Content Writer-SME | Updated On - Oct 19, 2024
Any number in the real world which is in the form p/q where p and q are integers is called a real number. In other words, real numbers in the number system are defined as the set of rational and irrational numbers. These numbers are expressed on a number line on which arithmetic operations can also be performed.
Imaginary numbers are unreal numbers that cannot be expressed on the number line and are usually used to represent a complex number. Thus, any number in mathematics, except complex numbers, can be considered a real number such as 23, 7.99, \(\frac {9}{2} \), and π or \(\frac {22}{7} \). In this article we will learn about real numbers, their properties as well as some important laws associated with real numbers.
Read Also: Real Numbers Formula
What are Real Numbers?
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The set of rational and irrational numbers together forms real numbers. They can be positive as well as negative. Symbolically, real numbers are represented by the letter “R” in Mathematics Class 10. All the numbers except complex numbers falls under the category of real numbers. In the real world, where every law and theorem needs to be proved with the help of equation, hence we cannot imagine science and mathematics without real numbers.
Thus it includes fractions, irrational numbers, negative integers, positive integers, and negative integers. Real numbers can be expressed in form of indefinite decimal expansion. These numbers help in representing distance along the number line. The set of all real numbers is infinite.
On the whole,
- Real numbers are created by combining all rational and irrational numbers.
- The number line can be used to plot any real number.
- Real numbers are used to measure different quantities.
Solved ExampleQues. Find the real numbers among √7, √-8, ⅔, -6. Ans. Among all the other numbers √-8 is a complex number so it cannot be a real number. All the other numbers are either rational or irrational numbers. So the real numbers are √7, ⅔ , -6. Ques. Answer in yes or no whether the set of numbers given below are real numbers or not:
Ans.
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Set of Real Numbers
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Real numbers are divided into two main categories i.e. rational numbers and irrational numbers.
Rational Numbers: The numbers that can be written in the form of p/q where p and q are both integers and q ≠ 0. In a fraction, the lower value or denominator is represented by q while the upper value or numerator is represented by p. Rational Numbers can be further divided into-
- Natural Numbers: All Counting numbers starting from one are called natural numbers. (1,2,3,....)
- Whole Numbers: The set of all the natural numbers and zeroes are called whole numbers. (0,1,2,3,...)
- Integers: All positive and negative numbers (not fractions) including zero are called the integers. (-∞ to +∞)
Irrational numbers: All numbers that can not be expressed in the form of p/q are known as irrational numbers. (√2, √3, etc.)
- Even numbers: Even numbers are integers that are divisible by 2. Even numbers always end up with 0, 2, 4, 6, or 8 in the last digit. Even numbers include 2, 4, 6, 8, 10, 12, 14, 16, etc.
- Odd numbers: Odd numbers are numbers that cannot be divided by two in an exact manner. It cannot be divided evenly into two separate integers. We obtain a remainder when we divide an odd integer by two. Examples include 1, 3, 5, 7, and so forth.
- Prime numbers: Prime numbers are natural numbers higher than 1 that are divisible by 1 and the number itself. Prime numbers have two factors: 1 and the number itself. The first ten primes are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 respectively. 1 is not a prime number because it has only one factor.
- Composite numbers: Composite numbers are natural numbers that are divisible by 1, itself as well as any other integer or number. 1 does not belong to prime numbers as well as composite numbers. For example, 4, 6, 8, 9, 10 etc.
Real Numbers Table
| Category | Definition | Example |
|---|---|---|
| Natural Numbers | All counting numbers start from 1. | All numbers such as 1, 2, 3, …... |
| Whole Numbers | Zero and natural number. | All numbers comprising 0 such as 0, 1, 2, 3, 4, 5, 6,…..… |
| Integers | Combination of Whole numbers and negative of all the natural numbers. | Includes: -infinity (-∞),…….. -2, -1, 0, 1, 2, ……+infinity (+∞) |
| Rational Numbers | Numbers that can be written in the p/q form, where q≠0. | Examples of rational numbers are ½, 5/4 and 12/6, etc. |
| Irrational Numbers | The numbers are not rational and can’t be written in the p/q form. | Non-terminating and non-repeating numbers such as √2 |
Some Important Points about Real Numbers
Some important points related to Real Numbers include:
- The square root of any positive real number is always there and can also be shown on a number line.
