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Real Numbers Class 10 Maths Notes




Table of Content
  • Real Numbers Important Definition
  • Euclid's Division Lemma
  • Proof of Euclid's Division Lemma
  • HCF (Highest common factor)
  • What is Prime Numbers
  • What is Composite Numbers
  • Fundamental Theorem of Arithmetic
  • HCF and LCM by prime factorization method
  • Irrational Numbers
  • How to prove the irrational numbers or Rational numbers

    • Real Numbers:Important Definition


      Natural and Whole Number


      What is Natural Numbers?
      Set of counting numbers is called the Natural Numbers
      N = {1,2,3,4,5,...}
      What is whole number?
      Set of Natural numbers plus Zero is called the Whole Numbers
      W= {0,1,2,3,4,5,....}
      Note:
      So all natural Number are whole number but all whole numbers are not natural numbers

      Integers


      What are Integers Numbers
      Integers is the set of all the whole number plus the negative of Natural Numbers
      Z={..,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,...}
      Note
      1) So ,integers contains all the whole number plus negative of all the natural numbers
      2)the natural numbers without zero are commonly referred to as positive integers
      3)The negative of a positive integer is defined as a number that produces 0 when it is added to the corresponding positive integer
      4)natural numbers with zero are referred to as non-negative integers
      5) The natural numbers form a subset of the integers.

      Rational and Irrational Numbers


      Rational Number

      : A number is called rational if it can be expressed in the form p/q where p and q are integers ( q> 0).
      Example : 1/2, 4/3 ,5/7 ,1 etc.
      Important Points to Note
      • every integers, natural and whole number is a rational number as they can be expressed in terms of p/q
      • There are infinite rational number between two rational number
      • They either have termination decimal expression or repeating non terminating decimal expression
      • The sum, difference and the product of two rational numbers is always a rational number. The quotient of a division of one rational number by a non-zero rational number is a rational number. Rational numbers satisfy the closure property under addition, subtraction, multiplication and division.


      Irrational Number

      : A number is called rational if it cannot be expressed in the form p/q where p and q are integers ( q> 0).
      Example : √3,√2,√5,p etc.
      Important Points to Note
      • Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Using this theorem we can represent the irrational numbers on the number line.
      • They have non-terminating and non-repeating decimal expression
      • The sum, difference, multiplication and division of irrational numbers are not always irrational. Irrational numbers do not satisfy the closure property under addition, subtraction, multiplication and division

      Real Numbers:


      • All rational and all irrational number makes the collection of real numbers. It is denoted by the letter R
      • We can represent real numbers on the number line. The square root of any positive real number exists and that also can be represented on number line
      • The sum or difference of a rational number and an irrational number is an irrational number.
      • The product or division of a rational number with an irrational number is an irrational number.
      • This process of visualization of representing a decimal expansion on the number line is known as the process of successive magnification

      Real numbers satisfy the commutative, associative and distributive laws. These can be stated as :
      Commutative Law of Addition:
      $a+b= b+a$
      Commutative Law of Multiplication:
      $a \times b=b \times a$
      Associative Law of Addition:
      $ a + (b+c)=(a+b) +c $
      Associative Law of Multiplication:
      $a \times (b \times c)=(a \times b) \times c $
      Distributive Law:
      $a \times (b + c)=(a \times b) + (a \times c)$
      or
      $(a + b) \times c=(a \times c) + (b \times c)$

      Real Numbers Examples
      1
      -2/3
      3/4
      √2
      √2 + 5

      Laws of exponents:


      Let a > 0 be a real number and p and q be rational numbers. Then, we have
      1. $a^p.a^q=a^{p+q}$
      2. $\frac {a^p}{a^q} =a^{p-q}$
      3. $(a^p)^q=a^{pq}$
      4. $a^p.b^p=(ab)^p$

      Euclid's Division Lemma

      For a and b any two positive integer, we can always find unique integer q and r such that
      $a=bq + r \; , \; 0 \leq r < b$

      It is basic concept and it is restatement of division
      • a is called dividend
      • b is called divisor
      • q is called quotient
      • r is called remainder.
      If Remainder r =0, then b is divisor of a.
      Example
      for 11 and 2
      $11= 2 \times 5 + 1$
      for 10 and 2
      $11= 2 \times 5$

      Proof of Euclid's Division Lemma

      Here we need to argue that q and r are no unique.So we are going to prove using contradiction method
      Let us assume q and r are not unique i.e. let there exists another pair $q_0$ and $r_0$ i.e. $a = bq_0 + r_0$, where $ 0 \leq r_0 < b$

      =>$ bq + r = bq_0 + r_0$
      => $b(q - q_0) = r - r_0$ ................ (I)

      Since $0 \leq r < b$ and $ 0 \leq r_0 < b$, thus $0 \leq r - r_0 < b$ ......... (II)

      The above equation (I) tells that b divides (r - r0) and Equation (II) tells us (r - r0) is an integer less than b. Both of these condition can be satisfied only when (r - r0) is 0.
      => $r - r_0 = 0$
      => $r = r_0$

      Equation (I) will be, $b(q - q_0) = 0$
      Since b > 0, => $(q - q_0) = 0$
      => $q = q_0$

      Since $r = r_0$ and $q = q_0$, Therefore q and r are unique.

