 Okay, so this lecture is designed to tell you a little bit about significant figures, or sig figs, which is something that you probably think you know something about, but the intent is to teach you something if you're the type of person who knows a little bit about sig figs and thinks about them when told to, but doesn't usually spend too much time thinking about them. The main reason for this is because, number one, sig figs do matter in certain types of problems, and chances are you're doing your sig fig rules wrong some of the time. So the basic idea of sig figs is that they are a way of expressing the certainty or uncertainty in a number. So for example, if you have a number 1.234, the fact that you've given three digits past the decimal, but no more or no fewer than that means that you are implying that you know that number to three digits past the decimal. So that third digit past the decimal is the last one in which you're certain. Likewise, if the number is just 1.2, you're implying that you only know that number to one digit past the decimal. The reason why we need to care about sig figs, especially if we're in an experimental situation or real world cases, is if you give more sig figs than you need or more than are justified in the real world that can have expensive consequences. For example, consider you need to weigh out some amount of some chemical. If you quote the weight to a tenth of a gram, then that's no problem. You can probably measure out to the tenth of a gram on an ordinary kitchen scale, but if you need an amount that's accurate out to the tenth of a milligram, then you're going to need an expensive analytical balance. Likewise, if you tell a machine shop that you want a part machine to within a tenth of an inch tolerance, that's not much of an issue. But if you want a machine to a thousandth of an inch tolerance, then they're going to need much more expensive equipment to do that. Likewise, using too few sig figs can also cause problems. If you have a solution whose concentration you need to know to the hundredth of a molar or an amount of substance that you need to know to the hundredth of a gram and you round off more aggressively than you should, then you end up with incorrect concentrations or amounts of material that can affect what happens in a chemical reaction. So start with some easy questions. How many sig figs are there in this number 8.314? That's probably an easy question if you've seen sig figs before. You count the number of digits. There's four sig figs in that answer. So sig figs are one of a couple of different ways of measuring the precision of a number. We can say that the absolute precision of that number is one one thousandth. We know the the thousandth digit of that number and no more. The relative precision is one one thousandth divided by the magnitude of the number. So that works out to be roughly one part in ten to the fourth. So if I if I really wanted to be accurate about the relative precision, I would say one part in 8.314. But roughly speaking, that's one part in about 10,000. One part in ten to the fourth. And then that number four is where we get four sig figs. Essentially that we assume that any number with four sig figs has the same relative precision as any other, which is an approximation. But it's what we typically do when we're counting sig figs. Only slightly trickier question. How many sig figs are there in 0.08206? Here we can't get away with just counting digits. We have to count only the significant figures. That's what the word significant means in significant figures. The leading zeros after the decimal point are not significant. And we only count the four digits 8206. And that's more obvious if we write the number in scientific notation. 8.206 times ten to the minus two has four sig figs. And the leading zeros just tell us how many powers of ten we have to include in the scientific notation. So again, the absolute precision of this number, we know that number to five digits past the decimal. The relative precision is 0.0001 divided by the magnitude of the number. Again, that comes out to be roughly one part in ten to the fourth. So that number also has four sig figs. So when this becomes an issue is when we need to do calculations and keep the appropriate number of sig figs. So if we have an addition problem, let's say 5.1 plus 12.76, plug that into your calculator and your calculator will give you a number. But as you know, the number of digits your calculator spits out is not always the number that you need to keep. So where should we round that value off to? If we think of that as an addition problem, we're adding 5.1 to 12.76. The hundredth digit, however, we can't do the addition because we don't know how many hundredths are in that number 5.1. So the answer that we get 17.8 something, we don't know the hundredth digit. And the hundredth digit is uncertain, so we can't keep it. The last digit we know anything about is the tenth digit. So we have to round that number off, keeping only the tenth digit and express it as 17.9. So the rule to remember, if you want to remember a rule, is that the absolute precision of the numbers that you're adding or subtracting determine the absolute precision of the final answer. Whichever number has, whichever the input numbers has the smallest number of digits past the decimal determines how many digits past the decimal there is in the final answer. Multiplication and division is a little bit different. So