**Natural Numbers:**

The group of numbers starting from 1 and including 1, 2, 3, 4, 5, and so on. Zero, negative numbers, and decimals are not included this group.

**EXAMPLE **

**1. **If n is an odd natural number, what is the highest number that always divides n(n^{2} – 1)?

Answer: n∙(n^{2} – 1) = (n – 1)∙n∙(n + 1), which is a product of three consecutive numbers. Since n is odd, the numbers (n – 1) and (n + 1) are both even. One of these numbers will be a multiple of 2 and the other a multiple of 4 as they are two consecutive even numbers. Hence, their product is a multiple of 8. Since one out of every three consecutive numbers is a multiple of 3, one of the three numbers will be a multiple of three. Hence, the product of three numbers will be a multiple of 8 ´ 3 = 24.

Hence, the highest number that always divides n∙(n^{2} – 1) is 24.

**2. **For every natural number n, the highest number that n∙(n^{2} – 1)∙(5n + 2) is always divisible by is

(a) 6 (b) 24 (c) 36 (d) 48

Answer:

**Case 1: **If n is odd, n∙(n^{2} – 1) is divisible by 24 as proved in the earlier question.

**Case 2: **If n is even, both (n – 1) and (n + 1) are odd. Since product of three consecutive natural numbers is always a multiple of 3 and n is even, the product n∙(n^{2} – 1) is divisible by 6. Since n is even 5n is even. If n is a multiple of 2, 5n is a multiple of 2 and hence 5n + 2 is a multiple of 4. If n is a multiple of 4, 5n + 2 is a multiple of 2. Hence, the product n∙(5n + 2) is a multiple of 8.

Hence, the product n∙(n^{2} – 1)∙(5n + 2) is a multiple of 24.

Hence, [b]

**Rule: **The product of n consecutive natural numbers is divisible by n!, where n! = 1 × 2 × 3 × 4 × 5…. × n

**EXAMPLE **

**3. **Prove that (2n)! is divisible by (n!)^{2}.

Answer: (2n)! = 1·2·3·4· … ·(n – 1)·n·(n + 1)· …·2n

= (n)!·(n + 1)·(n + 2)· …·2n.

Since (n + 1)·(n + 2)· …·2n is a product of n consecutive numbers, it is divisible by n!. Hence, the product (n)!·(n + 1)·(n + 2)· …·2n is divisible by n!·n! = (n!)^{2}.

**Whole Numbers:**

All Natural Numbers plus the number 0 are called as Whole Numbers.

**Integers:**

All Whole Numbers and their negatives are included in this group.

**Rational Numbers:**

Any number that can be expressed as a ratio of two integers is called a rational number.

This group contains decimal that either do not exist (as in 6 which is 6/1), or terminate (as in 3.4 which is 34/10), or repeat with a pattern (as in 2.333... which is 7/3).

**Irrational Numbers:**

Any number that can not be expressed as the ratio of two integers is called an irrational number (imaginary or complex numbers are not included in irrational numbers).

These numbers have decimals that never terminate and never repeat with a pattern.

Examples include pi, e, and √2. 2 + √3, 5 - √2 etc. are also irrational quantities called **Surds.**

**EXAMPLE **

**Real Numbers:**

This group is made up of all the Rational and Irrational Numbers. The ordinary number line encountered when studying algebra holds real numbers.

**Imaginary Numbers:**

These numbers are formed by the imaginary number i (i = √-1). Any real number times i is an imaginary number.

Examples include i, 3i, −9.3i, and (pi)i. Now i^{2} = −1, i^{3} = i^{2} × i = −i, i^{4} = 1.

**EXAMPLE **

**Complex Numbers:**

A Complex Numbers is a combination of a real number and an imaginary number in the form a + bi. a is called the real part and b is called the imaginary part.

Examples include 3 + 6i, 8 + (−5)i, (often written as 8 - 5i).

**Prime Numbers:**

All the numbers that have only two divisors, 1 and the number itself, are called prime numbers. Hence, a prime number can only be written as the product of 1 and itself. The numbers 2, 3, 5, 7, 11…37, etc. are prime numbers.

**Note: **1 is not a prime number.

**EXAMPLE **

- If x
^{2}– y^{2}= 101, find the value of x^{2}+ y^{2}, given that x and y are natural numbers.

Answer: x^{2} – y^{2} = (x + y)(x – y) = 101. But 101 is a prime number and cannot be written as product of two numbers unless one of the numbers is 1 and the other is 101 itself.

Hence, x + y = 101 and x – y = 1. -> x = 51, y = 50.

-> x^{2} + y^{2} = 51^{2} + 50^{2} = 5101.

- What numbers have exactly three divisors?

Answer: The squares of prime numbers have exactly three divisors, i.e. 1, the prime number, and the square itself.

**Odd and Even Numbers:**

All the numbers divisible by 2 are called even numbers whereas all the numbers not divisible by 2 are called odd numbers. 2, 4, 6, 8… etc. are even numbers and 1, 3, 5, 7.. etc. are odd numbers.

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