CBSE Real Numbers Class 10 Notes:-Download PDF Here
Class 10 Maths Chapter 1 Real Number Notes
CBSE Class 10 Maths Chapter 1 Real Numbers Notes are provided here in detail. As we all know, any number, excluding complex numbers is a real number. Positive and negative integers, irrational numbers, and fractions are all examples of real numbers. To put it another way, any number found in the real world is a real number. Numbers can be found all around us. Natural numbers are being used to count objects, integers are used to measure temperature, rational numbers are used to represent fractions, irrational numbers are used to calculate the square root of a number, etc. These various types of numbers form a collection of real numbers. Here, we are going to learn what is a real number? Euclid’s division algorithm, fundamental theorem of arithmetic, methods of finding LCM, HCF and the complete explanation on rational and irrational numbers with examples.
Positive integers, negative integers, irrational numbers, and fractions are all examples of real numbers. In other words, we can say, any number is a real number, except complex numbers. Examples of real numbers include -1, ½, 1.75, √2, and so on.
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Students can refer to the short notes and MCQ questions along with separate solution pdf of this chapter for quick revision from the links below:
Euclid’s Division Lemma
- Euclid’s Division Lemma states that given two integers a and b, there exists a unique pair of integers q and r such that a=b×q+r and 0≤r<b.
- This lemma is essentially equivalent to : dividend = divisor × quotient + remainder
- In other words, for a given pair of dividend and divisor, the quotient and remainder obtained are going to be unique.
For More Information On Euclid’s Division Lemma, Watch The Below Video.
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Euclid’s Division Algorithm
- Euclid’s Division Algorithm is a method used to find the H.C.F of two numbers, say a and b where a> b.
- We apply Euclid’s Division Lemma to find two integers q and r such that a=b×q+r and 0≤r<b.
- If r = 0, the H.C.F is b, else, we apply Euclid’s division Lemma to b (the divisor) and r (the remainder) to get another pair of quotient and remainder.
- The above method is repeated until a remainder of zero is obtained. The divisor in that step is the H.C.F of the given set of numbers.
For More Information On Euclid’s Division Algorithm, Watch The Below Video.
The Fundamental Theorem of Arithmetic
- Prime Factorisation is the method of expressing a natural number as a product of prime numbers.
- Example: 36=2×2×3×3 is the prime factorisation of 36.
Fundamental Theorem of Arithmetic
- The Fundamental Theorem of Arithmetic states that the prime factorisation for a given number is unique if the arrangement of the prime factors is ignored.
- Example: 36=2×2×3×3 OR, 36=2×3×2×3
- Therefore, 36 is represented as a product of prime factors (Two 2s and two 3s) ignoring the arrangement of the factors.
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Method of Finding LCM
As we know, the smallest of the common multiples of two or more numbers is called the lowest common multiple (LCM).
Example: To find the Least Common Multiple (L.C.M) of 36 and 56,
- 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
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Method of Finding HCF
We know that, the greatest number that divides each of the given numbers without leaving any remainder is the highest common factor (HCF) of two or more given numbers.
H.C.F can be found using two methods – Prime factorisation and Euclid’s division algorithm.
- Prime Factorisation:
- Given two numbers, we express both of them as products of their respective prime factors. Then, we select the prime factors that are common to both the numbers
- Example – To find the H.C.F of 20 and 24
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:
- It is the repeated use of Euclid’s division lemma to find the H.C.F of two numbers.
- Example: To find the HCF of 18 and 30
- The required HCF is 6.
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For More Information On HCF and LCM, Watch The Below Video.
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Product of Two Numbers = HCF X LCM of the Two Numbers
- For any two positive integers a and b,
- Example – For 36 and 56, the H.C.F is 4 and the L.C.M is 504
- Let us consider another example:
For 5 and 6, the H.C.F is 1 and the L.C.M is 30
5 × 6 = 30
1 × 30 =30
Thus, 5 × 6 = 1 × 30
- The above relationship, however, doesn’t hold true for 3 or more numbers
To know Relationship between LCM and HCF, visit here.
Applications of HCF & LCM in Real-World Problems
L.C.M can be used to find the points of common occurrence. For example,ringing of bells that ring with different frequencies, the time at which two persons running at different speeds meet, and so on.
For More Information On Applications Of LCM, Watch The Below Video.
