Integers

1 INTEGERS CONTENTS  Definition  Integers on Number Line  Addition of Integers  Subtraction of Integers  Limitations of the Number Line  Subtraction of Larger Number from Smaller Number       Properties of Addition & Subtraction  Multiplication of Integers  Properties of Multiplication  Distributive Property  Division of Integers  Sign system For Division  Rule of BODMAS  The Operation 'OF'  Absolute value of an integer   Types of Integers (1) Positive Integers : The numbers 1, 2, 3, 4, 5, .... i.e., the natural numbers are called positive integers. (2) Negative Integers : The numbers –1, –2, –3, –4, –5, .... are called negative integers. (3) Zero Integers : The number 0 is simply an integer. It is neither positive nor negative. Ex.1 5 etc. (2) The set of integers is denoted by I and I = {…… –3, –2, –1, 0, 1, 2, 3……..}  Write the predecessor and successor of the following numbers 4, –4, 6, 1, b, n2 Successor 5  3 7 2 b 1 n2 1 INTEGERS ON NUMBER LINE Positive numbers are always on right side of zero & negative numbers are on left side of zero. –3 –2 –1 –ve numbers 0 1 2 3 +ve numbers or we can say all integers are in ascending order from left to right. Ex.2 Whole numbers with + or – signs are called integers. Eg : –17, –5, 0, 1, 3, ……… Note : (1) Decimal numbers are not include in integers, like 5 , –11.97, 0.03, 7 EXAMPLE  Predecessor 3  5 5 0 b  1 n 2  1 Number 4 4 6 1 b n2 Sol. INTRODUCTION 0.3,  CHAPTER Sol. EXAMPLE  Fill the square by '<', '>' or '=' (i) 0 –2 (ii) –31 –21 (iii) –3 8 (iv) –7 7 (v) 11 –6 (vi) 3 3 (vii) –1132 –2 (viii) –1039 –2138 (i) > (ii) < (iii) < (iv) < (v) > (vi) = (vii) < (viii) > Sol. (i) –2 + 8 + (–9) = –3 ADDITION OF INTEGERS In order to add two integers on a number line, we follow the following steps : Step 1 : On the number line, mark one of the given integers. Step 2 : Move as many units as the second number to the : (i) right of the first, if the second integer is positive. (ii) left of the first, if the second integer is negative. Step 3 : The point thus we reach represents the sum of two given integers. Ex.3 EXAMPLES  Add the following integers : (i) 6 and –9 (ii) –3 and –4 Sol. (i) First we draw a number line and mark the integer 6 on it. Adding –9 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 To add –9 we move 9 steps to the left from 6. Thus, we reach at a point representing –3. Hence the sum of 6 and –9 is –3. That is, 6 + (–9) = –3. Note that if we represent the number –9 on the number line then to find 6 + (–9) we shall move 6 units to the right of –9. Obviously, we reach at –3. Adding –6 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 (ii) Draw a number line and mark the integer –3 on it. Adding –4 –11 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 Ex.4 To add –4 and –3 we have to move 4 steps to the left of –3. Thus, we arrive at –7. Hence, the required sum is –7. That is, (–3) + (–4) = –7. Note : No matter which number you choose as first and the other as second number, because in both the conditions you will get the same answer. Draw a number line and represent each of the following on it : (i) –2 + 8 + (–9) (ii) –2 + (–3) + (–5) Adding –9 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 Adding 8 (ii) –2 + (–3) + (–5) = –10 Adding – 5 Adding – 3 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 