3. NATURAL NUMBER
Natural numbers were first studied seriously by such Greek philosopher and
mathematics as Pythagoras (582-500BC) and Archimedes(287-212BC).
The natural numbers are those numbers use for counting and ordering.
Natural numbers are a part of the number system which includes all the positive
integers from 1till infinity and also used for counting purpose.
Natural number do not negative or zero.
1 is smallest Natural number.
It is denoted by N.
4. WHOLE NUMBER
Whole number is firstly discovered by Bob Sinclair in 1968.
The whole numbers start from 0,1,2,3,4 and so on.
All the natural numbers are considered as whole number,but all the whe numbers
are not natural numbers.
The whole number does not contain any decimal or fractional part.
It is denoted by W.
5. INTEGERS
Integers is discovered by Leopold kronecker
An integers is the number zero,a positive or a negative integers with a minus sign.
It can be represented in a number line excluding the fractional part.
It is denoted by Z.
6. RATIONAL NUMBERS
Rational numbers is discovered by Pythagoras.
A rational number that can be expressed as the quotient or fraction p/q.
0 is a rational number because it is an integer that can be written in 0/1,0/2 etc.
A rational number is a number whose decimal form is finite or recurring in nature
for e.g.2.67 and 5.66…
It is denoted by Q.
7. IRRATIONAL NUMBERS
Irrational numbers is discovered by Greek mathematician Hippasus in 5th century
Irrational number are those number that cannot be represented in the form of a
ratio.
The irrational number are all the real number that are not rational number.
8. EUCLID DIVISON ALGORITHM
It is the process of dividing one integer by another,In a way that produces an
integer quotient and an integer remainder smaller than the divisor.A
fundamental property is that the quotient and the remainder exist and are
unique,under some condition.
a = bq +r, where 0_<r<b.
9. EXAMPLES
EXAMPLE1. Use EUCLUD’S ALGORITHM TO FIND THE HCF OF 4052 AND 12576.
SOL. Step1. Since 12576>4052,we apply th division algorithm to 12576 and 4052 to get
12576 = 4052×3+ 420
Step 2. Since the remainder 420 is not equal to 0,we apply the division algorithm to 4052 And 420 to get.
4052=420×9+272
STEP 3. We consider the new divisor 420 and the new remainder 272 and apply the division algorithm to get
420 = 272×1+148
We consider the new division 272 and the new remainder 148 and apply the division algorithm to get
272 = 148×1+124
We consider the new divisor 148 and the new divisor 148 and the new remainder 124 and apply the divisor algorithm to get
148 = 124×1+124
We consider the new divisor 124 and the new remainder 24 and apply division algorithm to get
124 = 24×5+4
We consider the new divisor 24 and the new remainder 4 and apply division algorithm to get
24 = 4×6+0
NOTICE THAT 4 = HCF( 24,4) = HCF ( 124, 24) = HCF ( 148, 124) = HCF(272,148) = HCF ( 420,272) = HCF (4052,420) R HCF (12576,4052).
10. PRIME FACTORISATION METHOD
WE HAVE ALREADY LEARNT THAT HOW TO FIND THE HCF AND
LCM OF TWO POSITIVE INTEGER USING THE FUNDAMENTAL
THEORM OF ARTHMETIC IN EARLIER CLASSES WITHOUT
REALISING IT. THIS METHOD IS ALSO CALLED THR PRIME
FACTORISATION METHOD.
11. EXAMPLE
EXAMPLE:- Find the LCM and HCF of 6 and 20 by the prime
factorization method.
Sol. We have 6 = 2¹×3¹ and 20 = 2×2×5 = 2²×5¹.
We can find HCF(6,20) = 2 and LCM(6,20) = 2×2×3×5 = 60, as done
in your earlier classes.
NOTE THAT HCF(6,20)= 2¹ = product of the smallest power of each
common prime factor in the numbers.
LCM(6,20) = product of the greatest power of each prime factor
involved in the number.
12. EXAMPLE 2. Find the HCF of 96 and 404 by the prime
factorization method. Hence find the LCM.
Sol. The prime factorization of 96 and 404 gives:
96= 2⁵×3, 404 = 2²× 101
Therefore,the HCF of these two integer is 2² = 4.
LCM(96,404) = 96×404/HCF(96,404)= 96×404/4 = 9696.
13. REVISITING IRRATIONAL NUMBER
THEORM 1.3: Let p be a prime number.if p divides a²,the p divides a, where a is a positive
integers.
Proof:- let the prime factorization of a be as follows:
a = p¹p²….p,where p¹p² and p are primes , not necessary distinct.
Therefore,a² = (p¹p²…..p)(p¹p²…..p).
NOW, We are given that p divides a².Therefore from the fundamental theorem of arithmetic,
it follows that p is one of the prime factor of a². However using the uniqueness part of
fundamental theoerm of arithmetic we realize that only prime factor of a² are p²p²….p.
NOW, since a = p¹p²…p,p divides a.
14. 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.
