In this slide you get to know the all the detail and in depth knowledge of the chapter Real Number, 1st chapter of CBSE class 10th. Here you get all the variety of questions. You can watch the video lecture on YouTube- https://youtu.be/T2N-NObDf8w

1. 1ABHISHEK MISHRA

2. CONTENT
I. PRIOR KNOWLEDGE
II. EUCLID’S DIVISION ALGORITHM
III. FINDING HCF BY EUCLID’S DIVISION ALGORITHM
IV. THE FUNDAMENTAL THEOREM OF ARITHMETIC
V. RELATION OF LCM AND HCF WITH TWO NUMBERS
VI. REVISION OF THE CONCEPT OF IRRTAIONAL NUMBERS
VII. PROVE OF SOME OF IRRATIONAL NUMBER
VIII. RATIONAL NUMBER AND THEIR DECIMAL EXPANSIONS
IX. LIST OF ALL THE THEOREMS OF THE CHAPTER
X. VARIOUS TYPES OF QUESTIONS
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3. PRIOR -KNOWLEDGE
 Detail knowledge on LCM and HCF
 Knowledge of factorization of numbers
 Knowledge on the square and square roots of numbers.
 Some knowledge on Pythagoras theorem i,e
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4. EUCLID’S DIVISION ALGORITHM
It says that, “ Any positive integer a can be divisible by another positive integer
b in such a way that it leaves a reminder r that is smaller than b”.
THEOREM-1.1 (EUCLID’S DIVISION LEMMA)-
Given positive integers a and b, there exist unique integers q and r satisfying
Euclid’s division lemma and algorithm are so closely interlinked that people
often call former as the division algorithm also.
17 = 6 × 2 + 5
Dividend
Divisor Quotient
Reminder
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5. FINDING HCF BY EUCLID’S DIVISION
ALGORITHM
The HCF of two positive integers a and b is the largest positive integer d
that divides both a and b.
To obtained the HCF of two positive integers, say c and d, with c > d,
follow the steps below:
STEP-1: Apply Euclid’s division lemma, to c and d. So, we find whole
numbers, q and r such that c = dq + r, 0 ≤ r < d.
STEP-2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma
to d and r.
STEP-3: Continue the process till the remainder is zero. The divisor at
the stage will be the required HCF.
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6. THE FUNDAMENTAL THEOREM OF
ARITHMETIC
THEOREM 1.2 (Fundamental theorem of arithmetic)
Every composite number can be expressed (factorised) as a
product of primes, and this factorisation is unique, apart from
the order in which the prime factors occur.
The prime factorisation of a natural number is unique, except for
the order of its factors.
A composite number x, we factorise it as x = q1 × q2 × q3...qn where
q1, q2, q3...qn are primes and written in ascending order.
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7. RELATION OF LCM AND HCF WITH TWO
NUMBERS
HCF of two numbers: Product of the smallest power of each
common prime factor in the numbers.
LCM of two numbers: Product of the greatest power of each prime
factor involved in the numbers.
For any two positive integers a and b,
HCF(a, b) × LCM(a, b) = a × b
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8. REVISION OF THE CONCEPT OF
IRRTAIONAL NUMBERS
IRRATIONAL NUMBERS-The number which can’t be written in
the form of where p and q are integers and q≠0.
EXAMPLES-
π (Pi) is a famous irrational number. We cannot write down a simple
fraction that equals Pi.
The popular approximation of 22/7 = 3.1428571428571... is close but
not accurate. Another clue is that the decimal goes on forever
without repeating.
OTHER EXAMPLES-
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9. PROVE OF SOME OF IRRATIONAL
NUMBER
THEOREM-1.3: Let p be a prime number. If p divides , then p
divides a, where a is a positive integer.
PROOF:
Let the prime factorisation of a be as follows:
a=p1p2p3…pn , where p1, p2, p3…pn are primes, not necessarily distinct.
Therefore, = (p1p2p3…pn)(p1p2p3…pn)
Now, we are given that p divides . Therefore, from the Fundamental
Theorem of Arithmetic, it follows that p is one of the prime factor of
.
Clearly p is one of p1p2p3…pn.
Since a=p1p2p3…pn , p divides a.
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10. THEOREM-1.4- is irrational.
PROOF:
Let us assume, to the contrary, that is rational.
So, we can find integers a and b (≠ 0) such that where a and b are
co-prime numbers.
So,
Squaring both the sides and rearranging, we get
So, 2 divides . Now, by theorem 1.3, it follows that 2 divides a.
So, we can write for some integer c.
Substituting for a, we get
This means that 2 divides and so 2 divides b (Theorem 1.3)
Therefore, a and b have at least 2 as a common factor. But, this contradicts
the fact that a and b have no common factors other than 1.
This contradiction has arisen because of the incorrect assumption.
So, we conclude that is irrational.
⇒
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11. RATIONAL NUMBER AND THEIR
DECIMAL EXPANSIONS
THEOREM-1.5: Let x be a rational number whose decimal
expansion terminates. Then x can be expressed in the form of ,
where p and q are co-prime, and the prime factorisation of q is of
the form , where n, m are non-negative integers.
Any rational number of the form , where b is the power of 10, will
have a terminating decimal expression.
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12. THEOREM-1.6: Let x = be a rational number, such that
the prime factorization of q is of the form , where n, m
are non-negative integers. Then x has decimal expansion
which terminates.
THEOREM-1.7: Let x = be a rational number, such that
the prime factorization of q is not of the form , where
n, m are non-negative integers. Then x has decimal
expansion which is terminating or non terminating
repeating.
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13. LIST OF ALL THE THEOREMS OF THE
CHAPTER
THEOREM-1.1 (EUCLID’S DIVISION LEMMA)-
Given positive integers a and b, there exist unique integers q and r
satisfying
THEOREM 1.2 (Fundamental theorem of arithmetic)
Every composite number can be expressed (factorised) as a product
of primes, and this factorisation is unique, apart from the order in
which the prime factors occur.
THEOREM-1.3: Let p be a prime number. If p divides , then p
divides a, where a is a positive integer.
THEOREM-1.4 - is irrational.
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14. THEOREM-1.5: Let x be a rational number whose decimal expansion
terminates. Then x can be expressed in the form of , where p and q
are co-prime, and the prime factorisation of q is of the form ,
where n, m are non-negative integers.
THEOREM-1.6: Let x = be a rational number, such that the prime
factorization of q is of the form , where n, m are non-negative
integers. Then x has decimal expansion which terminates.
THEOREM-1.7: Let x = be a rational number, such that the prime
factorization of q is not of the form , where n, m are non-
negative integers. Then x has decimal expansion which is terminating
or non terminating repeating.
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15. VARIOUS TYPES OF QUESTIONS
TYPE-1 : Euclid’s Division Algorithm and their application
TYPE-2 : Some of the proves of Euclid’s Division Algorithm
TYPE-3 : Proves of The Fundamental theorem of Arithmetic
TYPE-4 : HCF and LCM type questions
TYPE-5 : Proves of the Irrational numbers
TYPE-6 : Decimal expansion type questions
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16. TYPE-1 : Euclid’s Division Algorithm and
their application
Q. Use Euclid’s Division Algorithm to find the HCF of
135 and 225
Ans- Since 225 > 135, we apply the division algorithm
to obtain 225 = 135 × 1 + 90
Again, 135 = 90 × 1 + 45
Also, 90 = 45 × 2 + 0
Now, the reminder is zero so the divisor is the HCF
So, HCF of 135 and 225 is 45.
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17. TYPE-2 : Some of the proves of Euclid’s
Division Algorithm
Q. Show that every positive even integer is of the form 2q,
and that every positive odd integer is of the form 2q + 1,
where q is some integers.
ANS- Let a be a positive integer and b = 2.
By Euclid’s division algorithm, a = 2q + r
Where q ≥ 0, and r = 0, or r = 1, as 0 ≤ r < 2
So, a = 2q or a = 2q + 1
Clearly if a integer is of the form 2q then it is even integer.
And the rest of all will be odd integer in the form 2q + 1
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18. TYPE-3 : Proves of The Fundamental
theorem of Arithmetic
Q. Consider the numbers , where n is a natural
number. Check whether there is any value of n for
which ends with the digit zero.
ANS- If the number , for any n, were to end with
zero, then it would be divisible by 5.
That is the prime factorization of would contain 5.
This is not possible because = , so the only prime
number is 2 present in .
So there is no natural number n for which ends
with the digit zero.
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19. TYPE-4 : HCF and LCM type questions
Q. Find the HCF of 96 and 404 by prime factorization
method. Hence find its LCM.
ANS- The prime factorisation of 96 and 404 are
Therefore, the HCF of these two integers is = 4
As, HCF(96, 404) × LCM(96, 404) = 96 × 404
so, 4 × LCM(96, 404) = 96 × 404
LCM(96, 404) = = 9696
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20. THANK YOU
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