SUBTRACTION OF INTEGERS We know that in the subtraction fact 7 – 2 = 5, 7 is the minuend, 2 is the subtrahend and 5 is the difference. Step 1 : First we draw a number line and mark (label) the minuend on it. Step 2 : (i) To subtract a positive integer, we move to the left from the minuend as many steps as the second integer is. (ii) To subtract a negative integer, we move to the right (not left) as many steps as the second integer is. Step 3 : The point thus we reach represents the difference of two integers. Ex.4 EXAMPLE  Subtract the following integers : (i) 4 – 8 (ii) – 5 – 4 (iii) – 3 – (–4) Sol. (i) First we draw a number line and mark the number 4 on it. Subtracting 8 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 To subtract 8, we move 8 steps to the left of 4, thus we reach at the point representing – 4. Hence, 4 – 8 = –4. (ii) Mark the integer –5 on a number line. Subtracting 4 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 To subtract 4, we move 4 steps to the left of –5, thus we reach at the point representing –9. Hence, –5 – 4 = –9. (iii) First we draw a number line and mark the integer –3 on it. Subtracting (–4) –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 LIMITATIONS OF THE NUMBER LINE Of course, addition and subtraction of integers on a number line would not work so well if we are dealing with large numbers. Eg, 465 – 739 or 465 + (–739). SUBTRACTION OF LARGER NUMBER FROM SMALLER NUMBER We subtract smaller number from the larger number and we put a negative sign before the difference so obtained. That is smaller natural number – Larger natural number = – [Larger natural number – Smaller natural number]. To add two negative numbers, we add the numbers without sign and then we put the negative sign (common sign) before the sum so obtained. Sol. Ex.7 Represent the following numbers on a number line : (i) +9 (ii) –3 (iii) +8 (iv) –5 Sol. Ex.8 7 To subtract a negative integer –4, we will move 4 steps to the right of –3, thus we reach at the point representing 1. Hence, (–3) – (–4) = –3 + 4 = 1 From the above example (iii) We observe that –3 – (–4) = 1 which is same as – 3 + 4. Note : Subtracting a negative is the same as adding a positive and subtracting a positive is the same as adding a negative. Ex.6 Edubull Sol. Signs (i) 1500 m above sea level + (iii) Depth of 500 m – A deposit of rupees thousand + (ii) (iv) (v) 15°C below 0°C temperature Withdrawal of rupees hundred – – 0 8 Srinagar Shimla –10 –5 0 9 5 Bhubaneswar 10 15 20 25 30 Nainital Chennai 35 40 45 50 Varanasi (i) Observe the number line and write the temperature of the cities marked on it. (ii) What is the difference of temperature between the hottest and the coldest places among the above ? (iii) Can we say temperature of Bhubaneswar is more than the temperature of Nainital and Srinagar together ? (i) Patnitop –9°C ; Srinagar –4°C ; Nainital 6°C ; Shimla 7°C ; Bhubaneswar 21°C ; Varanasi 32°C; Ex.9 Sol. Chennai 39°C (ii) 48°C (iii) Yes Draw a number line and represent each of the following : (i) –2 + 6 + (–7) (ii) –6 + 8 (i) –2 + 6 + (–7) = – 3 Subtracting 7 Adding 6 –7 –3 –2 0 (ii) –6 + 8 = 2 4 6 Adding 8 Represent the following numbers as integers with appropriate signs : Statement –3 A number line given below shows the temperature of different cities on a particular day: