squares and square roots

Squares and square roots

INTRODUCTION

•Numbers like ,1,4,9,16,25 are known as square numbers•If a natural number m can be expressed as n2 where n is also a natural number, then m is a square number.

If a number to be square has 1 or 9 in the units place then its square ends in 1.

When square number ends in 6 the number whose square it is will have either 4 or 6 in its units place.

PROPERTIES OF SQUARE NUMBERS

Any natural number n and (n+ 1), then (n+1)2 – n2 = (n2 + 2n +1)- n2 = 2n +1There are 2n non perfect square numbers

between the squares of the numbers n and (n+1).

NUMBERS BETWEEN SQUARE NUMBERS

Sum of the first n odd natural numbers is n2 .

If a natural number cannot be expressed as a sum of successive odd numbers starting with 1, then it is not a perfect square.

ADDING ODD NUMBERS

12 = 1112 = 1211112 = 12332111112 = 12344321111112 = 1234543211111112 = 123455432111111112 = 12345654321

PATTERNS IN SQUARE NUMBERS

72 = 49672 =44896672 =44488966672 =44448889666672 =44444888896666672 =444444888889

We can find out the square of a number without having two multiply the numbers.

For e.g. 232 = (20 +3)2 =20(20+3)+3(20+3) = 202 + 20*3 + 3*20 + 32

FINDING THE SQUARE OF A NUMBER

For any natural number m > 1, we have (2m)2 + (m2 – 1) = (m2 + 1)2 .

PYTHAGOREAN TRIPLETS

In mathematics, a square root of a number a is a number y such that y2= a, or, in other words, a number y whose square (the result of multiplying the number by itself, or y × y) is a.  For example, 4 and -4 are square roots of 16 because 42 = (-4)2 = 16.

SQUARE ROOTS

In Ancient India, the knowledge of theoretical and applied aspects of square and square root was at least as old as the Sulba Sutras, dated around 800–500 BC (possibly much earlier). A method for finding very good approximations to the square roots of 2 and 3 are given in the Baudhayana Sulba Sutra. Aryabhata in the Aryabhatiya (section 2.4), has given a method for finding the square root of numbers having many digits.

HISTORY

Finding square roots through repeated subtraction81−1=8080 − 3=7777 − 5=7272 − 7=6565 − 9=5656 − 11=4545 − 13=3232− 15=1717 − 17=0

FINDING SQUARE ROOTS

From 81 we have subtracted successive odd numbers starting from 1 and obtained 0 at 9th step. √81 = 9

Finding square root through prime factorizationEach prime factor in a prime factorization of the

square of a number, occurs twice the number of times it occurs in the prime factorization of a given square number, say 324.

324=2*2*3*3*3*3By pairing the prime factors, we get 324=2*2*3*3*3*3=22 * 32 * 32 = (2*3*3)2

So, 18332

1. There are 500 children in a school. For a P.T. drill they have to stand in such a manner that the numbers of rows is equal to number of columns. How many children would be left out in this arrangement?

2. Find the smallest square that is divisible by each of the numbers 4,9 and 10?3. Find the square root of 2.56? 4. Find the square root of 4489?5. Find the square root of 100 and 169 by repeated subtraction?6. Find the square of 71?

Comprehension check