 I am welcome to the session. Let us discuss the following question. The question says, where function f from x to y be an invertible function? Show that f has unique inverse. First of all, let us understand that if we are given a function f from a to b, such that it is invertible, then there exists a function g from b to a, such that g of f is equal to identity function on set a, and f of g is equal to identity function on set b, then j is called the inverse of f and is denoted by f inverse. This is the key idea to solve the given question. Let us now start the solution. We are given that function f from x to y is an invertible function. Now since f is an invertible function, so there exists a function g from y to x, such that g of f is equal to identity function on x for all x belonging to set x, f of g y is equal to identity function on y for all y belonging to set y. Also, f is one-one and all-two functions, since it is invertible. So, it must be one-one and all-two functions. So, we can write function f one-one. Let us assume that g-one and g-two are two inverses of f. f is the function f, and for all y belonging to set y, f for g and y is equal to identity function on y is equal to f for g-two y. We know function f is one-one and all-two functions, so the identity function of y is unique for g-one function f, which is possible only when g-one y is equal to g-two y. So, therefore, we get, it implies, function f has unique inverse. This completes the session. Hope you understood the session. Take care and goodbye.