      HCF (Highest common factor)

      Before starting on this topic, we need to prove an important theorem which will be used in finding HCF between two numbers

      Theorem

      If a and b are positive integers such that a = bq + r, then every common divisor of a and b is a common divisor of b and r, and vice-versa.

      Proof : Let m be a common divisor of a and b. Then,
      m| a => a = mx for some integer x
      m| b => b = mq2 for some integer y
      Now, a = bq + r
      => r = a - bq
      => r = mx - my q
      => r = m( x - yq)
      => m| r
      => m| r and m | b
      => m is a common divisor of b and r.
      Hence, a common divisor of a and b is a common divisor of b and r.

      How to find HCF (Highest common factor)

      Now HCF of two positive integers can be find using the Euclid's Division Lemma algorithm and above stated theorem

      We know that for any two integers a,b. we can write following expression
      $a=bq + r \; , \; 0 \leq r < b$
      If r=0 ,then
      HCF( a,b) =b
      If r >0 , then
      HCF (a,b) = HCF ( b,r) as already proved from above theorem

      Again expressing the integer b,r in Euclid's Division Lemma, we get
      b=pr + r1
      HCF ( b,r)=HCF ( r,r1)
      Similarly successive Euclid 's division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b

      Examples

      Use Euclid's algorithm to find the 65 and 117.

      Solution :
      Step:1 Since 117 > 65 we apply the division lemma to 117 and 65 to get ,
      $117 = 65 \times 1 + 52$
      Step:2 Since 52 > 0 , we apply the division lemma to 65 and 52 to get
      $65 = 52 \times 1 + 13$
      Step:3 Since 13 > 0 , we apply the division lemma to 52 and 13 to get
      $52 = 13 \times 4 + 0$
      The remainder has now become zero, so our procedure stops. Since the divisor at this Step is 13, the HCF of 117 and 52 is 13.

      Some other points to remember
      1.The HCF of three numbers can be calculated by first calculating the HCF of first two numbers ,then the calculating the HCF of the HCF of previous two numbers and third number.
      2. If HCF ( a,b) =1 ,the a and b are co primes.

      What is Prime Numbers

      A Prime Number is a number that cannot be evenly divided by any other number (except 1 or itself).

      Examples 2,3,5,7,11,13,17,19,23,29.......
      Prime numbers are interesting blocks in Mathematics.They are highly used in cryptography field

      Prime Number help
      PRIME NUMBERS to 100 ={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}
      Rules to find Prime Number
      Rule Number 1:
      Since even numbers are always divisible by 2, no even number (other than 2) can be a prime number because they'll always have more than two factors. For example, the factor pairs for 6 are: 1,2,3,6 So 6 has four factors and therefore is not a prime number.

      Rule Number 2:
      Numbers that end in 5 or 0 are always divisible by 5. So no numbers divisible by 5 (other than 5) can be prime numbers because they'll always have more than two factors. They'll have 1 and themselves, and they'll also have 5 and some other factor.

      Rule Number 3:
      0 and 1 are not prime numbers. It may seem at first glance that they are, but remember that a prime number has only 1 and itself as factors. But 0 has an infinite number of factors because 0 * 1 = 0, and 0 * 2 = 0, and 0 * 3 = 0, and so on. So 0 is far from prime.

      In the case of 1, it doesn't have two distinct factors because it's only factor is itself: 1 * 1 = 1. If you change one of those factors to any other integer, you no longer get a product of 1. Since it has only one factor, it's not prime.

      Rule Number 4:
      Add all of the digits composing the number. If the sum of the digits is divisible by 3, then so is the original number. Example: If the number is 111, add 1 + 1 + 1 = 3. The sum of the digits (3) is divisible by 3, so 111 is also divisible by 3.

