everything on the topic you need to know...the right ppt for your school

1. WELCOME

2. REAL
NUMBERS
TOPIC:

3. Presenting By-
Swapnil Yadav ,
X

4. rational numbers
Whole numbers are a subset of integers and
counting numbers are a subset of whole
numbers.
integers
whole numbers
counting
numbers
If we express a new set of numbers as the quotient of
two integers, we have the set of rational numbers
This means to
divide one integer
by another or
“make a fraction”

5. rational numbers
There are numbers that cannot be expressed as
the quotient of two integers. These are called
irrational numbers.
integers
whole numbers
counting
numbers
2

irrational
numbers
The rational numbers combined with the irrational
numbers make up the set of real numbers.
REAL NUMBERS

6. Real Number System
Irrational Numbers
pi, sqrt 7
Noninteger Rational Numbers
-14/5, 9/10, 30,13
Negative Integers
-20, -13, -1
Zero
0
Natural Numbers or
Positive Integers
1, 16, 170
Whole Numbers
0, 2, 56, 198
Integers
-10, 0, 8
Rational Numbers
-35, -7/8, 0, 5, 27/11
Real Number System
-18, -1/2, 0, sqrt 2, pi, 47/10

9. RATIONAL NUMBERS

11. IRRATIONAL NUMBERS

12. Theorem 1.1 (Euclid’s Division Lemma) :
Given positive integers a and b , there exist unique
integers q and r satisfying a = bq + r, 0 ≤ r < b.
• Example :
Prove that every positive even integer is of the form
2m and every positive odd integer is of the form 2m + 1,
where m is any integer.
Solution:
Let a be any positive integer and let b = 2.
According to Euclid’s division lemma, there exist two unique
integers m and r such that
a = bm + r = 2m + r, where 0 ≤ r < 2.
Thus, r = 0 or 1 If r = 0, i.e.,
if a = 2m, then the expression is divisible by 2. Thus, it is an
even number .
If r = 1, i.e., if a = 2m + 1, then the expression is not
divisible by 2.
Thus, it is an odd number . Thus, every positive even integer
is of the form 2m and every positive odd integer is of the
form 2m + 1.

13. Euclid’s division algorithm
To obtain the HCF of two positive integers, say c and
d, where 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 this stage will
be the required HCF.

14. Example : Find the HCF of 48 and 88.
Solution : Take a = 88, b = 48
Applying Euclid’s Algorithm, we get
88 = 48 × 1 + 40 (Here, 0 ≤ 40 < 48)
48 = 40 × 1 + 8 (Here, 0 ≤ 8 < 40)
40 = 8 × 5 + 0 (Here, r = 0)
Therefore , HCF (48, 88) = 8
For any positive integer a, b, HCF (a, b) × LCM (a, b) = a × b

15. 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 Fundamental Theorem of Arithmetic says that
every composite number can be factorised as a
product of primes.
• It also says that any composite number it can be
factorised as a product of prime numbers in a ‘unique’
way, except for the order in which the primes occur.
• That is, given any composite number there is one and
only one way to write it as a product of primes , as
long as we are not particular about the order in
which the primes occur.

16. Example:
Find the HCF of 300, 360 and 240 by the
prime factorisation method.
Solution:
300 = 22 × 3 × 52
360 = 23 × 32 × 5
240 = 24 × 3 × 5
Therefore , HCF (300, 360, 240) = 22 × 3 × 5 = 60

17. Theorem 1.3 : Let p be a prime number. If p divides
a₂, then p divides a, where a is a positive integer.
*Proof :
Let the prime factorisation of a be as follows :
a = p₁ p₂ . . . Pn, where p₁, p₂, . . ., pn are primes, not
necessarily distinct.
Therefore, a² = (p1p₂ . . . pn)(p₁p₂ . . . pn) = p²₁ p²₂. . . p²n.
Now, we are given that p divides a ₂.
Therefore, from the Fundamental Theorem of Arithmetic, it
follows that p is one of the prime factors of a².
However, using the uniqueness part of the Fundamental
Theorem of Arithmetic, we realise that the only prime factors
of a₂ are p₁, p₂, . . ., pn.
So p is one of p₁, p₂ , . . ., pn. Now, since a = p₁ p₂ . . . pn, p
divides a.

18. Theorem 1.4 : √2 is irrational.
• Proof :
Let us assume, to the contrary, that √2 is rational.
So, we can find integers r and s (≠ 0) such that √ 2 =r/s.
Suppose r and s have a common factor other than 1. Then, we
divide by the common factor to get √ 2 a/b,where a and b are
coprime.
So, b √ 2 = a.
Squaring on both sides and rearranging, we get 2b² = a².
Therefore, 2 divides a².
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b² = 4c², that is, b² = 2c²
This means that 2 divides b², and so 2 divides b (again using
Theorem 1.3 with p = 2).
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 our incorrect assumption
that √ 2 is rational.
So, we conclude that √ 2 is irrational.

19. Theorem 1.5 :
Let x be a rational number whose decimal
expansion terminates. Then x can be
expressed in the form p/q where p and q
are coprime, and the prime factorization
of q is of the form 2n5m, where n, m are
non-negative integers.
• Example

20. Theorem 1.6 :
Let x = p/q be a rational number, such
that the prime factorization of q is of the
form 2n5m, where n, m are non-negative
integers. Then x has a decimal expansion
which terminates.

21. • Example

22. Theorem 1.7 :
Let x =p/q be a rational number, such that
the prime factorization of q is not of the
form 2n5m, where n, m are non-negative
integers. Then, x has a decimal expansion
which is non-terminating repeating
(recurring).
Thus, we can conclude that the decimal
expansion of every rational number is
either terminating or non-terminating
repeating.

23. THANK
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