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### Maths Formulas Class X Ch -1 | Real Numbers

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# Maths Formulas Class X Ch -1 | Real Numbers

*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**

**"2" is the smallest prime number.**

**"2" is the only even prime number.**

**There are 10 prime number in a counting from 1 to 30**

**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. E**

**g:- 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, ....**

**"4" is the smallest composite number.**

**"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 (**2

^{m}x 5

^{n }

**)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 6^{n} can ends with digit 0 for any natural number n.

^{n}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 6

^{n}does not contain 5.

Hence 6

^{n}never ends with zero for any value of n

__Class 10 : Exercise 1.2__

__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**

**5b ^{2} = a^{2}**

**⇒**** b ^{2} = a^{2}/5 ……………..(1)**

**⇒**** 5 divides a ^{2} **

**⇒**

**5 divides a ………(2)**

**Let a = 5c , Squaring on both sides, we get**

**a ^{2} = 25c^{2} ………………(3)**

**Putting eqn.(3) in equation (1) we get**

**⇒**** b ^{2 }= 25c^{2}/5 , **

**⇒**** b ^{2} = 5c^{2} **

**⇒**** b ^{2}/5 = c^{2}**

** 5 divides b ^{2} **

**⇒**

**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.**

*******************************************************

## 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: **

**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 = 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 + 2

Possible values of r = 0, 1, 2

Possible values of a = 3q + 0, 3q + 1, 3q + 2

**Now taking a = 3q**

**Squaring on both side we get **

**(a) ^{2} = (3q)^{2} **

**⇒**** a ^{2} = 9q^{2 } **

**⇒**** a ^{2} = 3(3q^{2} ) **

**⇒**** a ^{2} = 3**

**m**

_{1}**… .......Where**

**m**

_{1}**= 3q**

^{2}**Now taking a = 3q +1**

** Squaring on both side we get **

**(a) ^{2} = (3q + 1)^{2} **

**⇒**** a ^{2} = 9q^{2} + 1 + 6q **

**⇒**** a ^{2} = 9q^{2} + 6q + 1 **

**⇒**** a ^{2} = 3(3q^{2} + 2q) + 1**

**⇒**** a ^{2} = 3**

**m**

_{2}**+1 .........Where**

**m**

_{2}**= 3q**

^{2}+ 2q**Now taking a = 3q + 2 **

**Squaring on both side we get **

**(a) ^{2} = (3q + 2)^{2} **

**⇒**** a ^{2} = 9q^{2} + 4 + 12q **

**⇒**** a ^{2} = 9q^{2} + 12q + 4 **

**⇒**** 9q ^{2} + 12q + 3 + 1 **

**⇒**** a ^{2} = 3(3q^{2} + 4q + 1) + 1 **

**⇒**** a ^{2} = 3**

**m**

_{3}**+ 1 ............Where**

**m**

_{3}**= 3q**

^{2}+ 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 = 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 + 2

Possible values of r = 0, 1, 2

Possible values of a = 3q + 0, 3q + 1, 3q + 2

**Now taking a = 3q**

**Cubing on both sides we get **

**(a) ^{3} = (3q)^{3} **

**⇒**** a ^{3} = 27q^{3 } **

**⇒**** a ^{3} = 9(3q^{3} ) **

**⇒**** a ^{3} = 9**

**m**

_{1}**… .......Where**

**m**

_{1}**= 3q**

^{3}**Now taking a = 3q +1**

**Cubing on both side we get **

**(a) ^{3} = (3q + 1)^{3} **

**⇒**** a ^{3} = 27q^{3} + 1 + 27q^{2} + 9q **

**⇒ **** a ^{3} = 27q^{3} + 27q^{2} + 9q +1 **

**⇒**** a ^{3} = 9(3q^{3} + 3q^{2} + q) +1**

**⇒**** a ^{3} = 9**

**m**

_{2}**+1 ...........Where**

**m**

_{2}**= 3q**

^{3}+ 3q^{2}+ q**Now taking a = 3q +2**

** (a) ^{3} = (3q + 2)^{3} **

**⇒**** a ^{3} = 27q^{3} + 8 + 54q^{2} + 36q **

**⇒ **** a ^{3} = 27q^{3} + 54q^{2} + 36q +8 **

**⇒**** a ^{3} = 9(3q^{3} + 6q^{2} + 4q) +8**

**⇒**** a ^{3} = 3**

**m**

_{3}**+8 ...........Where**

**m**

_{3}**= 3q**

^{3}+ 6q^{2}+ 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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Really it is very useful for we people.

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