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Rounding • Introduction • Rounding To Decimal Places • Rounding To Whole Numbers, Nearest 10, 100 And 1000 • Some Issues With Rounding • Significant Figures • Rounding To 1 Significant Figure • Rounding To 3 Significant Figures • Limits Of Accuracy, Upper And Lower Bounds • Maximum And Minimum Values Introduction Rounding is the process whereby we approximate an answer to some specified level of accuracy. Rounding To Decimal Places If I use an ordinary 30cm ruler it would be hard to get our answer to more than 1/2 mm of accuracy. It is more likely that we would only get our answers to 1 mm (i.e. 0.1 cm) of accuracy. For instance, I could measure a length to be 3.6 cm, but I would be unlikely to measure a length so that it was 3.63 or 3.63767 cm! I'd have to have very good eyesight (and a very good ruler)! However, when we work with numbers, especially division, we often get answers that have many decimal places, particularly when we use a calculator. Let's look at an example. I have a 35cm length of wood. I want to divide it into 7 pieces. Easy, just do the sum 35 ÷ 7 = 5cm. No problemo! Now what about a 30cm piece of wood split into 7 equal pieces. Well, do the sum 30 ÷ 7 = 4.285714286 on a calculator. How am I supposed to measure to that degree of accuracy! Even an industrial saw with laser control couldn't cut that accurately. What we need to do is approximate the answer so that it is reasonable. For a length using a ruler we have already seen that an answer with just one decimal position or place will often do. But before we do the sum let's just look closer at our numbers and the decimals: Look at this diagram. This diagram represents the number 1.1111. Each piece is 10 times smaller than the one to the left of it. It is easy to see that if you were to lose the last 1, you are not really losing that much. So we could say that 1.1111 to three decimal places was 1.111. Now if I look at the last bit now, it's still pretty small compared with the bits to the left of it. So I could ignore that piece as well and get 1.1111 to be 1.11 to two decimal places. Again I could lose the second decimal part without making too much difference and so 1.1111 would become 1.1 to one decimal place. I could even lose that bit, since it is so small in comparison with the whole number that 1.1111 will become 1 to the nearest whole number. We must always remember though that we have lost some of the original number and sometimes even though these parts are very small in comparison with the rounded number, they can be important. For instance if you imagine that the above number represented tonnes of gold, you might not want to lose even the smallest amount of gold! When we lose or leave off parts of the number in this way we say that we have truncated the number. Consider though the following number as a diagram: If we now truncate the number to one decimal place we will get 1.1. But look at the diagram. If we leave off the 0.09 part we are losing a quite substantial chunk of the number. If you think about it you should see that really the number is almost 1.2 really. There is only another 0.01 needed to make it 1.2 So really we should make this number equal to 1.2 when rounded to one decimal place. What about this diagram: It seems clear that we could ignore the 0.03 and get 1.1 to one decimal place. Consider this one: I think you will agree that this second decimal is quite large and so the number should be closer to 1.2 when rounded to one decimal place rather than 1.1. But what about this situation: We now have a bit of a dilemma. Should we lose the last bit and make it closer to 1.1 or should we think that it makes the number closer to 1.2? We could choose either. In fact mathematicians have agreed that it should make the number 1.2 rather than 1.1. Another way of looking at the situation is to think of a dial or meter with numbers on the scale: Obviously 1.17 is closer to 1.2 than to 1.1 In this example, 1.14 is closer to 1.1 Again for this example, mathematicians have agreed that it is closer to 1.2 than to 1.1. You might not agree, but we all have to accept it. Note: One of the difficulties that some have with these ideas is that they don't really understand the decimal system. When they see a number such as 1.17 they read it a "one point seventeen" instead as "one point one seven". If you read it as "one point seventeen" then you might think that this number is bigger than 1.2 "one point two". Actually you'd be wrong. Look at the diagram here to see why. Here we can see that 1.2 is actually larger than 1.17. The problem is that with decimals as you go to the right the numbers are representing ever smaller things. So we should never say "one point seventeen", but rather "one point one seven". The only exception to this is with money where we would say "one pound, seventeen pence" for £1.17. We would not however, see £1.2 (except on a calculator display), but we would add on the extra 0 to make it £1.20 and say it as "one pound, twenty (pence)". It's clear that this is now larger than £1.17! The other thing to note about rounding is that we don't need to even think about any of the other, smaller decimal places. For instance 1.149999999 is still 1.1 when rounded to one decimal place. 