If f : Q → Q, g : Q → Q are two functions defined by f(x) = 2x and g(x) = x + 2, show that f and g are bijective maps. Verify that (gof)-1 = f-1og-1.
We have f : Q → Q and f(x) = 2x.
Recall that a function is a bijection only if it is both one-one and onto.
First, we will prove that f is one-one.
Let x1, x2ϵ Q (domain) such that f(x1) = f(x2)
⇒ 2x1 = 2x2
∴ x1 = x2
So, we have f(x1) = f(x2) ⇒ x1 = x2.
Thus, function f is one-one.
Now, we will prove that f is onto.
Let y ϵ Q (co-domain) such that f(x) = y
⇒ 2x = y
Clearly, for every y ϵ Q, there exists x ϵ Q (domain) such that f(x) = y and hence, function f is onto.
Thus, the function f is a bijection and has an inverse.
We have f(x) = y ⇒ x = f-1(y)
But, we found f(x) = y ⇒
Hence,
Thus,
Now, we have g : Q → Q and g(x) = x + 2.
First, we will prove that g is one-one.
Let x1, x2ϵ Q (domain) such that g(x1) = g(x2)
⇒ x1 + 2 = x2 + 2
∴ x1 = x2
So, we have g(x1) = g(x2) ⇒ x1 = x2.
Thus, function g is one-one.
Now, we will prove that g is onto.
Let y ϵ Q (co-domain) such that g(x) = y
⇒ x + 2 = y
∴ x = y – 2
Clearly, for every y ϵ Q, there exists x ϵ Q (domain) such that g(x) = y and hence, function g is onto.
Thus, the function g is a bijection and has an inverse.
We have g(x) = y ⇒ x = g-1(y)
But, we found g(x) = y ⇒ x = y – 2
Hence, g-1(y) = y – 2
Thus, g-1(x) = x – 2
We have (f-1og-1)(x) = f-1(g-1(x))
We found and g-1(x) = x – 2
⇒ (f-1og-1)(x) = f-1(x – 2)
We know (gof)(x) = g(f(x)) and gof : Q → Q
⇒ (gof)(x) = g(2x)
∴ (gof)(x) = 2x + 2
Clearly, gof is a bijection and has an inverse.
Let y ϵ Q (co-domain) such that (gof)(x) = y
⇒ 2x + 2 = y
⇒ 2x = y – 2
We have (gof)(x) = y ⇒ x = (gof)-1(y)
But, we found (gof)(x) = y ⇒
Hence,
Thus,
So, it is verified that (gof)-1 = f-1og-1.