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Misc 2 - Let f(n) = n - 1, if is odd, f(n) = n + 1, if even - Finding Inverse

Misc 2  - Chapter 1 Class 12 Relation and Functions - Part 2
Misc 2  - Chapter 1 Class 12 Relation and Functions - Part 3
Misc 2  - Chapter 1 Class 12 Relation and Functions - Part 4
Misc 2  - Chapter 1 Class 12 Relation and Functions - Part 5
Misc 2  - Chapter 1 Class 12 Relation and Functions - Part 6
Misc 2  - Chapter 1 Class 12 Relation and Functions - Part 7
Misc 2  - Chapter 1 Class 12 Relation and Functions - Part 8
Misc 2  - Chapter 1 Class 12 Relation and Functions - Part 9

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Misc 2 (Method 1) Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers. f(n) = 𝑛−1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑﷮𝑛+1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Step 1 Let f(n) = y , such that y ∈ W n = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Let g(y) = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ where g: W → W Step 2: gof = g(f(n)) ∴ gof = n = IW Now, f(n) = 𝑛−1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑﷮𝑛+1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ & g(y) = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Step 3: fog = f(g(y)) ∴ fog = y = IW Since gof = IW and fog =IW f is invertible and inverse of f = g(y) = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Now g(y) = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Replacing y with n g(n) = 𝑛−1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑﷮𝑛+1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ = f(n) ∴ Inverse of f is f itself Misc 2 (Method 2) Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers. f(n) = 𝑛−1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑﷮𝑛+1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ f is invertible if f is one-one and onto Check one-one There can be 3 cases • x1 & x2 both are odd • x1 & x2 both are even • x1 is odd & x2 is even If x1 & x2 are both odd f(x1) = x1 + 1 f(x2) = x2 + 1 Putting f(x1) = f(x2) x1 + 1 = x2 + 1 x1 = x2 If x1 & x2 are both are even f(x1) = x1 – 1 f(x2) = x2 – 1 If f(x1) = f(x2) x1 – 1 = x2 – 1 x1 = x2 If x1 is odd and x2 is even f(x1) = x1 + 1 f(x2) = x2 – 1 If f(x1) = f(x2) x1 + 1 = x2 – 1 x2 – x1 = 2 which is impossible as difference between even and odd number can never be even Hence, if f(x1) = f(x2) , x1 = x2 ∴ function f is one-one Check onto f(n) = 𝑛−1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑﷮𝑛+1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Let f(n) = y , such that y ∈ W n = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Hence, if y is a whole number, n will also be a whole number i.e. n ∈ W Thus, f is onto. Finding inverse f(n) = 𝑛−1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑﷮𝑛+1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ For finding inverse, we put f(n) = y and find n in terms of y We have done that while proving onto n = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ ∴ Inverse of f = g(y) = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ where g: W → W Now g(y) = 𝑦−1, 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑﷮𝑦+1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ Replacing y with n g(n) = 𝑛−1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑﷮𝑛+1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛﷯﷯ = f(n) ∴ Inverse of f is f itself

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.