Answer to Question #175371 in Real Analysis for Anand

Question

Prove that a strictly decreasing function is always one-one.

Answer 1

$Solution: \\Proof~1: \\For~ a ~strictly~ decreasing ~function~ if ~x_1<x_2 ~then ~ f(x_1)>f(x_2). \\Here ~no ~ two ~ values ~ of ~ domain ~ can ~ have ~ same ~values ~ in ~ codomain. \\ One ~may ~ be ~ less ~than ~ or ~ greater ~than ~ other ~but ~ not ~equal .So~ a ~strictly ~ \\ decreasing ~ function ~ is ~always one-one. \\Proof ~ 2: \\Given: f:R\rightarrow~R~ is ~strictly~ decreasing ~function. \\To ~ prove: f(x)~ is ~ one-one. \\proof: f~ is ~strictly ~ decreasing ~ implies: if~ x<y, ~then ~f(x)>f(y) \\Let ~ us ~ assume ~ that~ f(a)=f(b) \\If ~ a<b, then ~by ~ the ~ definition ~ of ~ strictly~ increasing ~f(a)>f(b). \\Thus~ it ~ is ~ not ~possible ~that ~a<b~ when ~ f(a)=f(b). \\If ~ b<a, then ~by ~ the ~ definition ~ of ~ strictly~ increasing ~f(b)>f(a). \\Thus~ it ~ is ~ not ~possible ~that ~b<a~ when ~ f(a)=f(b). \\Since ~ a<b ~ is ~ not ~true ~ and ~ since ~ b<a ~ is ~ not ~ true ,a ~and~ b ~then ~ have ~ to ~be ~\\ the ~ same~: \\~~~~~~~~~~~~~~~~~~~~~~~~~~a=b \\By~ definition ~ of ~one-to-one~ function, ~we~ have ~ then ~ shown ~ that ~f~is ~ \\one-to-one .$