1.159999999 is still 1.2 when rounded to 1 decimal place. If you are not sure why, draw yourself some diagrams and you will see that the extra decimals never have as much influence as the one to the left of each one. One tricky area is when we get a number such as 1.99999. What is this rounded to 1 decimal place? Obviously when truncated the number is simply 1.9. But since the second decimal place is greater than 5 we need to round up. Well the next number to round up to is not 1.99, because that has two decimal places, so it can only be 2.0. Now that we have some understanding of what is happening let us work through some examples to see how we can do rounding quickly in practice. Look carefully through the following examples. Each one introduces a slightly different point. Example 1: Round 23.834 to one decimal place. First truncate the number to 1 d.p. That is 23.8 Now look at the digit in the second decimal place. It's a 3. Since this is smaller than 5 we don't need to round up. So our answer is 23.8 (to 1 d.p.) - Note how we must tell the reader of our answer that we have rounded the answer. Example 2: Round 34.7823 to one decimal place. First truncate the number to 1 d.p. That is 34.7 Now look at the digit in the second decimal place. It's an 8. This is lager than 5 so we need to round up. The next number after 34.7 is 34.8. So our answer is 34.8 (to 1 d.p.) Example 3: Round 287.653209 to one decimal place. Truncate to 1 d.p. i.e. 287.6 The second decimal is a 5. We use the agreed rule and round up. The answer is 287.7 (to 1.d.p) Example 4: Round 29.98087 to one decimal place. Truncate first, i.e. 29.9 The second decimal is an 8 so round up. The next number after 29.9 is 30.0 So the answer is 30.0 (to 1 d.p.) Example 5: Round the number 2.9876 to two decimal places. This time we truncate to 2 decimal places i.e. 2.98 We now look at the third decimal which is a 7, so we need to round up. So the answer is 2.99 (to 2 d.p.) Example 6: Round 0.0034 to two decimal places. Truncate to 2 decimals. We get 0.00. Look at the third decimal, it's a 3 so we don't round up. So the answer is 0.00 (to 2 d.p.) However consider the implications of this. If you divide by 0 you will cause the computer or calculator to come back with an error. This is because the answer is infinity. No computer or calculator can handle that. However the answer is not infinity. It might be large, but it isn't that large! Suppose your computer was controlling a nuclear power station and you decided to round the answers without thinking about divisions by 0. If you were lucky your computer might just crash. If you were unlucky the whole power station might crash! Example 7: Round 9,999.999999 to three decimal places. Truncate first, we get 9,999.999. Now look at the fourth decimal, it's a 9 so round up. The next number is 10,000.000 So the answer is 10,000.000 (to 3 d.p.) Rounding To Whole Numbers, Nearest 10, 100 And 1000. So far we have just looked at rounding to a certain number of decimal places. But we can also round to the nearest whole number, the nearest 10, the nearest 100, and nearest 1000 etc. We already have the basic ideas, so it will probably be easier to look at examples and see how it works in practice. But first let's make sure you understand how numbers work in terms of their position and relative value. Does it look familiar - it should, for it is the same diagram as for decimals, but with different numbers above. The number represented is 11,111 Rounded to the nearest 10 this number would be 11,110 Rounded to the nearest 100 this number would be 11,100 Rounded to the nearest 1000 this number would be 11,000 Rounded to the nearest 10,000 this number would be 10,000 Let's look at some examples to see how this rounding works: Example 1: Round 78.765 to the nearest whole number. A whole number doesn't have any decimals in it, so truncate the number to 78. Now we look at the first decimal place to see if it should be rounded. Since it is a 7 it should be rounded up. So the answer is 79 (to nearest whole number). Example 2: Round 768.9 to nearest 10 Truncate everything less than the tens column to get 760. Note we ignore any decimals. look at the units column to see if we should be rounding. The units column is an 8, so we should round up to the next number. So the answer is 770 (to the nearest 10). Example 3: Round 657,987 to the nearest 1000 We set everything in the units, tens and hundreds column to 0, so we get 657,000. Now we look in the hundreds column to see if we should round up and since it is a 9 we definitely should. So the answer is 658,000 (to the nearest 1000). Example 4: Round 9989 to the nearest 100. We set the tens and units column to zero to get 9900. Then look at the tens column to see if we need to round. It's an 8 so we need to round. So the answer is 10,000 (to the nearest 100). This because going one hundred up from 9900 gets us to 10,000 Example 5: Round 372 to the nearest 1000. We set the hundreds, tens and units column to 0, so we get 0 !. Now we look at the hundreds column to see if we should round. It is a 3 so we shouldn't. So the answer is 0 (to the nearest 1000). This is obviously strange, and possibly dangerous if we are going to divide by this number! Some Issues With Rounding We have already noted that when rounding we can get a zero, which in some situations can lead to unforeseen effects. It should be noted though that rounding too early in a calculation can also result in quite drastic errors. As a rule answers should only be rounded at the end of all calculations. Consider the following example: Question: Given a circle of circumference 234m work out its area. Answer: Firstly what value should we use for π (pi)? The true number is 3.1415926535897... Often in examinations you are told to use the number 3.14 which is π rounded to 2 decimal places. However on a calculator you might press the key for π and on my calculator this uses the value 3.141592654 which is π rounded to 9 decimal places. Will it make a difference. Well in the first answer I shall use the most accurate values possible at all times and in the second solution I shall keep rounding values to 2 decimal places. Solution 1: Solution 2: Notice the difference between the answers: 4359.29 - 4357.34 = 1.95 This might not seem like much, but it shows how answers can become quite