- The total sum or difference between a rational number and an irrational number shall be an irrational number.
- An irrational number is the product or division of a rational number with an irrational number.
- This process of representation of a decimal expansion on the number line is the process of successive magnification.
Properties of Real Numbers
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Like natural numbers and integers, real numbers also follow three main rules or properties – commutative, associative and distributive rules which are explained with the help of examples given below.
Commutative Property
Let a and b be two real numbers
Addition: So, by Commutative Law of Addition: a + b = b + a
For Example: 6 + 3 = 3 + 6; 7 + 1 = 1 + 7
Multiplication: Let a and b be two real numbers
So, Commutative Law of Multiplication: a × b = b × a
For Example: 2 × 6 = 6 × 2; 3 × 8 = 8 × 3
Associative Property
Let, a,b and c are three real numbers
Addition: So, by Associative Law of Addition: a + (b + c) = (a + b) + c
For Example: 3 + (4 + 9) = (3 + 4) + 9
Multiplication: By Associative Law of Multiplication-
Let, a,b and c are three real numbers
So, by Associative Law of Multiplication: a × (b × c) = (a × b) × c
For Example: 8 × (2 × 3) = (8 × 2) × 3
Distributive Property
Let a, b and c be three real numbers.
So, by distributive law-
- a × (b + c) = (a × b) + (a × c)
- (a + b) × c = (a × c) + (b × c)
For Example: 5(2 + 3) = 5 × 2 + 5 × 3.
Identity Law
In real numbers, there are two types of Identities i.e. Additive Identity and Multiplicative Identity.
- For Additive Identity: n + 0 = n (For a number, say n. Here, 0 is the additive identity)
- For Multiplicative Identity: n x 1= 1 x n = n (For a number, say n. Here, 1 is the multiplicative identity)
Positive integers have two important properties called Euclid’s division algorithm and the Fundamental Theorem of Arithmetic.
Read Also:- Complex Numbers and Quadratic Equations
Euclid’s division lemma
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The Euclid’s division lemma states that a positive integer ‘a’ can be divided by another positive integer ‘b’ in a way that it always leaves a remainder ‘r’ that is smaller than ‘b’. We use this algorithm to calculate the Highest Common Factor (HCF) of two given positive integers.
This is the method for determining the HCF (Highest Common Factor) of two positive integers q and r. The steps to calculate the HCF of two positive integers q and r with q > r are as follows:
- Step 1: Euclid's Division To get two integers q and r such that a=b×q+r and 0≤r
- Step 2: If r = 0, the H.C.F is b; otherwise, we use Euclid's division Lemma on b (the divisor) and r (the remainder) to generate a new pair of quotients and remainder.
- Step 3: The process described above is repeated until the remaining is zero. In that step, the divisor is the H.C.F of the provided set of numbers.
| For example, let’s try to find the HCF of 455 and 42. Using Euclid’s lemma, 455 = 42 × 10 + 35
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Fundamental Theorem of Arithmetic
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According to the Fundamental Theorem of Arithmetic, any integer greater than 1 will either be a prime number or can be written in the form of primes which means that all natural numbers can be the product of their prime factors. Let us note that prime factors are those numbers that can be divided by 1 and themselves only. Also note that the opposite, composite factors are those numbers that can be divided by numbers besides 1 and itself.
For example, the number 45 can be written in the form of its prime factors as:
45 = 9 × 5
Here, 9 and 5 are the prime factors of 45
Next, let’s take the number 114560 which may be represented as the product of its prime factors with the prime factorization method,
114560 = 27 × 5 × 179
Hence, the number 114560 has been factored as the product of the power of its primes.
Prime Factorisation
The method of representing a natural number as the product of prime numbers is known as prime factorisation.
The prime factorisation of 36, for example, is 36 = 2 × 2 × 3 × 3.
Method of Finding LCM
The methods for finding the LCM are as follows:
- The "smallest non-zero common number" that is a multiple of both numbers is the "least common multiple of two numbers."
- Example: To find the Least Common Multiple (L.C.M) of numbers 36 and 56 respectively,
- 36=2 × 2 × 3 × 3
56=2×2×2×7 - The common prime factors are 2×2.
- The uncommon prime factors are 3×3 for 36 and 2×7 for 56.
- LCM of 36 and 56 = 2×2×3×3×2×7 which is 504.
