Set of counting numbers is called the Natural Numbers

N = {1,2,3,4,5,...}

Set of Natural numbers plus Zero is called the Whole Numbers

W= {0,1,2,3,4,5,....}

So all natural Number are whole number but all whole numbers are not natural numbers

Integers is the set of all the whole number plus the negative of Natural Numbers

Z={..,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,...}

1) So ,integers contains all the whole number plus negative of all the natural numbers

2)the natural numbers without zero are commonly referred to as positive integers

3)The negative of a positive integer is defined as a number that produces 0 when it is added to the corresponding positive integer

4)natural numbers with zero are referred to as non-negative integers

5) The natural numbers form a subset of the integers.

Example : 1/2, 4/3 ,5/7 ,1 etc.

- every integers, natural and whole number is a rational number as they can be expressed in terms of p/q
- There are infinite rational number between two rational number
- They either have termination decimal expression or repeating non terminating decimal expression
- The sum, difference and the product of two rational numbers is always a rational number. The quotient of a division of one rational number by a non-zero rational number is a rational number. Rational numbers satisfy the closure property under addition, subtraction, multiplication and division.

Example : √3,√2,√5,p etc.

- Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Using this theorem we can represent the irrational numbers on the number line.
- They have non-terminating and non-repeating decimal expression
- The sum, difference, multiplication and division of irrational numbers are not always irrational. Irrational numbers do not satisfy the closure property under addition, subtraction, multiplication and division

- All rational and all irrational number makes the collection of real numbers. It is denoted by the letter R

- We can represent real numbers on the number line. The square root of any positive real number exists and that also can be represented on number line
- The sum or difference of a rational number and an irrational number is an irrational number.
- The product or division of a rational number with an irrational number is an irrational number.
- This process of visualization of representing a decimal expansion on the number line is known as the process of successive magnification

Real numbers satisfy the commutative, associative and distributive laws. These can be stated as :

Commutative Law of Addition:

$a+b= b+a$

Commutative Law of Multiplication:

$a \times b=b \times a$

Associative Law of Addition:

$ a + (b+c)=(a+b) +c $

Associative Law of Multiplication:

$a \times (b \times c)=(a \times b) \times c $

Distributive Law:

$a \times (b + c)=(a \times b) + (a \times c)$

or

$(a + b) \times c=(a \times c) + (b \times c)$

1

-2/3

3/4

√2

√2 + 5

Let a > 0 be a real number and p and q be rational numbers. Then, we have

- $a^p.a^q=a^{p+q}$
- $\frac {a^p}{a^q} =a^{p-q}$

- $(a^p)^q=a^{pq}$

- $a^p.b^p=(ab)^p$

$a=bq + r \; , \; 0 \leq r < b$

It is basic concept and it is restatement of division

- a is called dividend

- b is called divisor

- q is called quotient

- r is called remainder.

Let us assume q and r are not unique i.e. let there exists another pair $q_0$ and $r_0$ i.e. $a = bq_0 + r_0$, where $ 0 \leq r_0 < b$

=>$ bq + r = bq_0 + r_0$

=> $b(q - q_0) = r - r_0$ ................ (I)

Since $0 \leq r < b$ and $ 0 \leq r_0 < b$, thus $0 \leq r - r_0 < b$ ......... (II)

The above equation (I) tells that b divides (r - r

=> $r - r_0 = 0$

=> $r = r_0$

Equation (I) will be, $b(q - q_0) = 0$

Since b > 0, => $(q - q_0) = 0$

=> $q = q_0$

Since $r = r_0$ and $q = q_0$, Therefore q and r are unique.

Proof : Let m be a common divisor of a and b. Then,

m| a => a = mx for some integer x

m| b => b = mq2 for some integer y

Now, a = bq + r

=> r = a - bq

=> r = mx - my q

=> r = m( x - yq)

=> m| r

=> m| r and m | b

=> m is a common divisor of b and r.

Hence, a common divisor of a and b is a common divisor of b and r.

We know that for any two integers a,b. we can write following expression

$a=bq + r \; , \; 0 \leq r < b$

If r=0 ,then

HCF( a,b) =b

If r >0 , then

HCF ( a,b) = HCF ( b,r) as already proved from above theorem

Again expressing the integer b,r in Euclid's Division Lemma, we get

b=pr + r1

HCF ( b,r)=HCF ( r,r1)

Similarly successive Euclid 's division can be written until we get the remainder zero, the divisor at that point is called the HCF of the a and b

Use Euclid's algorithm to find the 65 and 117.