      Rule Number 5:
      Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule repeatedly as necessary. Example: 826. Twice 6 is 12. So, take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also

      Rule Number 6:
      If you are checking if a number is a prime number,then you don't need to check all the prime number below it. you need to check run for small primes(2,3,5,7) and you're not done factoring, then keep trying bigger and bigger primes (11, 13, 17, 19, 23, etc.) until you find something that works or until you reach primes whose squares are bigger than what you're dividing into. if you don't find then it is a prime Number Reason: If your prime doesn't divide in, then the only potential divisors are bigger primes. Since the square of your prime is bigger than the number, then a bigger prime must have as its remainder a smaller number than your prime. The only smaller number left, since all the smaller primes have been eliminated, is 1. So the number left must be prime, and you're done.
      Lets take it with an Example
      The number given is 1017
      We can observer that 1031 lies between square of 30 and 33. So we can just keep trying till 33
      a. 1031 is not divisible by 2
      b. 1031 is not divisible by 3
      c. 1031 is not divisible by 5
      d. 1031 is not divisible by 7
      e. 1031 is not divisible by 11
      f. 1031 is not divisible by 13
      g. 1031 is not divisible by 17
      h. 1031 is not divisible by 19
      i. 1031 is not divisible by 23
      h. 1031 is not divisible by 29
      j. 1031 is not divisible by 31
      The next prime is 37. So we are done factoring. We can say 1031 is a Prime Number
      List of Prime Number till 1000
      2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101
      103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197
      199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311
      313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431
      433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557
      569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673
      677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811
      821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941
      947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039

      What is Composite Numbers

      A whole number that can be divided evenly by numbers other than 1 or itself

      Examples
      4,6,8,9,10.....

      Since 4 is divided by 2,6 is divided by 2 and 3

      Fundamental Theorem of Arithmetic

      Every composite number can be written as the product of power of primes and this factorization is unique
      Composite number = Product of primes

      Example

      $16=2 \times 2 \times 2 \times 2$
      This is unique, We cannot express it in any other prime numbers

      Few things to take note
      • The prime Number can be repeated in the factorization
      • The order does not matter
      • The prime factorisation of a natural number is unique, except for the order of its factors.

      How to Factorize the Composite Numbers(Prime Factorisation)

      1. Few things we need to remember,
        a. If the number is even,then it will be divisible by 2
        b. If the sums of its digits is divisible by 3,then it is divisible by 3
        c. if the end of the number is 0 or 5,then it is divisible by 5
      2. We have to start with the small prime number with the rules given in step 1. Once we find the quotient, repeat the same process for the quotient. The last quotient will be a prime number itself

      Example

      Suppose the composite Number is 168
      1. Now 168 is even number.So, we know it will get divided by 2
      168 / 2 = 84
      2. Again 84 is even number.So, we know it will get divided by 2
      84 / 2 = 42
      3. Again 42 is even number.So, we know it will get divided by 2
      42 / 2 = 21
      4. Now sum of digits of 21 is 3.So, we know we can divide it by 3
      21 / 3 = 7
      5. 7 is a prime Number

      So prime factors = 2 X 2 X 2 X 3 X 7
      Prime Factorisation Tree Form
      Prime Factorisation tree for Real Numbers | Class 10 Maths notes for Real Numbers

      Here are the prime factors of the composite numbers between 1 and 30.
      $4 = 2 \times 2$
      $6 = 3 \times 2$
      $8 = 2 \times 2 \times 2$
      $9 = 3 \times 3$
      $10 = 5 \times 24$
      $12 = 3 \times 2 \times 2$
      $14 = 7 \times 2$
      $15 = 5 \times 3$
      $16 = 2 \times 2 \times 2 \times 2$
      $18 = 3 \times 3 \times 2$
      $20 = 5 \times 2 \times 2$
      $21= 3 \times 7$
      $22=2 \times 11$
      $24= 2 \times 2 \times 2 \times 3$
      $25= 5 \times 5$
      $26= 2 \times 13$
      $27= 3 \times 3 \times 3$
      $28= 2 \times 2 \times 7$
      $30= 2 \times 3 \times 5$

      HCF and LCM by prime factorization method


      We can find HCF and LCM between two number by the prime factorization method also
      HCF (Highest Common factor) = product of the smallest power of each common factor in the numbers

      Example: Suppose the number are 14,24
      1. Lets do prime factorization method for both the numbers
      $14=2 \times 7$
      $24= 2 \times 2 \times 2 \times 3= 2^3 \times 3$
      2. Now as per method 2 is the only factor and lowest power is 1
      HCF =2

      LCM( Lowest Common Multiple) = Product of the greatest power of each prime factor involved in the number
      Example: Suppose the number are 14,24
      1. Lets do prime factorization method for both the numbers
      $14=2 \times 7$
      $24= 2 \times 2 \times 2 \times 3= 2^3 \times 3$
      2. Now as per method
      $LCM =2^3 \times 7 \times 3=168$
      Also important Formula to remember
      If a and b are two number,then
      $HCF(a,b) \times LCM (a,b) =a \times b $
      So if HCF and both the numbers are given, we can calculate the LCM using the formula
      $LCM = \frac {a \times b}{HCF}$
      if LCM and both the numbers are given, we can calculate the HCF using the formula
      $HCF = \frac {a \times b}{LCM}$
      These Formula are Valid for two numbers only. This is not true for 3 or more Numbers

