2 Exploring numbers.2 Prime factor form You can write any number as the product of its prime factors. These factors are prime numbers: 720 = 2 60 = 2 2 80 = 2 2 2 90 = 2 2 2 2 45 = 2 2 2 2 5 = 2 2 2 2 5 This can be written as: 720 = 2 4 2 5 Here 720 is in 2,, and 5 are prime factor form prime factors of 720 A number written as the product of prime numbers is written in prime factor form.. Finding the Highest Common Factor (HCF) Sometimes you will need to find the largest factor two numbers have in common, called the Highest Common Factor or HCF. For example, to find the highest common factor of 720 and 84: First write each number in prime factor form: 720 = 2 2 2 2 5 84 = 2 2 7 Then pick out the common factors: those that appear in both numbers. These are: 2 2 The Highest Common Factor of 720 and 84 is 2 2 = 2 The Highest Common Factor (HCF) of two numbers is the highest factor common to both of them. Two numbers which have a Highest Common Factor of are called co-prime numbers. Exercise B Find the Highest Common Factor of: (a) 6 and 48 (b) 720 and 252 (c) 9 600 and 756 2 Find out whether these pairs of numbers are co-prime. Give a reason for each answer: (a) 6 and 5 (b) 8 and (c) 5 and 6
2 Exploring numbers.9 Powers of 2 and 0 You should be able to recall very quickly the powers of 2 and the powers of 0. Remember 2 means 2 2 2 and this equals 8 whilst 2 = 2 = 8 So the table for the powers of 2 from 2-5 to 2 5 is: 2-5 2-4 2-2 - 2 2-2 0 2 2 2 2 2 4 2 5 2 6 8 4 2 2 4 8 6 2 In a similar manner, 0 = 0 0 0 = 000 and 0 - = 0 = = 0.00 000 So the table of values for the powers of 0 from 0 5 to 0 5 is: 0-5 0-4 0-0 - 2 0-0 0 0 0 2 0 0 4 0 5 00000 0000 000 00 0 0 00 000 0000 00000.000 0.000.00.0. Example 7 Work out the values of: (a) 2 6 (b) 2 0 (c) 2 (d) 0 7 (e) 0 7 (f) 2 2 0 (a) 2 6 = 2 2 2 2 2 2 = 64 (b) 2 0 = 2 2 2 2 2 2 2 2 2 2 = 024 (c) 2-0 = 2 0 = 024 (d) 0 7 = 0 0 0 0 0 0 0 = 0 000 000 (e) 0-7 = 0 7 = 0.000 000 (f) 2 2 0 = (2 2) (0 0 0) = 4 000 = 4000
0 Exploring numbers Exercise G Work out each of the following giving your answer both in index form and without using indices where possible. 2 4 0 2 (a) 5 # 5 (b) 5 ' 5 (c) 0 ' 0 4 (d) 2 2 # 2 (e) # (f) 6 0 (g) (.) 0 (h) 25 0 9 (i) 6 ' 6 2 (j) 5 + 5 (k) 2 + 2 2 + 2 (l) (4 ) 2-2 2 (m) (5 ) (n) (2 5 ) 2 (o) 4 (p) - 2 4 (q) 6 2 6 (r) 5 ' 5-5 2 2 # 2 ( 4 ) -2 (s) 2 (t) 2 (u) (5 ) ( 2 ) 4 # 4 2 # 2 (v) ( 2 ) - 2 (w) - - ( ) 2 2 (x) ( 2 ) # 2 Investigate to find possible solutions of n m = m n..8 Fractional indices You should know that: 9 = so: D e m o 9 # 9 = # = 9 = 9 We can use this to evaluate 9 2 Using the multiplication rule: Similarly: In general: a n = n a D e m o But + D e 92 m # o 92 = 9 2 2 = 9 9 # 9 = 9 so 92 = 9 2 or 9 = + + D e 8 m # o 8 # 8 = 8 = 8 But D e m o 8 # 8 # 8 = 8 so 8 = 8 or 8 = 2 2 2-2 -6 cube roots The cube root of 8 is 2 because 2 2 2 = 8 The cube root is written: 8