Patnitop EXAMPLES  S.No. –5 Ex.10 Sol. Ex.11 –6 0 2 Find the difference between the following pairs of integers : (i) –20 and –40 (ii) –19 and 30 (iii) 45 and –36 (i) –20 – (–40) = –20 + 40 = 20 (ii) –19 – 30 = – 49 (iii) 45 – (–36) = 45 + 36 = 81 Draw a number line and answer the following : (i) Which number will we reach if we move 8 steps to the right of –15? Write this number with appropriate sign. (ii) If we are at –7 on a number line, in which direction should we move to reach –15 and how many steps? Sol. (i) 8 + (–15) = –7 Ex.14 –7 –6 –5 0 8 (ii) –15 – (–7) = –8 Left –15 Ex.12 Sol. S.No. (i) –7 0 Write all the integers between the given pairs in ascending and descending orders : (i) 0 and 5 (ii) –3 and 3 (iii) –8 and –15 (iv) –40 and – 32 Integers Ascending Order 0&5 1, 2, 3, 4 (ii) –3 & 3 (iv) –40 & –32 Ex.13 Complete the following table : (iii) –2, –1, 0, 1, 2 –8 & –15 + –3 Descending Order 4, 3, 2, 1 2, 1, 0, –1, –2 –14, –13, –12, – –9, –10, –11, – 11, –10, –9 12, –13, –14 –39, –38, –37, – –33, –34, –35, 36, –35, –34, – –36, –37, –38, 33 –39 3 –5 4 –4 0 0 7 –19 –12 4 –8 17 + 3 –5 –2 7 10 4 –4 0 –12 –5 –17 –1 –5 11 0 –7 –3 7 0 4 –9 3 7 –15 –8 4 1 8 –19 –16 –23 –19 –31 –15 17 20 13 17 5 21 –27 0 –5 –2 7 0 3 –3 Sol. (a) (i) 0 –3 Ex.15 Write true (T) or false (F) for the following statements. Also correct those which are false : (i) Sum of two positive integers is always positive. (ii) Sum of two negative integers is always positive. (iii) When a positive integer and a negative integer are added, the result is always a negative integer. (iv) The sum of an integer and its additive inverse is always zero. (v) When a positive integer and a negative integer are added, we take their difference and place the sign of bigger integer, ignoring the sign of both. (i) T (ii) F (Sum of two negative integers is always negative). (iii) F (When a positive and a negative integers are added, the result may be a positive or a negative integer). (iv) T (v) T (a) Check which of the following is a magic square. (If each row, column and diagonal have the equal sum.) (i) 5 –1 –4 (ii) 1 –10 0 –24 3 –31 –4 –27 0 –39 –12 –4 –3 –2 –6 4 –7 (b) Magic squares are given below, fill in the empty cells with appropriate integers: (i) 0 (ii) –5 4 –19 –27 Sol. Sol. Edubull –23 4 0 –2 –9 (ii) (b) (i) 1 5 –1 –4 –5 –2 7 0 3 –3 –6 0 0 0 1 –10 0 –4 –3 –2 –6 4 –7 –9 –9 0 –1 5 1 –2 3 –2 –6 1 –9 4 –3 2 No 0 0 0 –9 –9 –9 –9 Yes (ii) –5 2 –3 0 –2 –4 –1 –6 1 Ex.16 Sol. Neena has a loan of j 1200 to repay. Her brother gave j 2500. Describe Neena's financial position. Money of loan = j 1200 Money she has from her brother = left money after paying loan = j 2500 – 1200 Ex.17 Sol. = j 2500 Ex.18 Sol. j 1300 Ans. Find whether the given statements are true (T) or false (F) : (i) The smallest integer is 0. (ii) The opposite of zero on a number line is zero. (iii) Zero is not a positive integer. (iv) 0 is larger than every negative integer but less than every positive integer. (v) A positive integer is greater than its opposite. (vi) Every integer is less than every natural integer. (vii) –1 is the greatest negative integer. (viii) 0 is the smallest positive integer. (ix) The sum of greatest negative integer and smallest positive integer is zero. (x) The negative of a positive integer is a negative integer. (xi) The negative of a negative integer is positive. (xii) If a and b are two integers such that a < b then (b – a) is always a positive integer. (i) F (ii) T (iii) T (iv) T (v) T (vii) T (viii) F (x) T (xi) T (vi) F (ix) T (xii) T Ex.19 Sol. Ex.20 Sol. Ex.21 Sol. PROPERTIES OF ADDITION AND SUBTRACTION (1) Closure Addition (2) Commutative (3) Associative  (4) Additive Identity   Subtraction    × × × (5) Additive Inverse   Eg.