if we have this multiplication problem, 52.35 times 6.9, the calculator tells us 361.215. And that's clearly a lot of digits. Where would we round that number off? If we're uncertain in the values we're multiplying. This is the rule that you probably remember and use if you ever think to use a rule at all, which is count the sig figs. 52.35 has four sig figs. 6.9 has only two sig figs. So we keep only two sig figs in the answer and we round off to 360. So two comments here. Number one, this might be the type of problem where you're a little uncertain and you don't believe, you don't trust the sig fig rules. So you might keep more digits than just simply rounding off to 360. But that is the right thing to do. For example, if we are genuinely uncertain in that number 6.9 and the last digit could vary, it could be 6.8. Then when we vary that number in the last digit, our calculator will tell us that the answer begins to vary in the second digit. So again, two sig figs of accuracy in the inputs leads to only two sig figs worth of accuracy in the output. Likewise, if we vary the other direction, we start to see variation. So we can't trust the numbers in the ones digit and beyond and we're only allowed to keep two sig figs in our final answer. So the rule to consider here is the relative precision of the number of sig figs in your inputs determine the relative precision or the number of sig figs in your final answer. Those rules are fairly simple where students tend to get into trouble is when you need to do more than one of those things at the same time. So composite problem, for example, where you need to subtract two numbers and then divide by something. So in this case, you can plug those numbers in your calculator. Calculator will give you a number with a whole lot of digits. That's clearly too many. So where do we run this number off? It's important to take these problems step by step. So the first thing to do is first complete the subtraction. So 113.9 minus 109.47 is 4.33. That's the first step of the calculation. But since that's a subtraction, look at the absolute precision. The first number has only one digit past a decimal so we can really only trust one digit past a decimal in the numerator. And everything beyond that can't be trusted. So we'll just make a note of that, a mental note of that and move on. So we take that number divided by 109.47. That's when we get our long decimal. But now since what we've just done is a division, the numerator had only two sig figs, which is less than the five sig figs in the denominator. So we can only trust two sig figs in our final answer and everything else can't be trusted. So we have to round that number off to two sig figs and so we would keep 0.040. So again, that's difficult to determine looking just at the numbers you've put in. The numbers we started with have more than two sig figs each. So we have to consider each of those steps in the calculation to get the right answer. And if you don't trust yourself or if you don't believe the answer you've gotten, you can always use this trick of perturbing your inputs in the last digit and seeing what kind of effect it has on the output. So if I did the same problem, not with 113.9, but with 113.8, we can see that the calculator answer begins to differ in the second digit. So the second digit is the one we keep and everything beyond that is not trustworthy. So the key to remember for a composite problem are, first of all, perform the operations one at a time and then after each step, note how much precision you've got in your intermediate results. Go ahead and keep all the digits to plug into your calculator but make a note of how many digits are significant and then when you get to the end of the problem, you'll know how many digits to round off to. Another type of problem where students typically have problems is this type of problem. Suppose we've got an equation that wants us to calculate one minus delta times 24.935, specifically for the value of delta equals 0.0015. So this might be an equation that would come up, for example, if some theoretical prediction is 24.935 and the delta is telling us how much we're deviating from those theoretical conditions. So clearly we can plug into the calculator and get an answer and we want to know how much to round off. As before, we take this step by step. So one minus 0.0015 is 0.9985 and notice that all those digits are significant. This is a subtraction problem but the number one we're subtracting from is probably considered to be an exact value, 100%. We don't have any uncertainty in that value. One is an integer, it's got an infinite precision. So when we multiply those two numbers, we get the long decimal but since 0.9985 had four sig figs, we keep four sig figs in that answer and everything else we throw away. We'll round off, so we get 24.91 swing round. So the key here is that if you have a quantity in your equation that's an exact quantity like pi or one, doesn't matter how many digits you've chosen to write down when you wrote the number down. If it's an exact quantity, you treat it as if it has an infinite number of digits of precision. Note however that context affects this a little bit. We don't know where this equation came from that I gave you. If that number one wasn't in fact an exact quantity, if it was a measured quantity and if it was 1.0 for example, with a precision of one digit past the decimal, then we would use the normal rules for subtraction and that would