Shapes Euclid was a Greek mathematician who lived at Alexandria in Egypt about the year 00 BC. He wrote a book called The Elements explaining everything that was known about 2-D (plane) and -D (solid) shapes, like these: This two-dimensional (2-D) shape is a parallelogram. This three-dimensional (-D) shape is a cone. Euclid s book began with definitions of the mathematical meanings of everyday words such as straight, line and segment. This was followed by hundreds of theorems about 2-D and -D shapes. This unit will help you understand and use some of Euclid s theorems.. Angles When two lines meet at a point the amount of turn or rotation in moving one line to the other is called an angle. Angles can be measured in units called degrees. The turn can be made in a clockwise or anticlockwise direction. A complete turn or revolution is 60. You need to know the names of these types of angles: A quarter turn is called a right-angle. Two lines at right-angles are perpendicular. An angle between 0 and 90 is an acute angle. Half a complete turn (80 ) is a f lat or straight angle. An angle between 90 and 80 is called an obtuse angle. An angle between 80 and 60 is called a ref lex angle.
Discrete data 65 (e) the length of a road (f) the speed at which a car is travelling (g) the area of a f ield (h) the marks given by judges at an ice skating competition 4.2 Recording and presenting discrete data It is easy to lose count when you are collecting data. One way of avoiding this is to use tally marks to collect the data, then arrange it in a frequency distribution table (or frequency table for short). Example Fred is collecting data on snails for a science project. He counts the number of snails in 0 different sections of woodland, each with an area of m 2. Here is his data: 5 0 0 2 2 5 2 0 0 2 5 0 2 2 5 2 2 Record Fred s data in a tally chart and frequency table. This data is discrete with a maximum value of 5 snails in each area. Here is the frequency table. It includes a tally column. Number of snails in a section Tally Frequency 0 5 6 2 8 7 The frequency of a result is the number of times it occurred. Notice that 5 tallies are recorded as 4 0 5 4 Total 0 Fred could have collected his data directly in a blank frequency table. This is called a data capture sheet.
Grouped data 67 that is 0 8 9 is written 9 9, 7 22 is written 2 2 2 2, is written, 4 7 is then 9, 7 5 6 and 8 is 0 8 so the complete table becomes 0 8 9, 7, 2, 9, 5 2 2,,, 6, 8, 2, 6, 7 4 6, 8, 6, 7, 7, 5, 9, 2, 8, 4, 5, 6 2, 4 we now put the leaves in order 0 8 2, 5, 7, 9, 9 2, 2,, 6, 2, 6, 7, 8 4, 6, 6, 7, 7, 8 5, 2,, 4, 5, 8, 9 6 2, 4 A stem and leaf diagram can be used to work out the median. In this case there are fifteen numbers up to and including 8 and fifteen numbers at 4 or above. So the median is 8 + 4 8 D e m o = = 40. 5 2 2 4. Recording and presenting grouped data When quantitative data has a wide range of values it makes sense to group sets of values together. This makes it easier to record the data and to spot any patterns or trends. For example, scores out of 50 in a test might be grouped: 0 0, 20, 2 0, 40, 4 50 It often makes life easier if all the groups are the same size, but they do not have to be. In this case the first group has members and the others have 0. We call the groupings 0 0, 20 and so on class intervals. Class intervals are groupings of quantitative data.
