Consider `f : {1, 2, 3}->{a , b , c}`given by `f(1) = a`, `f(2) = b`and `f(3) = c`. Find `f^(-1)`and show that `(f^(-1))^(-1)= f`.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
`f: {1, 2, 3} to {a, b, c}`
and `f(1) = a, f(2) = b and f(3) = c `
Let, in `g: {a, b, c} to { 1, 2, 3}, g(a)=1, g(b) = 2 and g(c ) = 3`
`therefore ” ” (fog) (a) = f{g(a)} = f(1) =a`
`” ” (fog) (b) = f{g(b)} = f(2) = b`
`” ” (fog) (c ) = f{g(c)} = f(3) = c`
and `” ” (gof) (1) = g{f(1)} = g(a) =1 `
`” ” (gof)(2) = g{f(2)}= g(b) = 2`
`” “(gof) (3) = g{f(3)} = g(c) = 3`
`therefore ” ” gof = I_X and fog = I_Y`
where `X = { 1, 2, 3} and Y = {a, b, c}`
`therefore ” ” f^(-1) = g `
and `f^(-1) : {a, b, c} to {1, 2, 3 }`
and `f^(-1)(a) = 1, f^(-1) (b) = 2, f^(-1) (c) = 3`
Again let, in `h: {1, 2, 3} to {a, b, c}`
`” ” h(1) = a, h(2) = b, h(3) = c`
`” ” (goh) (1) = g{h(1)} = g(a)= 1`
`” ” (goh)(2) = g{h(2)} = g(b) = 2 `
`” ” (goh)(3) = g{h(3)}= g(c) = 3`
and `” ” (hog)(a) = h{g(a)} = h(1) =a`
`” ” (hog )(b) = h{g(b)} = h(2) = b`
`” ” (hog) (c) = h{g(c)} = h(3) =c `
`therefore ” ” goh = I_X and hog = I_Y`
where `” ” X= {1, 2, 3} and y = {a, b, c}`
`thererfore ” “g^(-1) = h`
`rArr ” ” (f^(-1)) = f`