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Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (i) (−5)/( 7) ⎕ 2/3 (−5)/7 ⎕ 2/3 Since negative number is always smaller than positive (−5)/7 < 2/3 Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (ii) (−4)/5 ⎕ (−5)/( 7) (−4)/5 ⎕ (−5)/( 7) Since both are negative, We ignore the signs 4/5 & 5/7 We compare these two numbers. 4/5 & 5/7 Since denominator in different, We make it same Common denominator = L.C.M of 5 & 7 = 5 × 7 = 35 So, 4/5 = 5/7 = Therefore, 28/35 > 25/35 i.e. 4/5 > 5/7 Multiplying –1 both sides 4/5 × –1 < 5/7 × –1 (−𝟒)/𝟓 < (−𝟓)/𝟕 Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (iii) (−7)/( 8) ⎕ 14/(−16) (−7)/8 ⎕ 14/(−16) Writing 14/(−16) in standard form 14/(−16) = (−14)/16 = (−14)/16 = (−7)/8 So, both numbers are (−7)/8 ∴ They are same (−𝟕)/𝟖 = 𝟏𝟒/(−𝟏𝟔) Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (iv) (−8)/5 ⎕ (−7)/4(−8)/5 ⎕ (−7)/( 4) Since both are negative, We ignore the signs 8/5 & 7/4 We compare these two numbers. 8/5 & 7/4 Since denominator in different, We make it same Common denominator = L.C.M of 4 & 5 = 2 × 2 × 5 = 20 So, 8/5 = 7/4 = Therefore, 32/20 < 35/20 i.e. 8/5 < 7/4 Multiplying –1 both sides 8/5 × –1 > 7/4 × –1 (−𝟖)/𝟓 > (−𝟕)/𝟒 Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (v) 1/(−3) ⎕ (−1)/41/(−3) ⎕ (−1)/( 4) Since both are negative, We ignore the signs 1/3 & 1/4 We compare these two numbers. 1/3 & 1/4 Since denominator in different, We make it same Common denominator = L.C.M of 3 & 4 = 2 × 2 × 3 = 12 So, 1/3 = 1/4 = Therefore, 4/12 > 3/12 i.e. 1/3 > 1/4 Multiplying –1 both sides 1/3 × –1 < 1/4 × –1 (−𝟏)/𝟑 < (−𝟏)/𝟒 Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (vi) 5/( −11) ⎕ (−5)/11 5/(−11) ⎕ (−5)/( 11) (−5)/11 ⎕ (−5)/( 11) (−𝟓)/𝟏𝟏 = (−𝟓)/( 𝟏𝟏) Ex 8.1, 8 Fill in the boxes with the correct symbol out of >, <, and = (vii) 0 ⎕ (−7)/6 0 ⎕ (−7)/( 6) Since negative number is always less than 0 ∴ 0 is greater than any negative number 0 > (−𝟕)/𝟔

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo