**Square and square root:**

** **A natural number m is said to be square number or perfect square if it can be expressed In terms of n^{2}, where n is also a natural number. It is necessary to know about square and square root to solve number system related problems.

**Example: **4 is a square number as 4 = 2^{2}

## Properties of perfect square:

- All perfect square numbers end with 0, 1, 4, 5, 6 or 9 at unit’s place.
- None of these end with 2, 3, 7 or 8 at unit’s place.
- If a number has 1 or 9 in the unit’s place, then it’s square ends in 1.
- when a perfect square number ends in 6, the number whose square it is will have either 4 or 6 in unit’s place.
**There are 2n non-perfect square numbers between n**between 2^{2}and (n + 1)^{2}

Example:^{2 }and 3^{2}the no. of non-perfect square no. are 2×2 =4 as (5, 6, 7, 8)- sum of first n odd natural numbers is n
^{2}also Remember it for square and square root related terms.

**Example:**1+3+5 = 9 = 3^{2} - Any perfect square number can be represented as sum of odd natural numbers starting from 1.
- Also, Any odd perfect square number can be represented as sum of two consecutive natural number.

**Example:**49 = 24+25, 121 = 60 + 61 - For any natural number (n + 1) × (n – 1) = n
^{2 }– 1.

**Example:**6 x 4 = 5^{2 }– 1 = 24 - Square numbers can only have
**even number of 0’s**at the end. - If a perfect square is of n-digits, then its square root will have \(\frac{n}{2}\) digits if n is even, or \(\frac{n + 1}{2}\) if n is odd?

**Pythagorean triplet:**

For any natural number a > 1, we have (2a)^{2}+(a^{2} – 1)^{2}=(a^{2 }+ 1)^{2}

**So, 2a, a ^{2 }– 1 and a^{2}+ 1 forms a Pythagorean triplet**.

By this rule, if any one of the members of Pythagorean triplet is given then you can find the rest. this also necessary for getting knowledge of square and square root.

**Example:
**1. 3

^{2}+

^{ }4

^{2 }= 5

^{2}so, 3, 4, 5 are Pythagorean triplet.

*If 5 is one of the members of the Pythagorean triplet then find the rest.*

**Solution:
**Try 2a = 5 then a will not be an integer so take next

a

^{2}– 1 = 5 then a

^{2}= 6 again not an integer

try a

^{2}+ 1 = 5 then a

^{2}= 4 thus, a = 2

2a = 4

a^{2} – 1 = 3

Thus, Pythagorean triplet are 3, 4, 5.

**Square root:**

It is simply the inverse operation of square. Square and square root are in converse of each other.

**There are many ways to find the square root of any number:**

- By just inversion of square.

**Example:**3^{2 }= 9 and √9 = 3 - By repeated subtraction of odd numbers.

**Example:**√9,

9 – 1 = 8, 8 – 3 = 5, 5 – 5 = 0, by 3 times consecutive subtraction we got zero thus √9 = 3. - By prime factorization.

√36 = 2x2x3x3 = 2^{2}x 3^{2 }thus square root of 36 is 2×3 i.e. 6 - By division method.

Try to solve examples based on square and square root. So that it will be easy for you to solve problems related to it.

## Be First to Comment