Chapter 2 Reinforcement in Number and Arithmetic
CHAPTER 1 SOLUTIONS TO REINFORCEMENT EXERCISES IN NUMBER AND ARITHMETIC
1.3.1 Types of number
1.3.1A.
Say what you can about the type and nature of the following numbers:-
i)2ii)– 3iii)11
iv)21v)– 0vi)
vii)viii)ix)–
x)xi)0.0xii)0.2
xiii)– 0.31xiv)6.3xv)
xvi)3xvii)exviii)e2
xix)– xx)– 1.371
Solution
i)2 is a positive integer. It is the only even prime
ii)- 3 is a negative integer
iii)11 is a positive integer. It is odd and prime
iv)21 is a positive integer, not prime, odd
v)– 0 is zero and both positive and negative
vi) is a positive fraction or rational number in lowest terms. It is a proper fraction
vii) is an improper fraction – a positive rational number
viii)1is a positive mixed fraction that can be expressed as the improper fraction
ix)– is a negative proper fraction
x) is an improper fraction which can be cancelled down to lowest terms as the integer 2
xi)0.0 is a decimal representation to one decimal place, of zero
xii)0.2 is a decimal fraction expressible as the proper fraction
0.2 = =
xiii)– 0.31 is a negative decimal fraction expressible as the negative proper fraction
–
xiv)6.3 is a decimal number, expressible as the mixed fraction
6
or the improper fraction
xv) is an irrational number, the square root of 3
xvi)3 is irrational – a transcendental number
xvii)e is irrational and transcendental
xviii)So is e2
xix)– is a negative irrational number
xx)– 1.371 is a negative decimal fraction expressible as a negative improper fraction
1.3.1B.
Say all you can about the following expressions
i)0 3ii)iii)0 – 2
iv)v)vi)04
vii)0 viii)– 1 0 + ix)
x)40xi)0!xii)
Solution
i)0 3 = 0
ii) does not exist – it is not defined, we cannot divide by zero
iii)0 2 = 2, a negative integer
iv) = 0
v) is not defined
vi)04 = 0 0 0 0 = 0
vii)0 is not defined, again because we cannot divide by zero and therefore we cannot ‘cancel’ the zeros on top and bottom
viii) 1 0 + = 0 + 0 = 0
ix)is not defined
x)40 = 1 by definition
xi)0! = 1 by definition (factorials are covered in Section 1.2.6 if you have not seen them before)
xii)= from xi) = 4! = 24
1.3.2 Use of inequality signs
1.3.2A.
Using inequality signs, order all of the numbers in Q1.3.1A
Solution
Points to note:
- positive numbers are greater than negative numbers.
- some cases of comparing irrational and rational numbers may be difficult, and for this purpose both are best put in decimal form – for example, how can you determine that 11 > 3 otherwise?
The result is:
21 > 11 > 3 > e2 > 6.3 > e > > 2 = > 1 > > 0.2 > 0.0 = 0 > 0.31 > > > 1.371 > 3
1.3.2B.
Suppose a, b and c are three non-zero positive numbers satisfying
a < b c
What can you say about:-
i)a2, b2, c2ii), , iii)a + b, 2c
iv)– a, – b, – cv), , ?
Solution
i) Squaring will not change the order of the inequalities so
a2 < b2c2
ii)Remembering that a, b, c are all positive and non-zero we have
if a < b then >
if b c then
So
iii)a < b c , so a < c and b c, and so
a + b < c + c = 2c
and therefore
a + b < 2c
iv)If a < b then a > b and if b c then c b, so
c b < a
v)Consider a < b then (both defined since a, b, both positive)
Similarly if b c then .
So
1.3.3Highest common factor and lowest common multiple.
1.3.3A.
Express in terms of prime factors
i)2ii)– 6iii)21
iv)24v)– 72vi)81
vii)viii)143ix)391
x)205
Solution
i)2 is a prime already
ii) 6 = 1 3 2
iii)21 = 3 7
iv)24 = 8 3 = 23 3
v) 72 = 9 8 = 1 23 32
vi)81 = 92 = 34
vii) = 33 21 71
viii)143 = 11 13
ix)391 = 17 23
x)205 = 5 41
1.3.3B.
