Different types of numbers & decimals mathematics , HCF & LCM by Euclid's Division Algorithm & Fundamental Theorem of Arithmetic's, Irrational numbers
Common topics for the students of IX and X standard
1. NATURAL NUMBERS :- Counting numbers are called natural numbers.
Eg:- N = {1, 2, 3, 4, …………}
2. WHOLE NUMBERS :– Natural numbers along with zero are called whole numbers. Eg:- W = {0,1, 2, 3, 4,…….} [N ⊂ W]
3. INTEGERS – All whole number and natural numbers with negative sign are called integers. Eg :- Z = {….-3, -2, -1, 0, 1, 2, 3…..} [N ⊂ W ⊂ Z]
4. RATIONAL NUMBERS – The number which can be put in the form of P/q (Where q ≠ 0) are called rational numbers.
Eg :- Q = -5, 0, 7, 3/5, -7/8 etc. [N ⊂ W ⊂ Z ⊂ Q]
5. FRACTIONS – Positive rational number are called fractions.
(OR)
Fraction is the ratio of two natural numbers. Eg:- 7/5, 1/2, 5/7 ….etc.
6. IRRATIONAL NUMBERS – The number which cannot be put in the form of p/q are called irrational numbers.
For Example : 
all are irrational numbers
7. REAL NUMBERS – All rational and irrational numbers are called real numbers.
R = Q ∪ S
8. PRIME NUMBERS:- The numbers which has only two factors one and itself are called prime numbers. Eg:- 2, 3, 5, 7, 11, 13 There are 25 prime numbers in a counting from 1 to 100
9. CO-PRIMES :- If H.C.F of two numbers is 1 then numbers are called co-prime numbers. Eg:- (5, 7), (13, 27), (15, 16)…….
Note: HCF of any two consecutive Natural Numbers is always 1. So any two consecutive natural numbers are always a pair of coprime.
10. TWIN PRIMES :- Consecutive prime numbers which are differ by 2 are called twin primes. Eg :- (11, 13), (17, 19), (29, 31)…….etc.
11. EVEN NUMBERS:- Natural numbers which are divisible by two are called even numbers. eg 2, 4, 6, 8, 10, ..........
12. ODD NUMBERS:- Natural numbers which are not divisible by 2 are called odd numbers. Eg:- 1, 3, 5, 7, 9, ........
13. COMPOSITE NUMBERS:- The numbers which has more than two factors are called composite numbers. Eg:- 4, 6, 9, 10, .... "9" is the smallest odd composite number
Note:- "1" is neither a prime number nor a composite number. It is a unit.
14. IMAGINARY NUMBER or Non-Real Number :- Negative square root of a natural number is called Imaginary number Or non-real numbers.
For Example
are irrational numbers.
DIFFERENT TYPES OF DECIMALS
Common topics for the students of IX and X standard
Decimals are of three types
a.
Terminating Decimal :-
If prime factors of denominator of
a rational number is either the power of 2 or 5 or both (2m x 5n )then
the decimal form is known as terminating decimal.
Example: 2.5, 3.6, 7.895
b. Non-Terminating But Repeating Decimal:- If prime factors of denominator of a rational number contain the factors other
than power of 2 or 5 or both then the decimal form is called non terminating
but repeating decimal.
Eg :-
3.6767…. 14.367367…....
c.
Non-Terminating Non-Repeating Decimal :–
Decimal expansion of the numbers like 
is called non terminating no-repeating decimals Eg:- 1.1010010001......, 2.5050050005......, 3.6060060006.......
12. RATIONAL
DECIMALS :-
Terminating or non- terminating but repeating
decimals are called rational decimals.
13. IRRATIONAL
DECIMALS:-
Non-terminating, Non-Repeating decimals are
called Irrational decimals
SPECIFIC TOPICS FOR X STANDARD ONLY
FUNDAMENTAL THEOREM OF ARITHMETIC –
Every composite number can be expressed as a product of primes, and this factorization is unique irrespective of their order.
HCF – Product of common factors with smallest power.
LCM – Product of all factors with greatest power.
When two numbers are given then
HCF x LCM = Product of two numbers
This rule is not valid for more than two numbers.
NCERT Exercise 1.1 (New Book)
Q. 5 Check whether 6n can ends with digit 0 for any natural number n.
Solution: If 6n ends with 0 then its prime factorization must contain the power of 2 and 5. Prime factors of 6 are 2 x 3 and by the fundamental theorem of arithmetic this factorization is unique. This shows that prime factors of 6n does not contain 5.
