Example 24 - Chapter 1 Class 12 Relation and Functions

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Example 24 (Method 1) Let Y = {n2 : n ∈ N} ⊂ N. Consider f : N → Y as f (n) = n2. Show that f is invertible. Find the inverse of f f(n) = n2 Step 1 Put f(n) = y y = n2 n2 = y n = ± ﷮𝑦﷯ Since f : N → Y, n ∈ N, So, n is positive ∴ n = ﷮𝑦﷯ Let g(y) = ﷮𝑦﷯ where g: Y → N Now, f(x) = n2 & g(y) = ﷮𝑦﷯ Step 2: gof = g(f(n)) = g(n2) = ﷮(𝑛2)﷯ = 𝑛﷮ 1﷮2﷯ × 2﷯ = 𝑛﷮1﷯ = n Hence, gof = n = IN Step 3: fog = f(g(y)) = f( ﷮𝑦﷯ ) = ( ﷮𝑦﷯)2 = 𝑦﷮ 1﷮2﷯ × 2﷯ = 𝑦﷮1﷯ = y Hence, fog(y) = y = IY Since gof = IN and fog = IY, f is invertible & Inverse of f = g(y) = ﷮𝒚﷯ Example 24 (Method 2) Let Y = {n2 : n ∈ N} ⊂ N. Consider f : N → Y as f (n) = n2. Show that f is invertible. Find the inverse of f f(n) = n2 f is invertible if it is one-one and onto Check one-one f(n1) = n12 f(n2) = n22 Put f(n1) = f(n2) n12 = n22 ⇒ n1 = n2 & n1 = – n2 As n ∈ N, it is positive So, n1 ≠ – n2 ∴ n1 = n2 So, if f(n1) = f(n2) , then n1 = n2 ∴ f is one-one Check onto f(n) = n2 Let f(x) = y , where y ∈ Y y = n2 n2 = y n = ± ﷮𝑦﷯ Since f : N → Y, n ∈ N, So, n is positive ∴ n = ﷮𝑦﷯ For all values of y, y ∈ Y, n is a natural number i.e. n ∈ N So, f is onto Finding inverse f(n) = n2 For finding inverse, we put f(n) = y and find n in terms of y We have done that while proving onto n = ﷮𝑦﷯ Let g(y) = ﷮𝑦﷯ where g: N → Y ∴ Inverse of f = g(y) = ﷮𝒚﷯