Determine the highest common factor of each of the following sets of numbers
i)11, 88ii)28, 40iii) 25, 1001
iv)20, 45, 90v)14, 63, 95vi)24, 72, 96
vii)36, 42, 54
Solution
Here we only use trial and error and splitting into primes to find the HCF.
i)11 divides 11 and 88 so the HCF is 11
ii)28 = 4 7 40 = 4 10 so the HCF is 4
iii)25 and 1001 ( a prime) have no factors in common, other than 1, so that is the highest common factor –we say they are relatively prime. Such sets of numbers are important in coding theory.
iv)20 = 22 5
45 = 32 5
90 = 2 32 5
By inspection the HCF is 5
v)14 = 2 7
63 = 9 7
95 = 5 19
So, apart from 1, these have no factors in common, they are relativelyprime
vi)24 = 23 3
72 = 23 32
96 = 8 12 = 23 22 3 = 25 3
So by inspection the HCF is 23 3 = 24
vii)36 = 62 = 22 32
42 = 6 7
54 = 6 9
So the HCF is 6
1.3.3C.
Find the lowest common multiple of each of the following sets of numbers
i)2, 4ii)5, 8iii)12, 15
iv)6, 9, 27v)12, 42, 60, 70vi)66, 144
Solution
i)2 divides 4 so the LCM is 4
ii)5 and 8 have no common factors besides 1 and so the LCM is their product, 40
iii)5 12 = 60 and 4 15 = 60
There are no lower multiples of 12 and 15 that are equal to each other, so
LCM = 60
iv)9 divides 27 but 6 does not. However, 6 9 = 54 = 2 27, so the LCM is 54
v)12 = 2 6
42 = 6 7
60 = 6 10
70 = 10 7
The LCM must be divisible by 7, 6 and 10, so try 420 this is divisible by 12 as well as 42, 60 and 70. (420/12 = 35). There is no lower multiple of 70 with this property, so the LCM is 420.
vi)66 = 6 11
144 = 12 12
11 144 is divisible by 66 and is the lowest multiple of 144 that is so. Therefore the LCM is 11 144 = 1584
1.3.4 Manipulation of numbers
1.3.4A.
Evaluate
i)3 – 6 7ii)3(4 – 1) – 2 iii) (4 – 1) (6 – 3)
iv)6 – (3 – 6) 4v)24 (6 2)vi)(24 6) 2
vii)(2 (3 – 1)) (7 – 2(3 – 1))
Solution
i)3 6 7 = 3 42 = 39
ii)3(4 1) 2 = 3(3) 2 = 9 2 = 7
iii)(4 1) (6 3) = 3 3 = 1
iv)6 (3 6) 4 = 6 ( 3) 4 = 6 ( 12) = 6 + 12 = 18
v)24 (6 2) = 24 (3) = = 8
vi)(24 6) 2 = (4) 2 = 2
vii)(2 (3 1)) (7 2(3 1)) = (2 2) (7 2(2))
= 4 (7 4) = 4 3 =
which we leave as it is
1.3.4B.
Evaluate
i)10 – (2– 3 42)ii)100 – 3(7 – 10)3 iii) – ( (–(– ( ( – 1) ) ) ))
iv)– 3(2–(3 + 1) (– 4 + 2) + 4 3)v)1 – 4 (– 2)
Solution
i)10 (2 3 42) = 10 (2 3 16)
= 10 (2 48) = 10 ( 46)
= 10 + 46 = 56
ii)100 3 (7 10)3 = 100 3 ( 3)3 = 100 + 81 (( 3)3 = 27)
= 181.
iii) (( ( ( ( 1))))) = (( ( ( 1)))) = (( 1)) = 1
requires a steady hand not all the brackets are necessary – but in fact all you have to do is count the minus signs
iv) 3(2 (3 + 1)( 4 + 2) + 4 3) = 3(2 4( 2) + 12)
= 3(2 + 8 + 12) = 3 22 = 66
v)1 4( 2) = 1 + 8 = 9
1.3.4C.