Step:1 Since 117 > 65 we apply the division lemma to 117 and 65 to get ,

$117 = 65 \times 1 + 52$

Step:2 Since 52 > 0 , we apply the division lemma to 65 and 52 to get

$65 = 52 \times 1 + 13$

Step:3 Since 13 > 0 , we apply the division lemma to 52 and 13 to get

$52 = 13 \times 4 + 0$

The remainder has now become zero, so our procedure stops. Since the divisor at this Step is 13, the HCF of 117 and 52 is 13.

1.The HCF of three numbers can be calculated by first calculating the HCF of first two numbers ,then the calculating the HCF of the HCF of previous two numbers and third number.

2. If HCF ( a,b) =1 ,the a and b are co primes.

Prime numbers are interesting blocks in Mathematics.They are highly used in cryptography field

Prime Number help

PRIME NUMBERS to 100 ={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}

PRIME NUMBERS to 100 ={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}

Since even numbers are always divisible by 2, no even number (other than 2) can be a prime number because they'll always have more than two factors. For example, the factor pairs for 6 are: 1,2,3,6 So 6 has four factors and therefore is not a prime number.

Numbers that end in 5 or 0 are always divisible by 5. So no numbers divisible by 5 (other than 5) can be prime numbers because they'll always have more than two factors. They'll have 1 and themselves, and they'll also have 5 and some other factor.

0 and 1 are not prime numbers. It may seem at first glance that they are, but remember that a prime number has only 1 and itself as factors. But 0 has an infinite number of factors because 0 * 1 = 0, and 0 * 2 = 0, and 0 * 3 = 0, and so on. So 0 is far from prime.

In the case of 1, it doesn't have two distinct factors because it's only factor is itself: 1 * 1 = 1. If you change one of those factors to any other integer, you no longer get a product of 1. Since it has only one factor, it's not prime.

Add all of the digits composing the number. If the sum of the digits is divisible by 3, then so is the original number. Example: If the number is 111, add 1 + 1 + 1 = 3. The sum of the digits (3) is divisible by 3, so 111 is also divisible by 3.

Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule repeatedly as necessary. Example: 826. Twice 6 is 12. So, take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also

If you are checking if a number is a prime number,then you don't need to check all the prime number below it. you need to check run for small primes(2,3,5,7) and you're not done factoring, then keep trying bigger and bigger primes (11, 13, 17, 19, 23, etc.) until you find something that works or until you reach primes whose squares are bigger than what you're dividing into. if you don't find then it is a prime Number Reason: If your prime doesn't divide in, then the only potential divisors are bigger primes. Since the square of your prime is bigger than the number, then a bigger prime must have as its remainder a smaller number than your prime. The only smaller number left, since all the smaller primes have been eliminated, is 1. So the number left must be prime, and you're done.

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101

103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197

199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311

313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431

433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557

569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673

677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811

821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941

947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039

,

4,6,8,9,10.....

Since 4 is divided by 2,6 is divided by 2 and 3

Composite number = Product of primes

$16=2 \times 2 \times 2 \times 2$

This is unique, We cannot express it in any other prime numbers

- The prime Number can be repeated in the factorization

- The order does not matter

- The prime factorisation of a natural number is unique, except for the order of its factors.

- Few things we need to remember,

a. If the number is even,then it will be divisible by 2

b. If the sums of its digits is divisible by 3,then it is divisible by 3

c. if the end of the number is 0 or 5,then it is divisible by 5 - We have to start with the small prime number with the rules given in step 1. Once we find the quotient, repeat the same process for the quotient. The
last quotient will be a prime number itself

Suppose the composite Number is 168

1. Now 168 is even number.So, we know it will get divided by 2

168 / 2 = 84

2. Again 84 is even number.So, we know it will get divided by 2

84 / 2 = 42

3. Again 42 is even number.So, we know it will get divided by 2

42 / 2 = 21

4. Now sum of digits of 21 is 3.So, we know we can divide it by 3

21 / 3 = 7

5. 7 is a prime Number

So prime factors = 2 X 2 X 2 X 3 X 7

Here are the prime factors of the composite numbers between 1 and 30.