82 Collecting and presenting datagcontinuous data This table shows the time intervals between cars on the Lucea High Road: The signs < (less than) and (less than or equal to) show that a time of 0 seconds is in the f irst class interval. The cumulative frequency table is: Time (seconds) Cumulative frequency up to 0 up to 20 26 up to 0 up to 40 4 up to 50 46 up to 60 50 Here is the cumulative frequency curve: Time t (seconds) Frequency 0 < t 0 0 < t 20 20 < t 0 7 0 < t 40 8 40 < t 50 5 50 < t 60 4 You plot the points (0, ) (20, 26) and so on, and join them with a smooth curve. Cumulative frequency 50 40 0 Remember to plot the point (0, 0) even though this is not a value in the table. 20 0 0 0 20 0 40 50 60 Time (Seconds)
Expanding brackets 20 9 Evaluate and compare: (a) (4 + 5) + 2 with 4 + (5 + 2) (b) (9 5) with 9 (5 ) (c) (5 2) with 5 (2 ) (d) (2 6) 2 with 2 (6 2) 20 Evaluate and compare: (a) 5 (6 + 2) with (5 6) + (5 2) (b) 7 (9 5) with (7 9) (7 5) 0.2 Expanding brackets You need to be able to expand algebraic expressions. Here expand means multiplying terms to remove brackets from expressions like 2(r + 2b). If r represents a red square and b represents a blue square, then r + 2b represents red squares plus 2 blue squares. So, an expression like 2(r + 2b) represents 6 red squares plus 4 blue squares: 2 (r + 2b) = 6r + 4b 2 ( r r r + b b ) = r r r + b b r r r + b b To expand an expression multiply each term inside the brackets by the term outside. # # 2(r + 2b) = 6r + 4b Multiplication can also be seen as repeated addition: 4 = 4 + 4 + 4 = 2 (x + 5) = x + 5 + x + 5 + x + 5 = x + 5 This diagram helps show that (x + 5) = x + 5: x 5 x A x B 5 Area of whole rectangle = (x + 5) Area of rectangle A = x Area of rectangle B = 5 = 5
226 Using and applying mathematics Since the nth triangular number is 2 n (n + ) the (V 2)th triangular number is: 2 (V 2) ((V 2) + ) = 2 (V 2) (V ) = 2 (V 2 V 2V + 2) = 2 (V 2 V + 2) so for V vertices, the number of diagonals D, is: D = 2 (V 2 V + 2) = 2 V 2 2 V + = 2 V 2 2 V D = 2 V (V ) This has justified the general rule in a very strong way because it has been related back to the geometry of the physical structure..4 Proving the result The more or less final stage in the process is to prove the general result. Proofs are never easy. To do this we will relate back to a picture of a general regular polygon with V vertices, labelled from to ṿ Starting at any vertex, for instance, the number of diagonals will be (V ), because the lines drawn will go to all but itself and the two vertices either side of, that is and ṿ The same will be true for every other vertex. So for each vertex there are (V ) diagonals and there are V vertices. So the total number of diagonals is: V (V ) But this gives the diagonals going in both directions. So it gives double the number of actual diagonals. Hence the actual number of diagonals is: 2 V (V ) This proves the result. Replace n by v 2 There is more about multiplying brackets in Unit 0.5 Extending the investigation The process outlined in this chapter is sufficient to guarantee the top grade at GCSE. However, unlike other textbook exercises and examination questions, a mathematical investigation is never finished. One of the things you can always do is extend the investigation. There are two extensions of Diagonals which you might like to see as a future challenge.
Rounding 257 You can round (or correct) numbers to a given number of decimal places (d.p.). The first decimal place is the first number (zero or non-zero) after the decimal point. Example 2 Round 46.82 to: (a) decimal place (b) 2 decimal places. (a) Rounding to d.p. ( digit after the decimal point): 46.82 The digit after the first decimal place is 8, so round up. 46.82 rounded to d.p. is 46.4 (b) Rounding to 2 d.p. 46.82 The digit after the second decimal place is 2, so round down. Notice that when you round down to 2 d.p. these two decimal places remain the same. Exercise 2B 46.82 rounded to 2 d.p. is 46.8 Copy this interval diagram. (a) Mark the position of 46.826 on the diagram. (b) Write 46.8267 rounded to d.p. 2 This table shows the times recorded for the first three athletes in an 800 m race: Name Time R. Grey min 45.4826 s M. Hobson min 45.4768 s T. Knight min 45.487 s 46.82 46.9 46.85 46.8 46.82 is closer to 46.8 than to 46.9 46.8 46.82 (a) Write these times correct to 2 d.p. (b) Write the names of the athletes in the order in which they finished the race.