inaccurate very easily. Think if there were even more steps involved. Moral: round only at the end of all your calculations. Significant Figures Another form of rounding is where we round to a certain number of significant figures. Take any number and the digit at the left hand side is the most significant, or most important part of it. The digit at the right hand is the least significant digit. e.g. Consider 3,456 The three is the most significant digit since it represents 3000. The 6 is the least significant because it only represents 6 units. e.g. Consider 0.003456 The 3 is again the most significant digit since it represents 0.003 or 3 thousandths. Whereas the 6 is the least significant digit because it represents only 0.000006 or 6 millionths. The usual forms of rounding to significant figures are rounding to one significant figure or rounding to three significant figures. Rounding To One Significant Figure In this we just keep the most significant figure and set all other digits to zero. Example 1: Round 345098 to 1 significant figure. Firstly keep the most significant digit i.e. 300000. Now check the second significant digit to see if we should round up. Its a 4 so we don't round up. So the answer is 300,000 (to 1 s.f.) Example 2: Round 0.065709 to one significant figure. Firstly keep the most significant figure and ignore all the other decimals. i.e. 0.06. Now check the second significant figure, which is a 5 so we round up. So the answer is 0.07 (to 1 s.f.) Example 3: Round 96.987 to 1 sig. fig. Firstly keep the most significant digit i.e. 90. Check the second significant figure, which is a 6 so we round up. The answer is 100 (to 1 s.f.) Example 4: Round 0.009609999 to one sig. fig. Keep the most significant digit, i.e. 0.009. The second significant digit is a 6 so round up. The answer is 0.01 ( to 1 s.f.) Rounding To Three Significant Figures Here we keep the three most significant figures, using the fourth to decide whether to round up or not. Example 1: Round 65487 to 3 sig. figs. Keep the three most significant digits. i.e. 65400. The fourth significant digit is an 8 so round up. The answer is 65,500 (to 3 s.f.) Example 2: Round 0.098734 to 3 sig. figs. We get 0.0987, the fourth digit is a 3 so don't round up. The answer is 0.0987 (to 3 s.f.) Example 3: Round 0.00999898 to 3 sig figs. We get 0.00999, the fourth digit is an 8 so round up. The answer is 0.01 (to 3 s.f.) The great advantage of rounding to significant figures is that it keeps the relative size of the number clear, without it becoming a zero. Limits Of Accuracy And Upper And Lower Bounds If we are give a number, say 25 cm and we are told that this value has been rounded to the nearest cm, then it’s true value could actually be a little bit less than 25cm, or a little bit more than 25cm. How can we know what the lower bound is (i.e. the smallest value) and what the upper bound is (i.e. the largest value)? Obviously, the lower bound cannot be 24 cm. Because 24 cm to the nearest cm is 24 cm. What about 24.1 cm. Well if we round this we get 24 cm. What about 24.5 cm? Well this rounded is 25cm. Can we get smaller than this. No. 24.4999 is still only 24 cm to the nearest cm. So our lower bound is 24.5 cm. What about the maximum value it could be. 26 cm is too large, 25.6 cm would round to 26cm. What about 25.4 cm. Well this would round to 25cm. So, now we look at 25.5 cm. If we rounded this we would get 26cm. So our true value must be less than this. So 25.5 cm is the upper bound of possible values. Clearly the actual value cannot equal 25.5 cm, but we cannot easily write a number that is just less than this, so we allow 25.5 to be the maximum possible value. This does cause a little confusion with people, because 25.5 cm would actually round to 26cm, so how can it be the maximum value. Well, if you tried to write the biggest number that is just less than 25.5, you will find you can’t do it. 25.49 can be made a little larger by using 24.99, which can be a bit larger as 24.999 and so on. But, you can’t say 24.999 recurring, because if you go to an infinite number of decimal places, all being 9, you are actually really writing 25.5 anyway. (This is because 0.9999 recurring actually equals 1!) It is easy to think of the lower and upper bounds as being part of a class interval or inequality. 24.5 Notice, that, since we have rounded to the nearest cm, then our lower bound is the rounded value (25cm) minus ½ of 1cm and the upper bound is 25cm + ½ of 1cm. Consider another example. The length of a football pitch is 110 m to the nearest 5m. What is the smallest possible value and the largest possible value for the length of the pitch? Well if the pitch has been rounded to the nearest 5m, then it could be ½ of 5m on either side of 110. So we have 110 – 2.5m or 110 + 2.5m i.e. the pitch could vary from 107.5 m to 112.5 m It’s very simple really. Just find half of the thing you are rounding to. 1.34 is rounded to 2 d.p. so it could vary from 1.34 – 0.005 to 1.34 + 0.005, i.e. from 1.335 to 1.345 If the number has been rounded to significant figures then you have a little more thinking to do. The number 1300 has been rounded to 2 significant figures. What is the smallest and largest values it could have? Well, being rounded to 2 s.f. implies for this number that it is rounded to the nearest 100, so we need 50 on each side. So it could vary from 1250 to 1350. Maximum And Minimum Values We often use these upper and lower bounds to work out the minimum and maximum values of an expression. The usual situation is if there is a division involved, such as when calculating speed or density. Suppose we have A = x y To obtain a maximum value for A, I need a maximum value of x, but a minimum value for y! To obtain a minimum value for A, I need a minimum value of x, but a maximum value for y.