(i) 5 + 3 = 8 (integer), –7 + 3 = –4 (integer) Eg.(ii) 3 + 7 = 10 = 7 + 3, 4 – 5 = –1 & 5 – 4 = 1 1  (7  9)  1  (2) 2  (3  5)  2  8  10 ,  1  2  3 Eg.(iii) ( 2  3)  5 (1  7)  9  6  9  5  5  10 ,  15  Ex.22 Sol. EXAMPLES  Find the integer for the following integers so that sum is zero. 23, –3, 0, –1, 7, 10, 253, –497 23 + (–23) = 0 ; –1 + (1) = 0; 253 + (–253) = 0 –3 + (3) = 0 ; 7 + (–7) = 0; –497 + (497) = 0 0+0=0 ; 10 + (–10) = 0; Note : (i) Sum of the given two integers in each of the given pairs is zero i.e. the additive identity for integers. (ii) To find the additive inverse, we change the + sign into – sign (except in case of 0) of the given integer and vice-versa. (iii) Each of the integer in such a pair is called the additive inverse of the other e.g. –8 is the additive inverse of 8. Write a pair of integers whose (i) sum is –7 and (ii) difference is –9. (i) –9 + 2 = –4 + (–3) = –7 (ii) 1 – 10 = 2 – 11 = –9 Write a pair of integers whose difference is : (i) a negative number (ii) an integer greater than only one of the integers. (i) –14 – (–5) = –9 (Negative integer) (ii) (–11) – (–3) = –8 (It is greater than – 11 and less than –3) Verify : [–a – (–b)] – c –a – [–b – (c)] : if a = 3, b = 7, c = –9 LHS = [–a – (–b)] – c = [–3 – (–7)] – (–9) = [–3 + 7] + 9 = 4 + 9 = 13 RHS = –a – [–b – (c)] = – 3 – [– 7 – (–9)] = – 3 – [– 7 + 9] = – 3 – [2] = –5 LHS RHS Verify a – (–b) = a + b for the following : a = 117, b = –112 LHS = a – (–b) = 117 – [– (–112)] = 117 – (112) =5 RHS = a + b = 117 + (–112) = 117 – 112 = 5 LHS = RHS MULTIPLICATION OF INTEGERS (i) Two positive numbers. (ii) One positive and one negative number or negative to positive number. (iii) Two negative numbers. Eg. (i) 5 × 6 = 30 (ii) 7 × 9 = 63 (iii) 9 × 10 = 90 (iv) –3 × 1 = –3 (v) –7 × 9 = –63 (vi) –11 × 11 = –121 (vii)13 × –5 = – 65 (viii) 10 × –10 = –100 (ix) –40 × –20 = 800 (x) –5 × –1 = 5 Sign system for multiplication (+) × (+) = + (–) × (+) = – Negative × Positive = Negative (–) × (–) = + Negative × Negative = Positive (+) × (–) = – Note: Positive × Positive = Positive (iii) 7 × (5 – 7) = 7 × (–2) = –14  Eg : 7 × 1 = 7, –3 × 1 = –3, 9 × 0 = 0 Ex.23 In a class test containing 20 questions, 3 marks are given for every correct answer and –1 mark is given for every incorrect answer. (ii) One of her friends attempt 8 questions but only one answer is incorrect. What is her friend’s total score ? Sol. (i) Marks given for one correct answer = 3 So, Marks given for 11 correct answer = 3 × 11 = 33 Marks given for one incorrect answer = –1 So, Marks given for 9 incorrect answers = –1 × 9 = –9 Therefore, Ritu's total score = 33 – 9 = 24 (ii) Marks given for one correct answer = 3 So, Marks given for 7 correct answer = 3 × 7 = 21 Marks given for one incorrect answer  ( Associative) (2  3)  20  6  20  120 For any three integers a, b, c ; a × (b + c) = a × b + a × c Let us observe the following products : (i) 7 × (2 + 5) and 7×2+7×5 = 49 = 14 + 35 = 49 Thus, 7 × (2 + 5) =7×2+7×5 (ii) –2 (–3 + 1) and –2 × –3 + (–2) × (1) = –2 (–2) = (–2) × (–3) + (–2) × 1 = –2 × –2 =6–2 =4 =4 Thus –2 × (–3 + 1) = –2 × (–3) + (–2) × (1) This property of integers is known as the distributive property of multiplication over addition. EXAMPLES  (i) Ritu attempt all questions but only 11 of her answers are correct. What is her total score ? Eg. (i) 16 × 12 = 192 (integer) (ii) 17 × 10 = 170 = 10 × 17 (commutative) (iii) 2  (3  20)  2  60  120 DISTRIBUTIVE PROPERTY = –14 Note : Any number ‘a’ when multiply by 1 and 0, gives itself and 0 respectively.   1 reciprocal of given number = 35 – 49 This property of integers is known as the distributive property of multiplication over subtraction. PROPERTIES OF MULTIPLICATION (i) Closure (ii) Commutative  (iii) Associative identity (iv) Multiplicative identity (v) Multiplicative inverse 7×5–7×7 Thus, 7 × (5 – 7) = 7 × 5 – 7 × 7 Positive × Negative = Negative (i) If negative integers are multiplied even times, product is always a positive integer. (ii) If negative integers are multiplied odd times, product is always a negative integer. and Edubull = 1 × –1 = –1 Therefore, her friend’s total score Ex.24 = 21 – 1 = 20 Complete the following multiplication : × –7 –6 –5 –4 0 –7 –6 5 4 Edubull Sol. × –7 –6 42 –7 –5 –4 Ex.25 –6 Compare : 4 49 42 –35 –28 35 30 –25 –20 36 28 0 5 –30 24 0 –20 0 0 –24 –16 0 and (–4) – 6 × –7 = – 4 – 6 × –7 = – 4 + 42 = 38 Ex.27 [(–4 – 6)] × –2 < (–4) – 6 × –7 If a × (–1) = –25, is the integer a positive or negative ? Match the following : a is positive (i) (–7) + 9 = 9 + (–7) (ii) 6 + [3 + (–2)] = [(6 + 3)] + (–2) (a) property of multiplicative identity (b) Commutative property of (iii) (–8) (–5) = (–5) (–8) addition (c) Multiplicative property of zero (iv) 4[5 × (–5)] = (4 × 5) (–5) (d) Associative property of multiplication (v) 7 × 0 = 0 (e) Associative property of addition (vi) 13 × 1 = 13 Sol. (i) b (ii) e (iii) f (f) Commutative property of (iv) d multiplication (v) c (vi) a ÷ 3 = –8 –24 35 ÷ = 7 48÷ = –8 –24 ÷ = –48 (vi) –8 × (ii) [(–4 – 6)] × (–2) = –10 × –2 = 20 – a = – 25 a = 25 –40 ÷ –5 = 8 (v) –6 × 4 = – 24 (7 + 9) × 10 > (7 + 9 × 10) Ans. 24 ÷ 4 = 6 (iv) 7 × 5 = 35 (By BODMAS Rule) Sol. Eg : Observe the pattern and fill up the boxes. (iii) –8 × 3 = –24  7 + 9 × 10 = 16 × 10 = 160 = 7 + 90 = 97 Ex.26 For example, to divide 32 by –4 means to find a number by which – 4 should be multiplied such that it gives the product 32. The answer is –8. (ii) 8 × –5 = –40 (ii) [(–4 – 6)] × (–2) and (–4) – 6 × –7 and Division is the reverse process of multiplication. (i) 6 × 4 = 24 (i) (7 + 9) × 10 and 7 + 9 × 10 Sol. (i) (7 + 9) × 10 DIVISION OF INTEGERS = 4 5 –6 –6 SIGN SYSTEM FOR DIVISION (i) The quotient of two integers involving two like signs is positive or (+) ÷ (+) = + and (–) ÷ (–) = +. (ii) The quotient of two integers having opposite signs is negative or (+) ÷ (–) = – and (–) ÷ (+) = –. Properties of division (1) Closure No (divisor should be non zero) (3) Associative No (2) Commutative No Eg :(i) 25 ÷ 5 = 5 (integer) (ii) 20 ÷ 10 = 2 (integer) (iii) 30 ÷ 7 integer (iv) 20 ÷ 5 = 4  5 ÷ 20 (v) (36 ÷ 9) ÷ 2 = 4 ÷ 2 = 2 36 ÷ (9 ÷ 2) = 36 ÷ = 36  9 2 2 =4×2=8 9 Note : Thus, division of any non-zero integer by zero is an undefined operation. Ex.28 