Number Play
Number Play — Study Notes
NCERT-aligned · 7 notes · 3 shown free
Introduction
ExplanationIntroduction
The chapter 'Number Play' introduces students to the fascinating world of numbers and their properties. It begins by revisiting the concept of numbers that students have encountered in earlier classes, such as natural numbers, whole numbers, and integers, and then extends the discussion to include rational numbers and their decimal expansions. The chapter emphasizes the importance of understanding numbers beyond mere counting and arithmetic operations, focusing on patterns, properties, and the behavior of numbers in various forms. It also introduces the concept of irrational numbers, setting the stage for more advanced topics in mathematics. The chapter aims to develop a deeper appreciation of numbers through exploration and problem-solving, encouraging students to think critically about numerical relationships and patterns. This foundational knowledge is crucial for higher mathematics, including algebra, geometry, and number theory.
- Numbers are fundamental to mathematics and everyday life.
- The chapter revisits types of numbers: natural, whole, integers, rational, and irrational.
- Focus on decimal expansions of rational numbers and their patterns.
- Introduction to irrational numbers and their non-repeating, non-terminating decimal expansions.
- Encourages exploration of number properties and patterns.
- Sets foundation for advanced mathematical concepts.
- 📌 Natural Numbers: Counting numbers starting from 1, 2, 3, ...
- 📌 Whole Numbers: Natural numbers including zero (0, 1, 2, 3, ...)
- 📌 Integers: Whole numbers including negative numbers (... -3, -2, -1, 0, 1, 2, 3, ...)
Decimal Representation of Rational Numbers
ExplanationDecimal Representation of Rational Numbers
This section delves into the decimal expansions of rational numbers, explaining how every rational number can be expressed in decimal form, which either terminates or repeats. It begins by defining rational numbers as numbers that can be written as a fraction p/q, where p and q are integers and q ≠ 0. The decimal representation of such numbers is obtained by performing division of p by q. The section explains two types of decimal expansions: terminating decimals, where the division ends after a finite number of steps, and non-terminating repeating decimals, where a pattern of digits repeats indefinitely. For example, 1/4 = 0.25 is a terminating decimal, while 1/3 = 0.333... is a non-terminating repeating decimal. The section also discusses how to identify the repeating part of the decimal and how to write it using a bar notation over the repeating digits. It further explains that every terminating decimal can be expressed as a fraction with a denominator that is a power of 10, and every repeating decimal can be converted back into a rational number. The section includes methods to convert repeating decimals into fractions, reinforcing the concept that rational numbers and their decimal expansions are closely linked.
- Rational numbers can be expressed as decimals by dividing numerator by denominator.
- Decimal expansions of rational numbers are either terminating or non-terminating repeating.
- Terminating decimals end after a finite number of digits.
- Non-terminating repeating decimals have a repeating pattern of digits.
- Bar notation is used to denote the repeating part of a decimal.
- Every repeating decimal can be converted back into a rational number.
- 📌 Terminating Decimal: Decimal number with a finite number of digits after the decimal point.
- 📌 Non-terminating Repeating Decimal: Decimal number with infinite digits after the decimal point but with a repeating pattern.
- 📌 Bar Notation: A line placed over digits to indicate the repeating part in a decimal.
Irrational Numbers
ExplanationIrrational Numbers
This section introduces irrational numbers, which are numbers that cannot be expressed as a ratio of two integers. Unlike rational numbers, their decimal expansions neither terminate nor repeat. The section explains that irrational numbers fill the g
Practice Questions — Number Play
Includes NCERT exercise questions with answers
Q1.6. If 3p7q8 is divisible by 44, list all possible pairs of values for p and q.
Answer:
3p7q8 is divisible by 44, means divisible by 11 and 4. For divisibility by 4, last two digits must be divisible by 4. So possible q8 are 08, 28, 48, 68, 88. For divisibility by 11, difference between sum of the odd place digits and even place digits must be 0 or multiple of 11. Sum of odd place digits = 8 + 7 + 3 = 18. Sum of even place digits = p + q. Difference is 18 − (p + q). Let k = 18 − (p + q) = 0 or a multiple of 11. (i) if p + q = 18 Not possible for q (0, 2, 4, 6, 8), since p is a digit. (ii) if 18 – (p + q) = 11 then p + q = 7. Possible pairs: (p=7, q=0), (p=5, q=2), (p=3, q=4), (p=1, q=6). (iii) 18 – (p + q) cannot be other multiples of 11. Hence the possible pairs for p and q are (7,0), (5,2), (3,4), and (1,6).
