Page 20 - ICSE Math 7
P. 20
• Existence of multiplicative inverse
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Let a be any integer other than zero, then its multiplicative inverse will be , such that a × = 1
a
a
1
but is not an integer except for a = 1. Here multiplicative inverse does not exist under multiplication
a
of integers.
• Property of zero
For any integer a, we have a × 0 = 0 = 0 × a.
For example, –3 × 0 = 0 = 0 × 3, 4 × 0 = 0 = 0 × 4
• Cancellation law
If a, b and c are any three integers, then a × c = b × c ⇒ a = b.
For example, 11 × 9 = x × 9 ⇒ x = 11
• Distributive property of multiplication over addition and subtraction
If a, b and c are any three integers, then
a × (b + c) = (a × b) + (a × c) and a × (b – c) = (a × b) – (a × c)
For example, 5 × (10 + 13) = 5 × 23 = 115 and 5 × (10 + 13) = (5 × 10) + (5 × 13) = 50 + 65 = 115
Division
For division of integers, we have two cases:
(a) When both the dividend and divisor have the same sign, then the Maths Info
quotient is positive. (+) ÷ (+) = (+)
(b) When one of the dividend or divisor is negative, then the quotient (–) ÷ (–) = (+)
is negative. (–) ÷ (+) = (–)
(+) ÷ (–) = (–)
Points to remember
• Division by 0 is not defined.
• Any integer divided by 1 gives the integer itself.
Example 8: Find the quotient.
(a) (+90) ÷ (+15) (b) (+156) ÷ (–13) (c) (–224) ÷ (–16)
90
Solution: (a) (+90) ÷ (+15) = + (Since both the dividend and the divisor are
15 positive, the quotient is positive.)
= +6
156
(b) (+156) ÷ (–13) = – (Since the divisor is negative, the quotient
13 is negative.)
= –12
224
(c) (–224) ÷ (–16) = + 16 (Since both the dividend and the divisor are
negative, the quotient is positive.)
= +14
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