigfigscalculator.appspot.com/ 
In the 11th Edition textbook read pages 15 to 22 in Chapter 2 regarding measurements and significant figures. 
In the 12th Edition textbook read pages 15 to 22 in Chapter 2 regarding measurements and significant figures. 
In the 13th Edition textbook read pages 15 to 21 in Chapter 2 regarding measurements and significant figures. 
In the 14th Edition textbook read pages 15 to 21 in Chapter 2 regarding measurements and significant figures. 