Sol. EXAMPLES  The product of two integers is –120. If one number is –30, what is the other. Let the other number be ‘a’ Then according to questions (a) (–30) = –120 Ex.29 a = –120 ÷ –30 = 40 Ans. In a test +4 marks are given for every correct answer and –2 marks are given for every incorrect answer. (i) Neeta answered all the questions and scored 40 marks though she got 15 correct answers. (ii) Radhey also answered all the questions and scored –16 marks though he got 5 correct answers. How many incorrect answers had they attempted? Sol. (i) Marks given for one correct answer = 4 So, Marks given for 15 correct answers = 4 × 15 = 60 Neeta’s score = 40 Marks obtained for incorrect answers = 40 – 60 = –20 Marks given for one incorrect answer = –2 Therefore, number of incorrect answers = –20 ÷ –2 = 10 (ii) So, Marks given for 5 correct answers = 5 × 4 = 20 Radhey’s score = –16 Marks obtained for incorrect answers = –16 – 20 = –36 Marks given for one incorrect answers = –2 Therefore, number of incorrect answers Ex.30 = –36 ÷ –2 = 18 A shopkeeper earns a profit of j 2 by selling one pen and incurs a loss of 50 paise per pencil while selling pencils of her old stock. (i) In a particular month she incurs a loss of j 10. In this period, she sold 45 pens. How many pencils did she sell in this period ? (ii) In the next month, she earns neither profit nor loss. If she sold 80 pens, how many pencils did she sell ? Edubull Sol. (i) Profit earned by selling one pen = j 2 Profit earned by selling 45 pens = 2 × 45 = j 90 Total loss given = 10, which we denote by j 10 Profit earned + Loss incurred = Total loss Therefore, Loss incurred = Total loss – Profit earned = j (– 10 – 90) = j –100 = – 10000 paise So, Number of pencils sold = – 10000 ÷ –50 = 200 pencils (ii) In the next month, there is neither profit nor loss. So, Profit + Loss incurred = 0 It means profit earned = – Loss incurred Now, profit earned by selling 80 pens = 2 × 80 = j 160 Hence, loss incurred by selling pencils = j 160 Which we indicate by – j 160 or – 16000 paise Total number of pencils sold = (–16000) ÷ 50 = 320 pencils RULE OF BODMAS B stands for brackets, O for the operation ‘Of ’ D for division, M for multiplication, A for addition and S for subtraction. Types of bracket  Round brackets or parenthesis () Curly brackets or braces {} Square brackets [] bar or vinculum – Vinculum or bar is used as the innermost brackets and then ( ), then { }, and finally [ ]. Eg. (i) (8  2  2) means 8 ÷ 4 (ii) 10 + [5 × {48 ÷ (2 × 4)}] = 10 + [5 × {48 ÷ 8}] = 10 + [5 × 6] = 10 + 30 = 40 Ans. THE OPERATION ‘OF’ Eg. (i) 9 of half of 20 means 9 of (ii) One third of 213 means 20 = 9 × 10 = 90 2 1  213  71 3 Ex.31 Sol. EXAMPLES  Simplify : 57 – [28 – {16 + (5 – 3  1 )}]. 57 – [28 – {16 + (5 – 3  1 )}] = 57 – [28 – {16 + (5 – 2)}] [Removal of bar] = 57 – [28 – {16 + 3}] [Innermost brackets removed] = 57 – [28 – 19] [Next Innermost brackets removed] = 57 – 9 = 48 Ex.32 Simplify : (i) 7 – {13 – 2(4 of – 4)} (ii) 81 of [59 – {7 × 8 + (13 – 2 of 5)}] Sol. (i) 7 – {13 – 2(4 of – 4)} = 7 – {13 – 2(4 × – 4)} = 7 – {13 – 2(–16)} = 7 – {13 – (–32)} = 7 – {13 +32} = 7 – 45 = – 38 (ii) 81 of [59 – {7 × 8 + (13 – 2 of 5)}] = 81 × [59 – {7 × 8 + (13 – 2 × 5)}] = 81 × [59 – {7 × 8 + (13 – 10)}] = 81 × [59 – {56 + 3}] = 81 × [59 – 59] = 81 × 0 = 0 Edubull