Revisiting Irrational Numbers
Any number that cannot be expressed in the form of p/q (where p and q are integers and q≠0.) is an irrational number. Examples √2,π, e and so on.
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Number theory: Interesting results
- If a number p (a prime number) divides a2, then p divides a. Example: 3 divides 62 i.e 36, which implies that 3 divides 6.
- The sum or difference of a rational and an irrational number is irrational
- The product and quotient of a non-zero rational and irrational number are irrational.
- √p is irrational when ‘p’ is a prime. For example, 7 is a prime number and √7 is irrational. The above statement can be proved by the method of “Proof by contradiction”.
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Proof by Contradiction
In the method of contradiction, to check whether a statement is TRUE
(i) We assume that the given statement is TRUE.
(ii) We arrive at some result which contradicts our assumption, thereby proving the contrary.
Eg: Prove that √7 is irrational.
Assumption: √7 is rational.
Since it is rational √7 can be expressed as
√7 = a/b, where a and b are co-prime Integers, b ≠ 0.
On squaring, a2/b2=7
Hence, 7 divides a. Then, there exists a number c such that a=7c. Then, a2=49c2. Hence, 7b2=49c2 or b2=7c2.
Hence 7 divides b. Since 7 is a common factor for both a and b, it contradicts our assumption that a and b are coprime integers.
Hence, our initial assumption that √7 is rational is wrong. Therefore, √7 is irrational.
Revisiting Rational Numbers and Their Decimal Expansions
Rational numbers are numbers that can be written in the form p/q, where p and q are integers and q≠0.
Examples -1/2, 4/5, 1,0,−3 and so on.
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Terminating and nonterminating decimals
Terminating decimals are decimals that end at a certain point. Example: 0.2, 2.56 and so on.
Non-terminating decimals are decimals where the digits after the decimal point don’t terminate. Example: 0.333333….., 0.13135235343…
Non-terminating decimals can be :
a) Recurring – a part of the decimal repeats indefinitely (0.142857142857….)
b) Non-recurring – no part of the decimal repeats indefinitely. Example: π=3.1415926535…
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Check if a given rational number is terminating or not
If a/b is a rational number, then its decimal expansion would terminate if both of the following conditions are satisfied :
a) The H.C.F of a and b is 1.
b) b can be expressed as a prime factorisation of 2 and 5 i.e b=2m×5n where either m or n, or both can = 0.
If the prime factorisation of b contains any number other than 2 or 5, then the decimal expansion of that number will be recurring
1/40=0.025 is a terminating decimal, as the H.C.F of 1 and 40 is 1, and the denominator (40) can be expressed as 23×51.
3/7=0.428571 is a recurring decimal as the H.C.F of 3 and 7 is 1 and the denominator (7) is equal to 71
Real Numbers for Class 10 Examples
Find the largest number that divides 70 and 125 leaving the remainder 5 and 8 respectively.
First, subtract the remainder from the number.
(i.e) 70-5 = 65
125-8 = 117.
Thus, we need to find the largest number that divides 65 and 117 and leaves the remainder 0.
To find the largest number, take the HCF of 65 and 117.
Finding HCF of 65 and 117.
65 = 5×13
117 = 3×3×13.
Hence, HCF (65, 117) = 13.
Therefore, the largest number that divides 70 and 125 leaving the remainder 5 and 8 respectively is 13.
Find the LCM of 306 and 657, given that HCF (306, 657) = 9.
Given that, HCF (306, 657) = 9.
We know that HCF × LCM = Product of Numbers
Hence, 9×LCM = 306×657
9×LCM = 201042
LCM = 201042/9
LCM = 22338.
Therefore, LCM of 306 and 657 is 22338.
Prove that 1/√2 is an irrational number.
To prove 1/√2 is an irrational number.
Now, let us take the opposite assumption.
(i.e) Take 1/√2 is a rational number.
We know that rational numbers are the numbers that can be written in the form of p/q, where q is not equal to 0. (p and q are two co-prime numbers)
Hence, 1/√2 = p/q.
Now, simplify the above equation by multiplying √2 on both sides.
1 = (p√2)/q
q = p√2
Hence, we get q/p = √2.
Here, p and q are integers, and hence q/p is a rational number.
But, √2 is an irrational number.
Hence, our assumption is wrong.
Therefore, 1/√2 is an irrational number.
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