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Question
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.
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Solution
It is given that:
f: W → W is defined as f(n) = `{(n -1, "if n is odd"),(n+1, "if n is even"):}`
One-one:
Let f(n) = f(m).
It can be observed that if n is odd and m is even, then we will have n − 1 = m + 1.
⇒ n − m = 2
However, this is impossible.
Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.
∴Both n and m must be either odd or even.
Now, if both n and m are odd, then we have:
f(n) = f(m) ⇒ n − 1 = m − 1 ⇒ n = m
Again, if both n and m are even, then we have:
f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m
∴f is one-one.
It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any even number 2r in co-domain N is the image of 2r + 1 in domain N.
∴f is onto.
Hence, f is an invertible function.
Let us define g: W → W as:
g(m) = `{(m+1,"if m is even"),(m-1,"if m is odd"):}`
Now, when n is odd:
`gof (n) = g(f(n)) = g(n - 1) = n - 1 + 1 = n`
And, when n is even:
gof(n) = g(f(n)) = g(n+1) = n+1-1 = n
Similarly, when m is odd:
`fog(m) = f(g(m)) = f(m -1) = m - 1 + 1 = m`
When m is even:
`fog(m) = f(g(m)) = f(m+1) = m+ 1-1 = m`
`:. gof = I_W and fog = I_W`
Thus, f is invertible and the inverse of f is given by f—1 = g, which is the same as f.
Hence, the inverse of f is f itself.
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