      We can also LCM using Division algorithm .In this method divide the given numbers by common prime number until the remainder is a prime number or one. LCM will be the product obtained by multiplying all divisors and remaining prime numbers.
      Steps are
      1. We place number in the line
      2. We start dividing the number by least prime number which is common among all of them or group of them
      3. Keep dividing by least until we have 1's in the remainder
      4. LCM is the product of the divisors

      Example
      1. Find the LCM of 14 and 20
      Finding LCM using division method
      $LCM = 2 \times 2 \times 7 \times 5=140$
      2. Find the LCM of 12 ,18 and 44
      Finding LCM of more than 2 numbers using division method
      $LCM = 2 \times 2 \times 3 \times 3 \times 11=396$

      Irrational Numbers

      A number r is called irrational number if it cannot be expressed in the form p/q where p and q are integer and q≠ 0

      Important Theorem

      Let p be a prime number ,if p divides a2 ,the p divide a where a is positive number

      Rational numbers

      A number r is called rational number if it can be expressed in the form p/q where p and q are integer and q≠ 0
      Decimal expressions of rational number are of two types
      a. Terminating decimal expression
      b. Non-terminating repeating decimal expression

      Terminating decimal expression can be written in the form
      $\frac {p}{2^n5^m}$
      Non terminating repeating decimal expression can not be expressed in this form
      $\frac {p}{2^n5^m}$

      How to prove the irrational numbers or Rational numbers

      1. We know the definition of each of them,so first try to express the number in the form p/q.If it is not possible then proceed to next step
      2. If the proof is irrational number which will be the case always,assume the number to rational in the form p/q and then try to prove it wrong

      Solved Examples


      1.Prove that $\sqrt {2}$ is irrational number
      Solution: Since we cannot clearly express in p/q form,it is difficult to say,So lets us assume this is rational number
      then
      $\sqrt {2}=\frac {p}{q}$

      where p and q are co primes.
      or
      $q\sqrt {2}=p$
      squaring both sides
      $2q^2=p^2$

      So 2 divides p2,from theorem we know that,
      2 will divide p also. p=2c

      $2q^2=4c^2$
      or
      $q^2=2c^2$
      So q divided by 2 also

      So both p and q are divided by 2 which is contradiction from we assumed
      So $\sqrt {2}$ is irrational number
      2. Prove that $\sqrt {3}$ is irrational number
      Solution: Since we cannot clearly express in p/q form,it is difficult to say,So lets us assume this is rational number
      then
      $\sqrt {3}=\frac {p}{q}$

      where p and q are co primes.
      or
      $q\sqrt {3}=p$
      squaring both sides
      $3q^2=p^2$

      So 3 divides p2,from theorem we know that,
      2 will divide p also. p=3c

      $3q^2=9c^2$
      or
      $q^2=3c^2$
      So q divided by 3 also

      So both p and q are divided by 3 which is contradiction from we assumed
      So $\sqrt {3}$ is irrational number

      Quiz Time

      Question 1 What is the HCF of 12576, 4052 ?
      A. 4
      B. 32
      C. 196
      D. 146
      Question 2 HCF (p, q) x LCM (p, q) ?
      A. p+q
      B. pq
      C. p-q
      D. p/q
      Question 3 HCF (a, b, c) x LCM (a, b, c).
      A. a+b+c
      B. abc
      C. a
      D. None of these
      Question 4 if p a prime number, divides b2,then
      A. p also divides b
      B. p may divide b
      C. p does not divide b
      D. none of these
      Question 5 Which of these is a rational number?
      A. √5 + √2
      B. (√5 + √2)(√5 - √2)
      C.√5 - √2
      D. (√5 + √2)2
      Question 6 Which of these is not prime number?
      A. 877
      B. 101
      C.97
      D. 873

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      Also Read



      Reference Books for class 10

      Given below are the links of some of the reference books for class 10 math.

      1. Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
      2. Mathematics for Class 10 by R D Sharma
      3. Pearson IIT Foundation Maths Class 10
      4. Secondary School Mathematics for Class 10
      5. Xam Idea Complete Course Mathematics Class 10

      You can use above books for extra knowledge and practicing different questions.


      Class 10 Maths Class 10 Science

      Practice Question

      Question 1 What is $1 - \sqrt {3}$ ?
      A) Non terminating repeating
      B) Non terminating non repeating
      C) Terminating
      D) None of the above
      Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
      A) 19.4 cm3
      B) 12 cm3
      C) 78.6 cm3
      D) 58.2 cm3
      Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
      A) 2 ,21,11
      B) 1,10,19
      C) -1 ,8,17
      D) 2 ,11,20






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