Evaluate 4 + 5 23 in its conventional meaning. Without these conventions how many pairs of brackets would be needed to make the meaning of the expression clear?
Now insert one pair of brackets in as many different non-trivial ways as possible and evaluate the resulting expressions (retaining the other usual conventions). Can any other results be obtained by the insertion of a second pair of brackets?
Solution
4 + 5 23 = 4 + 5 8 = 4 + 40 = 44
We would need to indicate:-
- exponentiation first
- multiplication next
- addition last
4 + (5 (23))
would do, so we need 2 pairs of brackets.
We could insert one set of brackets as follows
(4 + 5) 23 = 9 8 = 72
4 + (5 2)3 = 4 + 103 = 4 + 1000 = 1004
(4 + 5 2)3 = 143 = 2744
A second pair of brackets could give results:-
((4 + 5) 2)3 = 183
(4 + (5 2))3 = (4 + 5 2)3 , not new
((4 + 5) 2)3 = 183, not new
1.3.5 Handling fractions
1.3.5A.
Find in simplest form as a fraction
i)ii)iii)
iv) v)+ vi)–
vii)– viii)1 – + ix) – +
x)xi)
Solution
i)
ii)= 4
iii)
iv)
v)
vi)
vii)
viii)1
ix) = =
x) =
= =
xi) = 12 2 = 24
1.3.5B.
a)If a: b = 7: 3 determine a for the following values of b: i) 4 ii) 3 iii) –7 iv) 15.
b)Repeat for b with the same values for a.
Solution
a)
i) =
ii)a = = 7
iii)a = =
iv)a = = 35
b) i) b =
So if a = 4 then b =
ii)a = 3, b =
iii)a = 7, b = 3
iv)a = 15, b =
1.3.5C.
If a : b = 5 : 2 evaluate as fractions
i) ii) iii) iv)
v) + vi) – vii) +
Solution
In each case we could substitute, for example, b = and we would then find that the as all cancel out. However, as preliminary practice for algebra we will adopt a somewhat more mature approach here.
i) =
ii)= = =
iii)= =
iv) = = = =
v) + = + =
vi) – = – = – = – = =
vii) + = using iv)
= =
1.3.5D.
If a is proportional to b and a = 6 when b = 4, what are the values of the following
i) a when b = 3 ii) + iii) iv) b when a = 21
Solution
Since a is proportional to b we have a = kb where k is some constant. But we know that when a = 6, b = 3 and so whatever k is, it must satisfy
6 = k4
and so k = from which we have =
i)When b = 3, a = =
ii)+ = + =
iii)= = = =
iv)When a = 21, b = = = 14
1.3.6 Factorial and combinatorial notation – permutations and combinations
1.3.6A.
Evaluate
i)5!ii)10!iii)
iv)v)vi)
vii)10! + 11!viii)–
Solution
i)5! = 5 4 3 2 1 = 120
ii)10! = 10 9 8 7 6 5!
= 90 8 42 120
= 3628800
Notice how rapidly the expression n! grows in value.
iii)= = 301
iv)= = 18 17 6!
= 18 17 6 5! = 18 17 6 120
= 2200320
v) =
=
=
= 11 3 8 = 264
vi) = 10 3 7 = 210
vii)10! + 11! = 10! + 10! 11
= 10! (1 + 11) = 12 10!
= 43545600 from ii)
viii)
= 3 4 7 2 7
= 10 7 = 70
1.3.6B.
Evaluate
i)C2ii)C7iii)C4
iv)C4v)C100vi)P3
vii)P3 C3
Solution
i) = = 36
ii) = = = 36
iii) = =
= 11 10 3 = 330
iv) = = 10 3 7
= 210
v)= = 1
vi) = = 7 6 5 = 210
vii) =
=
= 20 120 = 2400
1.3.6C.
How many combinations of 4 different letters can be chosen from ABCDEFG?
Solution
The number of combinations is
=
= = 35
1.3.7 Powers and indices
1.3.7A.
