Answer to Question #186454 in Real Analysis for Harshita

Question

Prove that a strictly decreasing function is always one-one

Answer 1

Solution :-

Given :- f(x) is strictly decreasing function

To prove :- f(x) is one-one

Proof:-

f is strictly decreasing implies : if x<y , than f(x) >f(y)

let us assume 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 also 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 same

a=b

By definition of one-one , we have now shown that f is one - one function.