Misc 14 - Let a * b = { a + b, a + b - 6. Show 0 is identity

Misc 14 - Chapter 1 Class 12 Relation and Functions - Part 2
Misc 14 - Chapter 1 Class 12 Relation and Functions - Part 3
Misc 14 - Chapter 1 Class 12 Relation and Functions - Part 4

Go Ad-free

Transcript

Question 10 Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as a * b = {█(𝑎+𝑏, 𝑖𝑓 𝑎+𝑏<6@&𝑎+𝑏 −6, 𝑖𝑓 𝑎+𝑏≥6)┤ Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a. e is the identity of * if a * e = e * a = a Checking if zero is identity for this operation If a + b < 6 Putting b = 0 a < 6 This is possible Now, a * 0 = a + 0 = a 0 * a = 0 + a = a Thus, a * 0 = 0 * a = a So, 0 is identity of * If a + b ≥ 6 Putting b = 0 a ≥ 6 This is not possible as value of a can be {0, 1, ,2, 3, 4, 5} Now, we need to show that each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a. a * b = {█(𝑎+𝑏, 𝑖𝑓 𝑎+𝑏<6@&𝑎+𝑏 −6, 𝑖𝑓 𝑎+𝑏≥6)┤ An element a in set is invertible if, there is an element in set such that , a * b = e = b * a Putting b = 6 – a So, a + b = a + (6 – a) = 6 Since a + b ≥ 6 a * b = a + b – 6 a * b = a * (6 – a) = a + (6 – a) – 6 = 0 b * a = (6 – a) * a = (6 – a) + a – 6 = 0 Since a * (6 – a) = (6 – a) * a = 0 Hence, each element a of the set is invertible with 6 − a being the inverse of a. s

Davneet Singh's photo - Co-founder, Teachoo

Made by

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