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Friday, 3 March 2017

CBSE Class 8 - Maths - Squares - Properties of A Perfect Square (#cbsenotes)

Properties of A Perfect Square

CBSE Class 8 - Maths - Squares - Properties of A Perfect Square (#cbsenotes)

Please look at the table of some squares given below. You will see interesting patterns here.

Square of number 0 = 0

Number Square Number Square Number Square
1 1 11 121 21 441
2 4 12 144 22 484
3 9 13 169 23 529
4 16 14 196 24 576
5 25 15 225 25 625
6 36 16 256 26 676
7 49 17 289 27 729
8 64 18 324 28 784
9 81 19 361 29 841
10 100 20 400 30 900


PROPERTY 1:
Look at the table above. Observe that none of the column of squares ends with 2, 3, 7, or 8. Therefore we can say that a number endings with 2, 3, 7, 8 can never be a perfect square.
Thus 228257 , 132457 , 189678 , 84453 are not perfect squares.

PROPERTY 2:
Look at the last row of the table shown above. We notice that all the perfect squares are ending with an even number or 0. In the first line  e.g. 10² = 100 , 20² = 400, 30² =900,
Similarly 80²  = 6400, 50² = 2500.

We can prime factorise and see that 10, 20, 30,  etc . are not perfect squares. Thus a number ending with odd  number of zeros can never be a perfect square.




☛NOTE But this does not mean that a number ending with even number of 0s is always a perfect squares. It may or may not be  perfect squares.


PROPERTY 3:
See the table carefully and you will find out that squares of the even  number  are  even and squares of the odd numbers are odd. e.g. 13² = 169, 14² = 196, 26² = 676 etc.


PROPERTY 4:
You'd find an interesting pattern,

  1² = 1 (1st natural odd number)
  2²= 4 = 1+3 (Sum of the 1st two odd natural numbers )
  3²= 9 = 1+3+5 (Sum of the 1st three odd natural numbers )

Thus, we conclude that the squares of the natural numbers n is equal to the sum of the 1st n  odd natural numbers. 8²  = 64 = 1 + 3 + 5 + 7+ 9 + 11 + 13 + 15.


PROPERTY 5:
From the table you know that :
      2²-1² = 4-1 = 3 = 2+1
      3²-2² = 9-4 = 5 = 3+2
      12²-11² = 144 -121 = 23 = 12 + 11                  
      25²- 24² = 625-576 = 49 = 25 + 24
Or    (n+1)²-n² = (n+1) + n

Thus, we find that the difference of the squares of the consecutive natural numbers is equal to the sum of the numbers.

PROPERTY 6:
Three natural numbers a, b and c are called a Pythagorean triplet if a² + b² = c² .
For  any natural number n > 1.
(2n, n² - 1, n² + 1)  is a Pythagorean triplet .


PROPERTY 7:
Squares of the  natural number other than the natural number 1 is either divisible by 3 or leaves a remainder 1 when divided by 3.
e.g.  2² = 4 and 4 = 3 × 1 + 1
      3² = 9 and 9 = 3 × 3 + 0
      4² = 16 and 16 = 3 × 5 + 1      

☛NOTE  It does not mean that any number divisible by 3 and or when divided by 3 or leaves remainder 1 than its is always a perfect square . e.g. 30 = 3 × 10 is divisible by 3 but is not a perfect square.
19 = 3 × 6 +1  leaves a remainder 1 when divided by a 3 but is not a perfect square.

PROPERTY  8:
Squares of a natural number other than the natural number 1 is either divisible by 4 or leaves a remainder 1 when divided by 4.
e.g.
  2² = 4 and 4 = 4 × 1 +0                        
  3² = 9 and 9 = 4 × 2 + 1          
  4² = 16 and 16 = 4 × 4 + 0

☛NOTE  This does not mean that any number divisible by 4 or leaves a remainder 1 then it is always  a perfect square.
E.g., 52 is divisible by 4 but is not a perfect square. 65 when divided by 4 leaves remainder 1 but is not a perfect square.

From the above mentioned property 7 and 8, we can conclude that a perfect square is either a multiple of 3 a multiple of 4 or n-1 should be a multiple of 4 or 3. Note this does not mean that any number satisfying this condition is always perfect square.


PROPERTY 9:
If n is a perfect square than its double can never be a perfect square.
e.g.    9 = 3² is a perfect square but 2 × 9 = 18 is not a perfect square.
       36 = 6² is a perfect square but 2 × 36 = 72 is not a perfect square.



PROPERTY 10:

Square of a proper fraction is always less than the given fraction, e.g.  

   = 4
9

and
4  < 2
9   3

PROPERTY 11:
Check the following:
10² = 100     0² = 0      15² = 225     5² = 25                                        
11² = 121     1² = 1      16² = 256     6² = 36  
12² = 144     2² = 4      17² = 289     7² = 49    
13² = 169     3² = 9      18² = 324     8² = 62
14² = 196     4² = 16     19² = 361     9²= 81

We find that the unit digit of the square of a natural number is the unit digit of the square of the unit digit of the given numbers.

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