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Question
| Column I | Column II |
| (a) a × 1 | (i) Additive inverse of a |
| (b) 1 | (ii) Additive identity |
| (c) (–a) ÷ (–b) | (iii) Multiplicative identity |
| (d) a × (–1) | (iv) a ÷ (–b) |
| (e) a × 0 | (v) a ÷ b |
| (f) (–a) ÷ b | vi) a |
| (g) 0 | (vii) –a |
| (h) a ÷ (–a) | (viii) 0 |
| (i) –a | (ix) –1 |
Solution
| Column I | Column II |
| (a) a × 1 | (vi) a |
| (b) 1 | (iii) Multiplicative identity |
| (c) (–a) ÷ (–b) | (v) a ÷ b |
| (d) a × (–1) | (vii) – a |
| (e) a × 0 | (viii) 0 |
| (f) (–a) ÷ b | (iv) a ÷ (–b) |
| (g) 0 | (ii) Additive identity |
| (h) a ÷ (–a) | (ix) –1 |
| (i) –a | (i) Additive inverse of a |
Explanation:
(a) a × 1 = a
(b) 1 = 1 is multiplicative identity.
(c) (–a) ÷ (–b) = –a ÷ (–b) .....[Both signs are cancelled with each other]
(d) a × (–1) = a × (–1) = –a
(e) a × 0 = 0 .....[Any value, when multiplies with 0 becomes zero]
(f) (–a) ÷ b = a ÷ (–b)
(g) 0 = 0 is an additive identity
(h) a ÷ (–a) = `a ÷ (-a) = a/(-a)` = – 1
(i) –a = –a is additive inverse of a.
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