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Example 17 (Method 1) Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse. Checking inverse Step 1 f(x) = 4x + 3 Let f(x) = y y = 4x + 3 y – 3 = 4x 4x = y – 3 x = (𝑦 − 3)/4 Rough Checking inverse of f:X → Y Step 1: Calculate g: Y → X Step 2: Prove gof = IX Step 3: Prove fog = IY g is the inverse of f Let g(y) = (𝑦 − 3)/4 where g: Y → N Step 2: gof = g(f(x)) = g(4x + 3) = ((4𝑥 + 3) − 3)/4 = (4𝑥 + 3 − 3)/4 = 4𝑥/4 = x = IN Step 3: fog = f(g(y)) = f((𝑦 − 3)/4) = 4 ((𝑦 − 3)/4) + 3 = y – 3 + 3 = y + 0 = y = IY Since gof = IN and fog = IY, f is invertible & Inverse of f = g(y) = (𝒚 − 𝟑)/𝟒 Rough Checking inverse of f:X → Y Step 1: Calculate g: Y → X Step 2: Prove gof = IX Step 3: Prove fog = IY g is the inverse of f Example 17 (Method 2) Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse. f is invertible if f is one-one and onto Checking one-one f(x1) = 4x1 + 3 f(x2) = 4x2 + 3 Putting f(x1) = f(x2) 4x1 + 3 = 4x2 + 3 4x1 = 4x2 x1 = x2Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 If f(x1) = f(x2) , then x1 = x2 ∴ f is one-one Checking onto f(x) = 4x + 3 Let f(x) = y, where y ∈ Y y = 4x + 3 y – 3 = 4x 4x = y – 3 x = (𝑦 − 3)/4 Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((𝑦 − 3)/4) = 4((𝑦 − 3)/4) + 3 = y − 3 + 3 = y For every y in Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. There is a value of x which is a natural number such that f(x) = y Thus, f is onto Since f is one-one and onto f is invertible Finding inverse Inverse of x = 𝑓^(−1) (𝑦) = (𝑦 − 3)/4 ∴ Inverse of f = g(y) = (𝒚 − 𝟑)/𝟒 where g: Y → N

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.