Method of Finding HCF
The methods for finding the HCF are as follows:
- The greatest number among all the common factors of two or more numbers is called the HCF (Highest Common Factor).
- Prime factorisation and Euclid's division algorithm are two ways for finding H.C.F.
- Prime factorization is the process of expressing two numbers as the products of their prime factors. Where we find the common prime factors of both numbers.
- To find the H.C.F of 20 and 24, for example.
- 20=2 × 2 × 5 and 24=2 × 2 × 2 × 3.
- The factor common to 20 and 24 is 2×2, which is 4, which in turn is the H.C.F of 20 and 24.
- Euclid’s Division Algorithm is the method for determining the HCF (Highest Common Factor) of two positive integers q and r.
Shortcut method to Euclid’s division algorithm method
Dividend = Divisor × Quotient + Remainder
HCF X LCM of the Two Numbers = Product of Two Numbers
a×b=H.C.F×L.C.M (for any two positive integers a and b)
For example, the H.C.F. for 36 and 56 is 4 and the L.C.M. is 504
36×56=2016
4×504=2016
Thus, 36×56=4×504
However, with three or more numbers, the above relationship does not hold true.
Solved ExamplesQues: The time taken by two racing cars X and Y to complete 1 lap of the track in a racing competition is 30 minutes and 45 minutes respectively. In how much time will X and Y meet again at the starting point? Ans. Since the time taken by car Y is more compared to that taken by X to complete one lap thus it can be assumed that X will reach first and the two cars will meet again when X has already reached the starting point. We can calculate this through the LCM of the time taken by each car. 30 = 2 × 3 × 5 45 = 3 × 3 × 5 LCM = 90. Therefore, the two cars will meet at the starting point after 90 minutes. Ques: Find HCF of 56 and 72. Ans. Steps:
Consider 56 as the new dividend and 16 as the new divisor since 16 0. 56 = 16 × 3 + 8 Once more, if 8 ≠ 0, apply 16 as the new dividend and 8 as the new divisor. 16 = 8 × 2 + 0 The remainder being zero, (8) is HCF's divisor. Euclid's Division Lemma can be extended to all integers except zero, i.e., b ≠ 0. This is true even if it is only stated for positive integers. |
Proof by Contradiction
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To determine whether a statement is TRUE using the method of contradiction,
(i) We first assume that the provided statement is TRUE.
(ii) We discover a finding that proves the opposite of our presumption and goes against it.
Show, for instance, why √7 is unreasonable.
The supposition that √7 is logical.
Given that it is rational, the expression for √7 is
√7 = a/b, where a and b are coprime integers and b ≠ 0.
a2/b2 = 7 (a2=7b2) upon squaring.
7, therefore, divides a. Following that, there is an integer c such that a=7c. So, a2 = 49c2. Therefore, b2 =7c2 or 7b2=49c2.
7 therefore divides b. It goes against our belief that a and b are coprime integers because 7 is a factor of both a and b.
Hence, our initial assumption about √7 being rational is wrong. Therefore, √7 is irrational.
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| Class 10 Chapter 1 Real Numbers Topics | ||
|---|---|---|
| Root 2 is an irrational number | Quadratic Equations Formula | Real Numbers Important Questions |
| What are Real Numbers? | Number Lines | MCQs for Real Numbers |
Things to Remember
- Real numbers are a combination of rational and irrational numbers. They can be either positive or negative.
- Real numbers comply with three properties- commutative, associative, and distributive properties.
- Euclid’s division lemma states that a positive integer can be divided by another positive integer to leave a remainder that is smaller than the second positive integer.
- Fundamental Theorem of Arithmetic – Any integer greater than 1 is either a prime number or a form of primes.
- All the natural numbers can be the product of their prime factors.
Sample Questions
Ques: Prove that the product of any three consecutive positive integers is divisible by 6. (All India- 2016)(5 Marks)
Ans. Let three consecutive positive numbers are n, n + 1, and n + 2
- Case 1: If n is even
This means n + 2 will also be even.
Hence n and n + 2 can be divisible by 2
Also, the product of n and (n + 2) will be divisible by 2.
Means, n(n + 2) will be divisible by 2.
