For the binary operation multiplication modulo 10 × 10 defined on the set S = 1 , 3 , 7 , 9 , write the inverse of 3 .

Here, 1 × 10 1 = Remainder obtained by dividing 1 × 1 by 10 = 1 3 × 10 1 = Remainder obtained by dividing 3 × 1 by 10 = 3 7 × 10 3 = Remainder obtained by dividing 7 × 3 by 10 = 1 3 × 10 3 = Remainder obtained by dividing 3 × 3 by 10 = 9 So, the composition table is as follows × 10 1 3 7 9 1 1 3 7 9 3 3 9 1 7 7 7 1 9 3 9 9 7 3 1 We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column. These two intersect at 1. ⇒ a * 1 = 1 * a = a , ∀ a ∈ S So, the identity element is 1 . Also, 3 × 10 7 = 1 ∴ The inverse of 3 is 3 - 1 = 7

For the binary operation multiplication modulo 5 ( × 5 ) defined on the set S = 1 , 2 , 3 , 4 . Write the value of 3 × 5 4 - 1 - 1 .

Here, 1 × 5 1 = Remainder obtained by dividing 1 × 1 by 5 = 1 3 × 5 4 = Remainder obtained by dividing 3 × 4 by 5 = 2 4 × 5 4 = Remainder obtained by dividing 4 × 4 by 5 = 1 So, the composition table is as follows : × 5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 We observe that the first row of the composition table coincides with the top-most row and the first column coincides with the left-most column. These two intersect at 1 . ⇒ a × 5 1 = 1 × 5 a = a , ∀ a ∈ S Thus 1 is the identity element. Now, 3 × 5 4 - 1 - 1 = 3 × 5 4 - 1 [ ∵ 4 × 5 4 = 1 ] = 2 - 1 = 3 [ ∵ 2 × 5 3 = 1 ]

Write the composition table for the binary operation × 5 (multiplication modulo 5 ) on the set S = 0 , 1 , 2 , 3 , 4 .

Here, 1 × 5 1 = Remainder obtained by dividing 1 × 1 by 5 = 1 3 × 5 4 = 1 Remainder obtained by dividing 3 × 4 by 5 = 2 4 × 5 4 = Remainder obtained by dividing 4 × 4 by 5 = 1 So, the composition table is as follows: × 5 0 1 2 3 4 0 0 0 0 0 0 1 0 1 2 3 4 2 0 2 4 1 3 3 0 2 1 4 2 4 0 4 3 2 1

A binary operation * is defined on the set R of all real numbers by the rule a * b = a 2 + b 2 for all a , b ∈ R . Write the identity element for * on R .

Let e be the identity element in R with respect to * such that a * e = a = e * a , ∀ a ∈ R a * e = a and e * a = a , ∀ a ∈ R Then, a 2 + e 2 = a and e 2 + a 2 = a , ∀ a ∈ R ⇒ a 2 + e = a and e + a 2 = a , ∀ a ∈ R [ ∵ e 2 = e ] ⇒ a 2 + e = a 2 and e + a 2 = a 2 , ∀ a ∈ R ⇒ e = 0 ∈ R , ∀ a ∈ R Thus, 0 is the identity element in R with respect to * .