Download REAL NUMBER LEC -1

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(Mathematics) Class X
Real Numbers
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IIT I NEET I BOARDS I FOUNDATION I OTHERS
REAL NUMBERS
LECTURE # 1
Real Numbers: Numbers which can represent actual physical quantities in a meaning full way are known as
real numbers.
Real Numbers
Rational Numbers
Integers
Irrational Numbers
Natural Numbers
Whole Numbers
Prime Numbers: All natural numbers that have one and itself only as their factors are called prime numbers.
ex: 2, 3, 5, 7...... etc.
Composite Numbers: All natural numbers which are not prime are composite numbers.
Co-prime Numbers: If the HCF of the given numbers is 1, then they are known as co-primes.
 1 is neither prime nor composite number.
 Any two consecutive numbers will always be co-prime
Divisibility: A non-zero integer ‘a’ is said to divide an integer ‘b’ if there exists an integer ‘c’ such that b = ac.
The integer ‘b’ is called divident, integer ‘a’ is called divisor and integer ‘c’ is known as quotient.
a/b read as ‘a divides b’
a/|b read as ‘b is not divisible by a’.
Euclid’s Division Lemma: For any two given positive integers a and b, there exists unique integers q and r
such that
a = bq + r, o  r  b
Ex. 1 : Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5, where q is some integer.
Sol. Let a be any positive odd integer and b = 6 then by Euclid’s division lemma
a = 6q + r, where o  r  6
 ‘a’ can be 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5
[ o  r < 6,  r = o, 1, 2, 3, 4, 5]
 a = 6q + 1, 6q + 3 or a = 6q + 5
[  a  6q, a  6q + 2, a  6q + 4, all are even numbers]
hence, any odd integer is of the form 6q + 1, 6q + 3, 6q + 5.
Ex. 2: With the help of Euclid’s division lemma, show that the square of any positive integer is either of the form
3m or 3m + 1, for some integer m.
Sol. let x be any positive integer and b = 3
 x = 3q + r, o  r < 3
so r can be 0, 1, 2,
ie. x can be of the form 3q, 3q + 1, or 3q + 2
if x = 3q
 x2 = 9q2 = 3 × 3q2 = 3m, where m = 3q2
x = 3q + 1
 x2 = 9q2 + 1 + 6q = 3 (3q2 + 2q) + 1 = 3m + 1, where m = 3q2 + 2q
x = 3q + 2
 x2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1) +1 = 3m + 1, where m = 3q2 + 4q + 1
If means x2 is of the form 3m or 3m + 1, where m is integer.
hence square of any positive integer is of the form 3m or 3m + 1.
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