Natural Numbers And Whole Numbers-INTEGERS

1. INTEGERS We have learnt about whole numbers and integers in previous class. Everyone know that integers form a bigger collection of numbers which contains whole numbers and negative numbers. In this chapter, we will study more about integers, their properties and operations.    Representation of Integers on Number Line: Natural Numbers and Whole Numbers Natural numbers are counting numbers but these set of numbers do not include zero. Example: 1,2,3,4,5,6……etc are all natural numbers. All natural numbers along with zero are called Whole numbers. For example: 0, 1, 2, 3, 4, 5, 6………etc are all whole numbers. Properties of Integers

2. Integers are closed under: Addition Subtraction Multiplication Division     It means that operation on integers will also give integers. Properties of Addition and Subtraction of Integers Closure under Addition and subtraction: Integers are closed under addition. oIn general, for any two integers a and b, a + b is an integer. Integers are closed under subtraction. oThus, if a and b are two integers then a – b is also an integer.   Commutativity Property for addition: In general, for any two integers a and b, we can say a + b = b + a Associativity Property for addition: In general for any integers a, b and c, we can say a + (b + c) = (a + b) + c Additive Identity: Zero is an additive identity for integers. oIn general, for any integer a: a + 0 = a = 0 + a  Additive Inverse: If we add numbers like (-7) and 7 then we get the result as zero. So these are called the Additive inverse of each other. If we add (-2) + (2), then first we move 2 steps to the left of zero then we move two steps to the right of (- 2). So, finally we reached to zero.

3. Hence, if we add the positive and negative of the same number then we get the zero. Properties of Multiplication of Integers Closure under Multiplication: Integers are closed under multiplication. So, a × b is an integer, for all integers a and b.   Commutative Property of Multiplication: Multiplication is commutative for integers. In general, for any two integers a and b, a × b = b × a   Multiplication by Zero: The product of a negative integer and zero is zero. So, a × 0 = 0 × a=0   Multiplicative Identity: 1 is the multiplicative identity for integers. a × 1 = 1 × a = a   Associative property of Multiplication: Multiplication is associative for integers. (a × b) × c = a × (b × c)   Distributive Property of Integers: The distributivity of multiplication over addition is true for integers. oa × (b + c) = a × b + a × c The distributivity of multiplication over subtraction is true for integers. oa × (b – c) = a × b – a × c   Addition and Subtraction of Integers Addition of Two Positive Integers If you have to add two positive integers then simply add them as natural numbers. o(+6) + (+7) = 6 + 7 = 13  Addition of Two Negative Integers If we have to add two negative integers then simply add them as natural numbers and then put a negative sign on the answer. o(-6) + (-7) = - (6+7) = -13 

4. Addition of One Negative and One Positive Integer If we have to add one negative and one positive integer then simply subtract the numbers and put the sign of the bigger integer. We will decide according to the bigger integer and ignoring the sign of the smaller integer. o (-6) + (7) = 1 (bigger integer 7 is positive integer) o(6) + (-7) = -1(bigger integer 7 is negative integer)   Multiplication of Integers Product of two positive integers is a positive integer. Example: (+4) x (+2) = +8 Product of two negative integers is a positive integer. Example:(−5) × (−2) = +10 Product of a positive and a negative integer is a negative integer. Example: (+6) × (−3) = −18 Example: (−2) × (+5) = −10 Product of even number of negative integers is positive. Example: (-4) × (−3) = 12 Product of odd number of negative integers is negative. Example: (-4) × (−3) × (−2) = −24 Division of Integers When we divide a positive integer by a negative integer: We first divide them as whole numbers. And then, put a minus sign (-) before the quotient. oGeneral Rule: a ÷ (-b) = (-a) ÷ bwhere b ≠ 0.   When we divide a negative integer by a negative integer: We first divide them as whole numbers. And then, put a positive sign (+). oGeneral Rule: (-a) ÷ (-b) = a ÷ bwhere b ≠ 0   Any integer divided by 1 gives the same number. a ÷ 1 = a 

5. For any integer a, we have a ÷ 0 is not defined. Rules of Addition of Integers Rules for subtraction of integers

6. Rules for Multiplication and Division of Integers How to Tackle Problem Solving Strategy

7. Exercise 1 1. Following number line shows the temperature in degree Celsius (C°) at different places on a particular day. (a) Observe this number line and write the temperature of the places marked on it. (b) What is the temperature difference between the hottest and the coldest places among the above? (c) What is the temperature difference between Lahulspriti and Srinagar? (d) Can we say temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also less than the temperature at Srinagar? Solution: (a) From the given number line, we observe the following temperatures. Cities Lahulspriti -8°C Srinagar -2°C Shimla Ooty Bengaluru 22°C Temperature 5°C 14°C (b) The temperature of the hottest place = 22°C The temperature of the coldest place = -8°C Difference = 22°C – (-8°C) = 22°C + 8°C = 30°C (c) Temperature of Lahulspriti = -8°C Temperature of Srinagar = -2°C Difference = -2°C – (-8°C)