Evaluate in terms of powers of primes
i)22 23ii)34/32iii)63 32/4
iv)62 2–2 32v)224 25vi)56/104
vii) 34 22 3–1 viii) 49 7/212
Solution
i)22 23 = 25
ii)34/32 = 32
iii)63 32/4 = = 2 35
iv)62 2–2 32 = 32 22 2–2 32 = 34
v)224 25 = 22 22 25 = 29
vi) 56/104 = = 522 4
vii)34 22 3–1 = 22 33
viii) 49 7/212 = = 7 3 2
1.3.7B.
Simplify (write as simplest products of powers of primes)
i)34 36 32ii)23 42/25 iii)
iv)(4 6)6/(32 42)v)275/95vi) (– 4)3/(– 12)4
vii)vii)
Solution
i)34 36 32 = 34+6+2 = 312
ii) = 23(22)2 25 = 23 24 25 = 22
iii) = ( 2 3)2 23 34 32 22 33
= 22 32 2 33
= 23 35
iv) = = (22)4 26 34
= 28+6 = 214 34
v) = 35
vi) = 41 34 = 2 2 34
vii)= 32 (22)6 33 2 = 213 35
1.3.7C.
Show that the following are all the same number
, , , , , , ,
Solution
Multiplying top and bottom by gives
=
Multiplying top and bottom of gives
Clearly
=
=
Also
So all the expressions are indeed equal.
1.3.7D.
Express in terms of simplest surds
i)ii)iii)iv)
v)vi)vii)viii)
Solution
i)
ii)
iii)
iv)
v)
=
vi) = 6
vii)
viii)
1.3.7E
Rationalize
i)ii)– iii)
iv)v)vi)
vii)+ + viii)+ +
Solution
i)
ii)
iii)In this case we multiply top and bottom by 1 +
=
= =
iv) = = =
v) =
= =
=
vi) =
=
=
=
= 3 8
vi)
=
=
=
=
=
viii)
= 4
= 7 .
1.3.8 Decimal notation
1.3.8A.
Express in decimal form, to four decimal places in each case
i) – ii)iii)iv)–
v)vi)
Solution
i) = 0.5000ii) = 3.5000
iii)= 0.6667iv) = 0.2222
v) = 0.0000vi) = 0.1250
1.3.8B.
Express as fractions
i)0.25ii)0.125iii)72.45
iv)– 0.312v)0.17.
Solution
i)0.25 =
ii)0.125 =
iii)72.45 = 72
iv) 0.312 =
v)0.17 =
1.3.8C.
Write the following numbers in scientific notation stating the mantissa and exponent
i) 21.3241ii)429.003iii)– 0.000321
iv) 0.00301v)1,000,100vi)300491.2
Solution
21.i)21.3241 = 2.13241 10
Mantissa = 2.13241
Exponent = 1
ii)429.003 = 4.29003 102
Mantissa = 4.29003
Exponent = 2
iii) 0.000321 = 3.21 104
Mantissa = 3.21
Exponent = 4
iv)0.00301 = 3.01 103
Mantissa = 3.01
Exponent = 3
v)1,000,100 = 1.0001 106
Mantissa = 1.0001
Exponent = 6
vi)300491.2 = 3.004912 105
Mantissa = 3.004912
Exponent = 5
1.3.8D.
Write the numbers in C to a) 3, b) 6 significant figures.
Solution
a)i) 21.3ii) 429iii) 0.000321
iv) 0.003v) 1,000,000vi)300,000
b)i) 21.3241ii) 429.003iii) 0.000321
iv) 0.003010v) 1,000,100vi) 300491
1.3.9 Estimation
1.3.9A.
Assuming at 3.142 and e 2.718 give approximate values for
i)2ii)e3iii)–2 iv)e–3
Solution
i) 2 10 ii) e320 iii) 2
ix)e3
1.3.9B.
Estimate the values of
i)ii)
iii)iv)
Solution
i) = = 0.1
ii) 13
iii)
= 6 106 + 2 105
= 62 105
iv)
10
Note the distinction made at each step between = and . Also, note that you may find alternative, equally good, ways of estimating the results. You can check your approximations with a calculator.