422 Simplifying algebraic expressions Step 4 Rewrite the bx term using these two = [2x 2 6x + 5x 5] factors. Step 5 Proceed as for factorizing by pairing. = [2x(x ) + 5(x )] = [(x )(2x + 5)] Example 20 Factorize 4x 6 4x 2. Step l 4x 6 4x 2 = 2[7x 2x 2 ] (2 is HCF) Step 2 = 2[ 2x 2 + 7x ] (rearrange) Step ( 2) ( ) = ( + 6) You need factors of +6 (+6) (+) = 6, (+ 6) + (+) = +7 which add to +7 4x 6 4x 2 = 2[ 2x 2 + 7x ] Step 4 = 2[ 2x 2 + 6x + x ] Step 5 = 2[ 2x(x ) + (x )] = 2[(x )( 2x + )] or = 2[(x )( l)(2x l)] = 2 (x )(2x ) Example 2 Find two numbers whose product P is 60 and whose sum S is 7. The product is negative, so the two numbers must have different signs. The sign of the greater number must be negative because their sum is negative. S is odd so one number must be odd and the other even. Try: +, 60 +, 20 +4, 5 +5, 2 Exercise 20 Factorize: (a) x 2 + x (b) 8x + 24 (c) x 2 + x + 8x + 24 (d) x 2 + 6x (e) 2x + 2 (f) x 2 + 6x + 2x + 2 (g) x 2 5x (h) 8x 40 (i) x 2 5x + 8x 40 (j) x 2 + 5x + x + 5 (k) x 2 + 4x 8x 2 (l) x 2 6x 2x + 2 (m) x 2 + 4x x 2 (n) x 2 + 5 + x + 5x (o) x 2 + 6 7x 9x Remember to look for an HCF as a first step to make the rest of the factorization easier.
Produce number sequences 585 Example 5 compound interest (see also Example 6) A student borrows 600 to buy a computer. Interest is added to the loan at the rate of 5% per annum. How much does the student owe after years if no repayments are made? Type 6 0 0 EXE Now type Ans + Ans x 5 0 0 EXE EXE EXE 600 600. Ans + Ans 5 00 60. 66.5 694.575 29.6 Drawing graphs on your calculator Mcl Choose the default ranges for x and y by pressing Range SHIFT DEL Range (or Range INIT Range Range ) Clear the graphics screen by pressing SHIFT Cls EXE. Have a look at the graphics screen by pressing G T. Both axes are marked off in intervals of. Example 6 Draw the graphs of y = x + and y = x. Press the Graph key, then type x + EXE. The first graph is drawn. To see the second graph, press the Graph key again and type x + EXE. Notice that the second graph is superimposed. It is drawn on the same axes as the first without erasing the first. This will continue to happen with further graphs until you clear the graphics screen. Clear the screen by pressing SHIFT Cls EXE. Example 7 Draw the graphs of y = mx + c for several different values of m and c. In this example m = 2 and c has values between 4 and 4. Type M? M C? C : Graph MX + C EXE. When the first? appears, enter your value for M, then EXE. When the second? appears, enter your value for C, then EXE. Wait until the graph is drawn. Now press EXE again to draw another graph on the same axes. Type: 2 EXE 4 EXE (wait until the graph is drawn) EXE 2 EXE 4 EXE (wait until the graph is drawn) EXE 2 EXE 0 EXE (wait until the graph is drawn) EXE 2 EXE 2 EXE (wait until the graph is drawn) EXE 2 EXE 4 EXE (wait until the graph is drawn) Clear the screen to finish. Try other values for m and c. Find out how to set the ranges of x and y to your own chosen values. If your calculator has a key marked x, o, T press it to get x.