end up being rounded off much more aggressively. Where things begin to get slightly more complicated is with exponentiation. If you're taking one number raised to the power of another number, as long as the exponent is an integer, it's not that difficult. So 2.4 raised to the third power for example, your calculator will tell you that's 13.824. Key is to remember that exponentiation is just repeated multiplication. So 2.4 to the third is just 2.4 times 2.4 times 2.4. And since 2.4 has three sig figs, we round our final answer off to two sig figs and that number would be rounded off to 14. So if you're exponentiating with integer exponents, it's just like multiplication and the number of sig figs in your base number is gonna be the number of sig figs in your final answer. Things do get more tricky when you have exponentiation with non-integer numbers as an exponent. So for example, e to the 13.6 calculator will give you a very large number with a lot of digits. Rounding this number off is a little bit tricky. I'll go ahead and give you the answer because it's a little bit counterintuitive and then I'll explain the rule. The right way to round this result off is to the number 800,000, which has only one sig fig to make it a little more obvious that it's only one sig fig. We could write it as eight times 10 to the fifth. So that to many students begins to seem fairly counterintuitive. Why does that number have only one sig fig? Maybe the best way to illustrate that is to perturb the inputs and see what happens to the output. So for example, if we weren't raising e to the 13.6, but we raise it to the 13.5, we get this result which differs already in the very first digit. So if I'm perturbing the first digit after the decimal in the exponent, I'm already making changes to the very first sig fig in the output. So the rule here is anytime you exponentiate a number, the absolute precision or the number of digits past the decimal in the exponent will tell you the relative precision or how many sig figs to keep in your final answer. So again, that's slightly counterintuitive. This is the sort of thing that most students will write down the wrong number of sig figs if they're not paying attention to the rules. So one, because that's counterintuitive, one way to help it make a little more sense is to think about it as base 10 exponents. Let's say we had 10 to the 3.5. I can think of that as 10 to the integer part and the fractional part. So 10 to the 0.5 multiplied by 10 to the 3. That fractional part, 10 to the 0.5, ends up giving me the prefix part of the scientific notation. The integer part just tells me how many powers of 10 to keep or where the decimal point is. Compare that to a different number like 10 to the 23.5. The only difference between these two numbers is the position of the decimal point. So they're both 3.2 times 10 to the something. It's the number of digits past the decimal, the 0.50, that tells me how many sig figs to keep in the prefix, 3.2. And the integer part of the exponent is just telling me the position of the decimal, just telling me how many zeros there are. So I don't care how many sig figs there are in that exponent. I only care how many digits past the decimal. Logarithms are the opposite of exponents. So the rules work in reverse. So let's say I want to take the logarithm, the natural logarithm of 6.022 times 10 to the 23rd. Calculator will give me a long number, where do I round that number off? Here, just remember that the exponentiation rule will work exactly in reverse if you're taking the logarithm. So here it's a relative precision or the number of sig figs in your argument that will tell you the absolute precision or the number of digits past the decimal in your result. So count the number of sig figs in the argument, keep that many digits past the decimal. We've got four sig figs in our argument of the logarithm. We keep four digits past the decimal in our result. So after rounding off, we would get 54.7549. Again, that seems a little counterintuitive. You only had four sig figs in your input. How do you end up with a number with six sig figs in the output? It's because of the numbers before the decimal place. The 54 essentially just tells you where the decimal point is. It's the .7549 that tells you the precision of that number. All right, so those are all the rules you need to work most problems. As a summary, the first few rules you already knew, most likely, when you're adding and subtracting, just count digits past the decimal, whichever of the inputs has the fewest digits past the decimal or the least absolute precision. We'll tell you the absolute precision or the number of digits past the decimal in your final answer. Multiplication and division, you use sig figs. So whichever number has fewer sig figs or least relative precision, we'll tell you how many sig figs or what relative precision to use in the final answer. Exponentiation and logarithms get complicated only because we start to mix and match. It's the absolute precision or the number of digits in the exponent that determines the relative precision of the answer or the number of sig figs in the answer. When you're taking a logarithm, it's the number of sig figs in the argument of the log that will determine the number of digits past the decimal in your final answer. So if you think about those rules and apply them when you're working problems, you won't make any mistakes that affect the number of sig figs in your problems.