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1 2 −3
Example 16: Verify (x – y) – z ≠ x – (y – z), if x = , y = and z = .
2 3 5
−  3
−  1
−
3
3
Solution: LHS = (x – y) – z =    1 − 2 −    − 5   =  34  −    − 5   =    6   −    5   = −1 + 3 = −+518 = 13


30
5
6 
3

30
6
2
1  2  − 3  1  2 3 1  10 9 1 19 15 38 − 23
+
−
RHS = x – (y – z) = −  −    = −  +  = −   = − = =
2  3  5   2  3 5 2  15  2 15 30 30
13 − 23
Clearly, ≠ . Hence, (x – y) – z ≠ x – (y – z)
30 30
Existence of Identity Property
Let us verify whether the identity element exists in the case of subtraction for rational numbers.
1 1 1 1 1
Consider the rational number . We have, 4 −= 4 . But, 0 − 4 = − .
0
4
4
1 1
So, −≠ 0 − , thus the identity element does not exist in the case of subtraction for rational
0
4 4
numbers.
In general,
p p p
if is a rational number, then −≠ −0 0 , i.e., identity property does not hold for the subtraction
q q q
of rational numbers.
Practice Time 1C

1. Subtract the following.
3 5 −1 −2 −3 −5 −2 −3
(a) from (b) from (c) from (d) from
8 4 2 5 8 9 7 −4
2. Simplify the following.

2 4 −7 1 −3  −  5 5  − 7 
(a) − (b) − (c) −   (d) −  
3 5 10 8 7  6  24  − 12
3. Verify that p – q is a rational number, when
5 −3 −2
(a) p = 1, q = (b) p = , q =
7 7 5
4. Solve the following.
−14
(a) The sum of two rational numbers is –2. If one number is , find the other number.
5
−5 −3
(b) What should be added to to get ?
8 22
−2 −1
(c) What should be subtracted from to get ?
8 6
5. Evaluate the following.

−5 7 5 −5 3  −  2 −2  −  4  −  3
(a) − + (b) + −   (c) −   −  
18 12 24 18 8  9  5  15   10 

21 Rational Numbers