Page 27 - Maths Class 06
P. 27

For example :   (i)                (4 5+  ) 6+  =  9 +  6 =  15
                                 Again,         4 +( 5 6+ )  =  4 11 15+  =
                                 So that,      (4 5+  ) 6+  =  4 + (5 6+  )

                             (ii) (15 16+  ) 17+  =  15 +  (16 +  17 )
                            (iii) (115 226+  ) 338+  =  115 +  (226 +  338 )

            II. Subtraction Properties
            Operations of addition and subtraction are inverse of each other.

            1. Closure Property
            The closure property does not hold good for subtraction of whole numbers. If a and b are whole numbers,
            then a – b is a whole number, when a > b or a = b. If a < b, then a – b is not a whole number.

            For example,

                                     5         –        3        =        2 (A whole number)

                                     8         –        8        =        0 (A whole number)


                                     2         –        3        =     -1 (Not a whole number)


            2. Commu ta tive Prop erty
            If a and b are two whole numbers, then in general, a – b is not equal to b – a i.e., a – b ¹ b –a. Hence, the
            commutative property is not true for subtraction of whole number.

            For example,         5 3- ¹  3 5- ;                      16 -  12 ¹  12 -  16


                NOTE

              We may restate the Commutative property as: If a and b are whole numbers and a ¹ b, then either a – b is a whole
              number or b – a is a whole number. Both are whole numbers only, if a = b.

            3. Associative Property
            If a, b, c are three whole numbers and c is not equal to 0, then a – (b – c) is not equal to (a – b) – c.

            Thus, the associative property for subtraction also does not hold for whole numbers.
            Study the following carefully:
                                 (5 – 2) – 1                         5 – (2 – 1)



                                 3 – 1 = 2                           5 – 1 = 4


                                                     different
            Thus,                 (5 – 2) – 1 ¹ 5 – (2 – 1)







                                                                                                    27
   22   23   24   25   26   27   28   29   30   31   32