8. = -2°C + 8°C = 6°C (d) Temperature of Srinagar = -2°C Temperature of Shimla = 5°C Temperature of the above cities taken together = -2°C + 5°C = 3°C Temperature of Shimla = 5°C Hence, the temperature of Srinagar and Shimla taken together is less than that of Shimla by 2°C. i.e., (5°C – 3°C) = 2°C 2. In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, –5, –10, 15 and 10, what was his total at the end? Solution: Given scores are 25, -5, -10, 15, 10 Marks given for correct answers = 25 + 15 + 10 = 50 Marks given for incorrect answers = (-5) + (-10) = -15 => Total marks given at the end = 50 + (-15) = 50 – 15 = 35 3. At Srinagar temperature was –5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day? Solution: Initial temperature of Srinagar on Monday = -5°C

9. Temperature on Tuesday = -5°C – 2°C = -7°C Temperature was increased by 4°C on Wednesday. => Temperature on Wednesday = -7°C + 4°C = -3°C Hence, the required temperature on Tuesday = -7°C and the temperature on Wednesday = 3°C 4. A plane is flying at the height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200 m below the sea level. What is the vertical distance between them? Solution: Height of the flying plane = 5000 m Depth of the submarine = -1200 m => Distance between them = + 5000 m – (-1200 m) = 5000 m + 1200 m = 6200 m Hence, the vertical distance = 6200 m 5. Mohan deposits Rs. 2,000 in his bank account and withdraws Rs. 1,642 from it, the next day. If withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal. Solution: The deposited amount will be represented by a positive integer i.e., Rs. 2000. Amount withdrawn = Rs. 1,642 => Balance in the account = Rs. 2,000 – Rs. 1,642 = Rs. 358 Hence, the balance in Mohan’s account after the withdrawal = Rs. 358.

10. 6. Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance traveled towards west? By which integer will you represent her final position from A? Solution: Distances traveled towards east from point A will be represented by positive integer i.e. +20 km. Distance traveled towards the west from point B will be represented by negative integer, i.e., –30 km. Final position of Rita from A = 20 km – 30 km = – 10 km Hence, the required position of Rita will be presented by a negative number, i.e., -10. 7. In a magic square each row, column and diagonal have the same sum. Check which of the following is a magic square. (i) (ii) Solution: (i) Row one R1 = 5 + (-1) + (-4) = 5 – 1 – 4 = 5 – 5 = 0 Row two R2 = (-5) + (-2 ) + 7 = -5 – 2 + 7 = -7 + 7 = 0 Row three R3 = 0 + 3 + (-3) = 0 + 3 - 3 = 0 Column one C1t = 5 + (-5) + 0 = 5 – 5 + 0 = 0

11. Column two C2 = (-1) + (-2) + (3) = -1 – 2 + 3 = -3 + 3 = 0 Column three C3 = (-4) + 7 + (-3) = -4 + 7 – 3 = 7 – 7 = 0 Diagonal d12 = 5 + (-2) + (-3) = 5 – 2-3 = 5 – 5 = 0 Diagonal d2 = (-4) + (-2) + 0 = -4 –2 + 0 = -6 + 0 = -6 Here, the sum of the integers of diagonal d2 is different from the others. Hence, it is not a magic square. (ii) Row one R1 = 1 + (-10) + 0 = 1 – 10 + 0 = -9 Row two R2 = (-4) + (-3) + (-2) = -4 – 3 – 2 = -9 Row three R3 = (-6) + (4) + (-7) = -6 + 4 –7 = -9 Column one C3 = 1 + (-4) + (-6) = 1 – 4 – 6 = -9 Column two C2 = (-10) + (-3) + 4 = -10 – 3 + 4 = -9 Column C3 = 0 + (-2) + (-7) = 0 – 2 – 7 = -9 Diagonal d1 = 1 + (-3) + (-7) = 1 – 3 – 7 = 1 – 10 = -9