$4 = 2 \times 2$

$6 = 3 \times 2$

$8 = 2 \times 2 \times 2$

$9 = 3 \times 3$

$10 = 5 \times 24$

$12 = 3 \times 2 \times 2$

$14 = 7 \times 2$

$15 = 5 \times 3$

$16 = 2 \times 2 \times 2 \times 2$

$18 = 3 \times 3 \times 2$

$20 = 5 \times 2 \times 2$

$21= 3 \times 7$

$22=2 \times 11$

$24= 2 \times 2 \times 2 \times 3$

$25= 5 \times 5$

$26= 2 \times 13$

$27= 3 \times 3 \times 3$

$28= 2 \times 2 \times 7$

$30= 2 \times 3 \times 5$

$4 = 2 \times 2$

$6 = 3 \times 2$

$8 = 2 \times 2 \times 2$

$9 = 3 \times 3$

$10 = 5 \times 24$

$12 = 3 \times 2 \times 2$

$14 = 7 \times 2$

$15 = 5 \times 3$

$16 = 2 \times 2 \times 2 \times 2$

$18 = 3 \times 3 \times 2$

$20 = 5 \times 2 \times 2$

$21= 3 \times 7$

$22=2 \times 11$

$24= 2 \times 2 \times 2 \times 3$

$25= 5 \times 5$

$26= 2 \times 13$

$27= 3 \times 3 \times 3$

$28= 2 \times 2 \times 7$

$30= 2 \times 3 \times 5$

We can find HCF and LCM between two number by the prime factorization method also

HCF (Highest Common factor) = product of the smallest power of each common factor in the numbers

Example: Suppose the number are 14,24

1. Lets do prime factorization method for both the numbers

$14=2 \times 7$

$24= 2 \times 2 \times 2 \times 3$

2. Now as per method

HCF =2

LCM( Lowest Common Multiple) = Product of the greatest power of each prime factor involved in the number

Example: Suppose the number are 14,24

1. Lets do prime factorization method for both the numbers

$14=2 \times 7$

$24= 2 \times 2 \times 2 \times 3$

2. Now as per method

$HCF =2 \times 2 \times 2 \times 7 \times 3=168$

Also important Formula to remember

If a and b are two number,then

**HCF(a,b) X LCM (a,b) =a X b **

If a and b are two number,then

We can also LCM using Division algorithm .In this method divide the given numbers by common prime number until the remainder is a prime number or one. LCM will be the product obtained by multiplying all divisors and remaining prime numbers.

Steps are

1. We place number in the line

2. We start dividing the number by least prime number which is common among all of them or group of them

3. Keep dividing by least until we have 1's in the remainder

4. LCM is the product of the divisors

Example

1. Find the LCM of 14 and 20

$LCM = 2 \times 2 \times 7 \times 5=140$

2. Find the LCM of 12 ,18 and 44

$LCM = 2 \times 2 \times 3 \times 3 \times 11=396$

Decimal expressions of rational number are of two types

a. Terminating decimal expression

b. Non-terminating repeating decimal expression

Terminating decimal expression can be written in the form

$\frac {p}{2^n5^m}$

Non terminating repeating decimal expression can not be expressed in this form

$\frac {p}{2^n5^m}$

2. If the proof is irrational number which will be the case always,assume the number to rational in the form p/q and then try to prove it wrong

1.Prove that $\sqrt {2}$ is irrational number

then

$\sqrt {2}=\frac {p}{q}$

where p and q are co primes.

or

$q\sqrt {2}=p$

squaring both sides

$2q^2=p^2$

So 2 divides p

2 will divide p also. p=2c

$2q^2=4c^2$

or

$q^2=2c^2$

So q divided by 2 also

So both p and q are divided by 2 which is contradiction from we assumed

So $\sqrt {2}$ is irrational number

2. Prove that $\sqrt {3}$ is irrational number

then

$\sqrt {3}=\frac {p}{q}$

where p and q are co primes.

or

$q\sqrt {3}=p$

squaring both sides

$3q^2=p^2$

So 3 divides p

2 will divide p also. p=3c

$3q^2=9c^2$

or

$q^2=3c^2$

So q divided by 3 also

So both p and q are divided by 3 which is contradiction from we assumed

So $\sqrt {3}$ is irrational number

- Real Numbers Important Definition
- |
- Euclid's Division Lemma
- |
- Proof of Euclid's Division Lemma
- |
- HCF (Highest common factor)
- |
- What is Prime Numbers
- |
- What is Composite Numbers
- |
- Fundamental Theorem of Arithmetic
- |
- HCF and LCM by prime factorization method
- |
- Irrational Numbers
- |
- How to prove the irrational numbers or Rational numbers

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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