हिंदी

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) - Mathematics

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प्रश्न

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
जोड़ियाँ मिलाइएँ

उत्तर

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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अध्याय 1: Integers - Exercise [पृष्ठ १९]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 7
अध्याय 1 Integers
Exercise | Q 117 | पृष्ठ १९

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