12. Diagonal d2 = 0 + (-3) + (-6) = 0 – 3-6 = -9 Here, sum of the integers column wise, row wise and diagonally is same i.e. -9. Hence, (ii) is a magic square. 8. Verify a – (-b) = a + b for the following values of a and b. (i) a = 21, b = 18 (ii) a = 118, b = 125 (iii) a = 75, b = 84 (iv) a = 28, b = 11 Solution: (i) a – (-b) = a + b LHS = 21 – (-18) = 21 + 18 = 39 RHS = 21 + 18 = 39 LHS = RHS Hence, verified. (ii) a – (-b) = a + b LHS = 118 – (-125) = 118 + 125 = 243 RHS = 118 + 125 = 243 LHS = RHS Hence, verified. (iii) a – (-b) = a + b LHS = 75 – (-84) = 75 + 84 = 159 RHS = 75 + 84 = 159 LHS = RHS Hence, verified. (iv) a – (-b) = a + b

13. LHS = 28 – (-11) = 28 + 11 = 39 RHS = 28 + 11 = 28 + 11 = 39 LHS= RHS Hence, verified. 9. Use the sign of >, < or = in the box to make the statements true. (a) (-8) + (-4) [] (-8) – (-4) (b) (-3) + 7 – (19) [] 15 – 8 + (-9) (c) 23 – 41 + 11 [] 23 – 41 – 11 (d) 39 + (-24) – (15) [] 36 + (-52) – (-36) (e) -231 + 79 + 51 [] -399 + 159 + 81 Solution: (a) (-8) + (-4) [] (-8) – (-4) LHS = (-8) + (-4) = -8 – 4 = -12 RHS = (-8) – (-4) = -8 + 4 = -4 Here – 12 < -4 Hence, (-8) + (-4) [<] (-8) – (-4) (b) (-3) + 7 – (19) [] 15 – 8 + (-9) LHS = (-3) + 7 – (19) = -3 + 7-19 = -3 – 19 + 7 = -22 + 1 = -15 RHS = 15 – 8 + (-9) = 15-8-9 = 15 – 17 = -2 Here – 15 < -2

14. Hence, (-3) + 7 – (19) [<] 15 – 8 +(-9) (c) 23 – 41 + 11 [] 23 – 41 – 11 LHS = 23 – 41 + 11 = 23 + 11 – 41 = 34 – 41 = -7 RHS = 23 – 41 – 11 = 23 – 52 = -29 Here, -7 > -29 Hence, 23 – 41 + 11 [>] 23 – 41 – 11 (d) 39 + (-24) – (15) [] 36 + (-52) – (-36) LHS = 39 + (-24) – (15) = 39 – 24 – 15 = 39 – 39 = 0 RHS = 36 + (-52) – (-36) = 36 – 52 + 36 = 36 + 36 – 52 = 72 – 52 = 20 Here 0 < 20 Hence, 39 + (-24) – (15) [<] 36 + (-52) – (-36) (e) -231 + 79 + 51 [] -399 + 159 + 81 LHS = -231 + 79 + 51 = -231 + 130 = -101 RHS = -399 + 159 + 81 = 399 + 240 = -159 Here, -101 > -159 Hence, -231 + 79 + 51 [>] -399 + 159 + 81 10. A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step. (i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level? (ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?

15. (iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his move in part (t) and (ii) by completing the following: (a) – 3 + 2 –… = -8 (b) 4 –2 + … = 8. In (a) the sum (-8) represents going down by eight steps. So, what will the sum 8 in (b) represent? Solution: (i) The position of monkey after the 1st jump J1 is at 4th step ¯ 2nd jump J2 is at 2nd step

16. 3rd jump J3 is at 5th step ¯ 4th jump J4 is at 3rd step 5th jump J5 is at 6th step ¯ 6th jump J6 is at 4th step 7th jump J7 is at 7th step ¯ 8th jump J8 is at 5th step 9th jump J9 is at 8th step ¯ 10th jump J10 is at 6th step 11th jump J11 is at 9th step ¯ (Water level) Hence the required number of jumps = 11. (ii)Monkey’s position after the 1st jump J1 is at 5th step 2nd jump J2 is at 7nd step 3rd jump J3 is at 3rd step 4th jump J4 is at 5th step ¯ 5th jump J5 is at 1st step

17. Hence, the required number of jumps = 5. (iii) According to the given conditions we have the following tables Jumps Number of steps -3 +2 -3 +2 -3 +2 -3 +2 -3 +2 -3 J1 J2 J3 J4 J5 J6 J7 J8 J9 J10 J11 Therefore (a) Total number of steps = -3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 = -8 which represents the monkey goes down by 8 steps. In case (ii), we get Jumps Number of steps +4 -2 +4 -2 +4 J1 J2 J3 J4 J5 Therefore (b) Total number of steps. = +4 – 2 + 4 – 2 + 4 = 8 Here, the monkey is going up by 8 steps.