Hence 6n never ends with zero for any value of n
Class 10 : Exercise 1.2
Question 1 Prove that √5 is an irrational number
Solution:Let √5 is a rational number
Then there exist co-prime a and b such that
√5 = a/b
Cross multiply here we get
√5b = a
Now squaring on both side we get
5b2 = a2
⇒ b2 = a2/5 ……………..(1)
⇒ 5 divides a2 ⇒ 5 divides a ………(2)
Let a = 5c , Squaring on both sides, we get
a2 = 25c2 ………………(3)
Putting eqn.(3) in equation (1) we get
⇒ b2 = 25c2/5 ,
⇒ b2 = 5c2
⇒ b2/5 = c2
5 divides b2 ⇒ 5 divides b…..(4)
From (2) and (4)
5 divides a and b both
⇒ a and b are not co-prime
But it is given in the starting that a and b are co-prime numbers.
So there is a contradiction and our assumption is wrong
Hence √5 is an irrational number
Question: Prove that 6 - 2√5 is an irrational number
Solution:
Let 6 - 2√5 is a rational number.
Then there exist unique integers a and b such that
a, b, 2 and 6 all are integers
⇒ 
is a rational number.
But √5 is an irrational number.
So there is a contradiction and our assumption is wrong
Hence 6 - 2√5 is an irrational number.
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DELETED TOPICS FROM CBSE SYLLABUS
EUCLID DIVISION LEMMA :
For given positive integers a and b there exist unique integer q and r such that
a = bq + r , ................ where 0 ≤ r <b.
LEMMA :
It is the proven statement used for proving another statement.
ALGORITHM:
A series of well defined steps which gives a procedure for solving a type of problem.
Question
Prove that the square of any positive integer is of the
form 3m or 3m + 1
Solution: For two given positive integers a and b there exist unique integers q and r such that
a = bq + r, where 0 ≤ r < b
Let b = 3 ⇒ a = 3q + r
Possible values of r = 0, 1, 2
Possible values of a = 3q + 0, 3q + 1, 3q + 2Now taking a = 3q
Squaring on both side we get
(a)2 = (3q)2
⇒ a2 = 9q2
⇒ a2 = 3(3q2 )
⇒ a2 = 3m1 … .......Where m1 = 3q2
Now taking a = 3q +1
Squaring on both side we get
(a)2 = (3q + 1)2
⇒ a2 = 9q2 + 1 + 6q
⇒ a2 = 9q2 + 6q + 1
⇒ a2 = 3(3q2 + 2q) + 1
⇒ a2 = 3m2 +1 .........Where m2 = 3q2 + 2q
Now taking a = 3q + 2
Squaring on both side we get
(a)2 = (3q + 2)2
⇒ a2 = 9q2 + 4 + 12q
⇒ a2 = 9q2 + 12q + 4
⇒ 9q2 + 12q + 3 + 1
⇒ a2 = 3(3q2 + 4q + 1) + 1
⇒ a2 = 3m3 + 1 ............Where m3 = 3q2 + 4q + 1
Here we prove that the square of any positive integer is of the form 3m or 3m + 1
Question
Prove that the cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8
Solution: For two given positive integers a and b there exist unique integers q and r such that
a = bq + r, where 0 ≤ r < b
Let b = 3 ⇒ a = 3q + r
Possible values of r = 0, 1, 2
Possible values of a = 3q + 0, 3q + 1, 3q + 2Now taking a = 3q
Cubing on both sides we get
(a)3 = (3q)3
⇒ a3 = 27q3
⇒ a3 = 9(3q3 )
⇒ a3 = 9m1 … .......Where m1 = 3q3
Now taking a = 3q +1
Cubing on both side we get
(a)3 = (3q + 1)3
⇒ a3 = 27q3 + 1 + 27q2 + 9q
⇒ a3 = 27q3 + 27q2 + 9q +1
⇒ a3 = 9(3q3 + 3q2 + q) +1
⇒ a3 = 9m2 +1 ...........Where m2 = 3q3 + 3q2 + q
Now taking a = 3q +2
(a)3 = (3q + 2)3
⇒ a3 = 27q3 + 8 + 54q2 + 36q
⇒ a3 = 27q3 + 54q2 + 36q +8
⇒ a3 = 9(3q3 + 6q2 + 4q) +8
⇒ a3 = 3m3 +8 ...........Where m3 = 3q3 + 6q2 + 4q
Here we prove that the cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8
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