This means that-
n(n + 2) (n + 1) will be divisible by 2 …(i)
Since,
n, n + 1 and n + 2 are three consecutive numbers. So, the product of these three numbers i.e. n(n + 1) (n + 2) will be a multiple of 3.
This means that-
n(n + 1) (n + 2) will be divisible by 3. …(ii)
On solving Eq. (i) and (ii) we can say
n(n + 1) (n + 2) can be divisible by 2 and 3 both.
Hence, we can conclude that-
n(n + 1) (n + 2) is divisible by 6.
- Case 2: When n is odd.
This means (n + 1) is even
Hence (n + 1) is divisible by 2. …(iii)
This concludes that n(n + 1) (n + 2) is an even number and can be divisible by 2.
Also, we know that-
The product of three consecutive numbers is a multiple of 3.
n(n + 1)(n + 2) is divisible by 3. …(iv)
From Equation (iii) and (iv) we can say
n(n + 1) (n + 2) can be divisible by both 2 and 3.
Hence, we can conclude that-
n(n + 1)(n + 2) is divisible by 6.
Ques: Find HCF of 378,180 and 420 by prime factorisation method. Is HCF x LCM of three numbers equal to the product of the three numbers? (All India- 2014)(4 Marks)
Ans. By Prime Factorization Method, the HCF of 378, 180 and 420 can be written as-
378 = 2 x 33 x 7
180 = 22 x 32 x 5
420 = 22 x 3 x 5 x 7
So, the HCF (378, 180, 420) = 2 x 3 = 6.
No, the HCF (p, q, r) x LCM (p, q, r) ≠ p x q x r.
Where p, q and r are positive integers.
Ques: Explain whether the number 3 x 5 x 13 x 46 + 23 is a prime number or a composite number. (All India- 2016)(3 Marks)
Ans. We have 3 x 5 x 13 x 46 + 23 = 23 x (3 x 5 x 13 x 2) + 23
= 23 x (3 x 5 x 13 x 2) + 23 = 23 x 390 + 23 = 8,993
From the above equation, it is clear that the (3 x 5 x 13 x 2) + 23 is a multiple of 23. Therefore, (3 x 5 x 13 x 2) + 23 is not a prime number.
Ques: Three bells toll at intervals of 12 minutes, 15 minutes, and 18 minutes respectively. If they start tolling together, after what time will they next toll together? (All India- 2015)(3 Marks)
Ans. LCM of 12, 15, 18
= 2² x 3² x 5
= 4 x 9 x 5 = 180
So, we can say that after 180 minutes the bells will ring together next time.
Ques: By using Euclid's algorithm, find the largest number which divides 650 and 1170. (All India- 2014) (3 Marks)
Ans. Given numbers – 650 and 1170.
On applying Euclid’s division algorithm,
we get 1170 = 650 x 1 + 520650 = 520 x 1 + 130
520 = 130 x 4 + 0
Since, at the last stage, the divisor is 130.
So, the HCF of 650 and 1170 is 130.
Ques: Show that the reciprocal of 3+2√2 is an irrational number. (All India- 2014) (3 Marks)
Ans. We have to prove that \({1 \over 3+2\sqrt2} = 3 - \sqrt{2}\) is an irrational number
Let’s assume that \(3 - 2\sqrt{2}\) is rational
Therefore there exist coprime integers a and b (b is not equal to 0) such that
\(3 - 2\sqrt{2}\) = a/b = \(2\sqrt{2}\) = 3 – a/b
\(2\sqrt{2}\) = (3b- a)/b
\(\sqrt{2}\) = 3/2 – a/2b
Since a and b are integers, we get 32- 22bis rational. So, 2 is rational.
But this contradicts the fact that 2 is irrational.
And this contradiction is due to our incorrect assumption that 3 - 2√2 is a rational number.
Hence, we can say that 3 - 2√2 is an irrational number.
Ques: Determine the values y and q so that the prime factorization of 2520 is expressible as 23 X y X q x 7. (All India- 2014) (2 Marks)
Ans. Prime factorization of 2520 can be given by
2520 = 23 x 32 x 5 x 7
Given that the factor of 2520 = 23 x 3 x y x q x 7
On comparing both factorisations we get y = 2 and q = 5.
