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2 − 4
Example 14: Verify the closure property of subtraction for rational numbers and .
3 7
2  − 4 2 4 14 12 26
+
Solution: −   = + = = , which is a rational number.
3  7  3 7 21 21
Hence, the closure property of subtraction is verified.
Commutative Property

Let us check whether the commutative property exists for the subtraction of rational numbers.

3
5
Consider the pair of rational numbers and .
9 9
5 3 2 3 5 − 2 2 − 2 5 3 3 5
Since, − = and − = . Clearly, ≠ , therefore − ≠ −
9 9 9 9 9 9 9 9 9 9 9 9
This shows that subtraction is not commutative for rational numbers. In other words, the
commutative property does not hold for the subtraction of rational numbers.

p r p r r p
In general, for any two distinct rational numbers and , − ≠ − .
q s q s s q

2 −4
Example 15: Verify that the subtraction for rational numbers x = and y = is not commutative.
3 7

2  − 4 2 4 14 12 26
+
Solution: x – y = −   = + = = (Q LCM of 3 and 7 is 21)
3  7  3 7 21 21
−
−  4
y – x =    7   − 2 = −4 − 2 = −12 14 = −26
3
3
7
21
21
26 − 26
Clearly, ≠
21 21
2  − 4  − 4 2
Thus, −   ≠   − , i.e., x – y ≠ y – x
3  7   7  3
Hence, subtraction of rational numbers is not commutative.
Associative Property

Let us check whether the associative property exists for the subtraction of rational numbers.

Let us take three rational numbers −5 , 3 and 5 .
7 14 14
−5  3 5  −5  −  2 −10 2 −8 −4
We have −  −  = −   = + = or
7  14 14  7  14  14 14 14 7
 −5 3  5  −10 3  5 −13 5 −18 −9
Again,   7 − 14  − 14 =   14 − 14  − 14 = 14 − 14 = 14 or 7


Clearly, −4 ≠ −9 , therefore −5 −  3 − 5   ≠   −5 − 3   − 5

7 7 7  14 14   7 14  14
This shows that subtraction is not associative for rational numbers. In other words, the associative
property does not hold for the subtraction of rational numbers.

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