Explanation:
Step 1: Divisibility by 4 requires last two digits q8 to be divisible by 4, so q8 can be 08, 28, 48, 68, 88. Step 2: Divisibility by 11 requires difference between sum of digits in odd and even positions to be 0 or multiple of 11. Sum odd = 3 + 7 + 8 = 18. Sum even = p + q. Difference = 18 - (p + q). Check for difference = 0 or ±11, ±22,... Only difference = 11 is possible with p + q = 7. Check all q from possible q values and find p accordingly. Thus, pairs are (7,0), (5,2), (3,4), (1,6).
Q2.7. Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Answer:
Let the three consecutive numbers be n, n+1, n+2. Given: n is multiple of 2, n+1 is multiple of 3, n+2 is multiple of 4. One such set is 2, 3, 4. Since the numbers are consecutive, the pattern repeats every LCM of 2, 3, and 4, which is 12. So, the next such set is 14, 15, 16. Hence, such numbers occur every 12 numbers.
Explanation:
Step 1: Assign variables to consecutive numbers. Step 2: Apply divisibility conditions. Step 3: Find one example: 2 (even), 3 (multiple of 3), 4 (multiple of 4). Step 4: The pattern repeats every LCM(2,3,4) = 12. Step 5: Next set is 14,15,16 and so on.
Q3.8. Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Answer:
Since 36 = 4 × 9, a number divisible by 36 must be divisible by both 4 and 9. For divisibility by 4, last two digits must be divisible by 4. For divisibility by 9, sum of digits must be divisible by 9. Between 45000 and 47000, multiples of 36 are: 45036, 45072, 45108, 45144, 45180. Approach: Check numbers ending with last two digits divisible by 4 and sum of digits divisible by 9.
Explanation:
Step 1: Understand divisibility rules for 4 and 9. Step 2: Check numbers in given range ending with digits divisible by 4. Step 3: Check sum of digits for divisibility by 9. Step 4: List first five such numbers.
Q4.9. The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
Answer:
Let the five consecutive even numbers be: 5p - 4, 5p - 2, 5p, 5p + 2, 5p + 4. Since the middle number is 5p, the two numbers before it are 2 and 4 less, and the two numbers after it are 2 and 4 more, respectively.
Explanation:
Step 1: Consecutive even numbers differ by 2. Step 2: Middle number is 5p. Step 3: Numbers before middle are 5p - 2 and 5p - 4. Step 4: Numbers after middle are 5p + 2 and 5p + 4.
Q5.10. Write a 6-digit number that is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Answer:
A number divisible by 15 must be divisible by 3 and 5. So, it must end with 0 or 5. If it ends with 0, reversed number starts with 0, which is not a 6-digit number. So, last digit f = 5. For reversed number to be divisible by 6, it must be divisible by 2 and 3. So, first digit a of original number (which is last digit of reversed) must be even and non-zero. Possible a values: 2, 4, 6, 8. Examples: 200025, 200055, 200085, 202005, etc., where sum of digits is divisible by 3. Hence, such numbers exist with these conditions.
Explanation:
Step 1: Divisibility by 15 requires divisibility by 3 and 5. Step 2: Number ends with 0 or 5. Step 3: If ends with 0, reversed number not 6-digit. Step 4: So ends with 5. Step 5: Reversed number divisible by 6 requires even first digit. Step 6: Choose a = 2,4,6,8. Step 7: Check sum of digits divisible by 3. Step 8: Construct numbers accordingly.
Q6.11. Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11. But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true; explain your conclusion.
Answer:
Deepak’s conjecture is false. Let n be a multiple of 11, so n = 11k. When doubled, 2n = 2 × 11k = 11 × (2k), which is also a multiple of 11. Hence, all multiples of 11 when doubled remain multiples of 11.
Explanation:
Step 1: Represent multiple of 11 as 11k. Step 2: Double it: 2 × 11k = 11 × 2k. Step 3: Since 2k is an integer, 2n is multiple of 11. Step 4: Therefore, Deepak’s claim that some multiples of 11 when doubled are not multiples of 11 is incorrect.
Q7.12. Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning. (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9. (ii) The sum of three consecutive even numbers will be divisible by 6. (iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6. (iv) 8 (7b – 3) – 4 (11b + 1) is a multiple of 12.
Answer:
(i) Always true. (ii) Always true. (iii) Always true. (iv) Never true.
Explanation:
(i) Product of multiples of 6 and 3 includes at least two 3s and one 2, so multiple of 9. (ii) Sum of three consecutive even numbers is divisible by 6 because sum is 6 times the middle number. (iii) Rearranging digits does not affect divisibility by 6 if digits sum and last digit conditions hold. (iv) Expression simplifies to a form not divisible by 12 for all b, so never true.
All 7 Chapters in Ganita Prakash Part-I
Mathematics · Class 8