Ques. There are two pieces of cloth. The width of one section is 72 inches, while the width of the other piece is 90 inches. To make strips out of both pieces that are the same width and as long as possible. How broad should the strips be cut? (2 marks)
Ans. Cutting or "dividing" the strips of cloth into smaller pieces (Factor) of 72 and 90 (Common) and looking for the widest possible strips, this problem can be solved using H.C.F. (Highest).
So, since the H.C.F. of 72 and 90 is 18, Priyanka should cut each piece to be 18 inches wide.
Ques. Ram exercises every 12 days, whereas Deepika exercises every 8 days. Both Ram and Deepika worked out today. How long will it take before they work out together again? (2 marks)
Ans. So, we can utilize Least Common Multiple to solve this problem since we're trying to figure out when the earliest (Least) time will be so that when the exercise event proceeds (Multiple), it will all happen at the same time (Common).
The L.C.M. of 12 and 8 equals 24.
Ques. Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively. (1 mark)
Ans. 398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527
HCF of 391, 425, 527 = 17
Ques. HCF and LCM of two numbers are 9 and 459 respectively. If one of the numbers is 27, find the other number. (2 marks)
Ans. 1st number × 2nd number = HCF × LCM
Or, 27 × 2nd number = 9 × 459
Or, 2nd number = (9×459)/27 = 153
Ques. Find the largest number which divides 70 and 125 leaving the remainder 5 and 8 respectively. (2 marks)
Ans. It is given that there is a remainder of 5 after dividing 70 by the required number.
Therefore, 70 - 5 = 65 is exactly divisible by the specified number.
Likewise, 125 minus 8 equals 117 is exactly divisible by the required number.
65 = 5× 13
117 = 32 × 13
HCF = 13
Required Number = 13
Ques. Given that HCF (306, 657) = 9, find LCM (306, 657). (3 marks)
Ans. We know,
HCF × LCM = Product of the two given numbers
So,
9 × LCM = 306 × 657
LCM = (306 × 657)/9 = 22338
Therefore, LCM(306,657) = 22338
Ques. Two tankers contain 850 litres and 680 litres of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker an exact number of times. (3 marks)
Ans. Maximum capacity of a container that can measure the exact quantity of gasoline.
= HCF of (850 and 680)
850 = 2*5*5*17
680 = 2*2*2*5*17
HCF = (850 and 680) = 2*5*17 = 170 liters
Ques. The length, breadth, and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly. (3 marks)
Ans. To determine the length of the longest rod that can precisely measure the room's dimensions, we must calculate HCF.
L, Length = 8 m 50 cm = 850 cm = 21 × 52 × 17
B, Breadth = 6 m 25 cm = 625 cm = 54
H, Height = 4 m 75 cm = 475 cm = 52 × 19
HCF of L, B and H is 52 = 25 cm
Length of the longest rod = 25 cm
Ques. In a school, there are two Sections A and B of class X. There are 48 students in Section A and 60 students in Section B. Determine the least number of books required for the library of the school so that the books can be distributed equally among all students of each Section. (3 marks)
Ans. Since the books are to be distributed equally to Section A and Section B students. Consequently, the number of books must also be a multiple of 48 and 60.
Therefore, the required quantity of books is the least common multiple of 48 and 60.
48 = 24 × 3
60 = 22 × 3 × 5
LCM = 24 × 3 × 5 = 16 × 15 = 240
Hence, the required number of books is 240.
Ques. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. (5 marks)
Ans. Let x be any positive integer and y = 3.
Using Euclid’s division algorithm;
x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)
Therefore,
x = 3q, 3q + 1 and 3q + 2
According to the question, if we take the squares on both sides, we obtain;
x2 = (3q)2 = 9q2 = 3.3q2
Let 3q2 = m
Therefore,
x2 = 3m ………………….(1)
x2 = (3q + 1)2
= (3q)2 + 12 + 2 × 3q × 1
= 9q2 + 1 + 6q
= 3(3q2 + 2q) + 1
Substituting 3q2+2q = m we get,
x2 = 3m + 1 … (2)
x2 = (3q + 2)2
= (3q)2 + 22 + 2 × 3q × 2
= 9q2 + 4 + 12q
= 3(3q2 + 4q + 1) + 1
Again, substituting 3q2 + 4q + 1 = m, we get,
x2 = 3m + 1…(3)
From equations 1, 2, and 3, we can conclude that the square of any positive integer is either of the form 3m or 3m + 1, where m is an integer.
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