Page 12 - Maths Skills - 7

10 Maths

4. Existence of Additive Identity: If a is any integer, then a + 0 = a = 0 + a. We say 0 is an additive identity for
integers.

Since 5 is an integer, \ 5 + 0 = 5 = 0 + 5
Since – 5 is an integer, \ (– 5) + 0 = – 5 = 0 + (– 5 )
Hence 0 is the additive identity for integers.

5. Existence of Additive Inverse: If a is any integer, then a + (– a) = 0 = (– a) + a. We say – a is an additive
inverse of a.
Since 6 + (– 6) = 0 = (– 6) + 6, \ additive inverse of 6 is – 6.

SUBTRACTION OF INTEGERS

Subtraction is the inverse process of addition, therefore subtraction can be explained in terms of addition.
e.g., 9 – 5 = 4 is the same as
9 = 4 + 5 (or what should be added to 5 to get 9)
Using number line, we can justify in the following manner:

1. From 0, we move 5 steps toward right to reach + 5. 1 2 3 4 (number of
2. Now count the number of units that we need to units moved)
shift to reach 9. We need to move 4 units.
1 2 3 4 6 7 8
So, 5 + 4 = 9 or 9 – 5 = 4 Fig.

SUBTRACTION WITHOUT USING NUMBER LINES Fact-o-meter
If x and y are two integers, to subtract y from x; change the sign of
y and add it to x. � (+) × (+) = +
e.g., x – y = x + (– y) � (+) × (–) = –
4 – 3 = 4 + (– 3) = 1 � (–) × (–) = +
� (–) × (+) = –
PROPERTIES OF SUBTRACTION OF INTEGERS
1. Closure Property: If we subtract any two integers the result is again an integer.
Since 7 and 3 are integers, \ 7 – 3 = 4 and 3 – 7 = – 4 are also integers.

Since – 2 and – 5 are integers, \ – 2 – (– 5) = – 2 + 5 = 3 is an integer.
Also, – 5 – (– 2) = – 5 + 2 = –3 is also an integer.

2. Commutative Property: If a and b are any two integers, then a – b ≠ b – a i.e., commutative property does
not hold for subtraction of integers.
For integers – 3 and 5, we have
– 3 – (+ 5) = – (3 + 5) = – 8 and 5 – (– 3) = 5 + 3 = 8

\ – 3 – (+ 5) ≠ 5 – (– 3)
3. Associative Property: If a, b and c are any three integers, then (a – b) – c ≠ a – (b – c). Associative property
does not hold for subtraction of integers.

For integers – 1, 6 and 7, we have (– 1 – 6) – 7 = – 7 – 7 = – 14
and – 1 – (6 – 7) = – 1 – (–1) = – 1 + 1 = 0
\ (– 1 – 6) – 7 ≠ – 1 – (6 – 7)