Give am example of a function f.N N with the property that there exists a function g. N N such that the composition got is the identity function on N but for no function h. N N does it hold that fog is the identity function on N.Prove all claims
Sure! Let's define the function f : N -> N as follows:
f(n) = n + 1
Now, let's define another function g : N -> N as follows:
g(n) = n - 1
To prove that the composition gof is the identity function on N, we need to show that (gof)(n) = n for all n in N.
So, let's compute (gof)(n):
(gof)(n) = g(f(n))
= g(n + 1)
= (n + 1) - 1
= n
Therefore, (gof)(n) = n, which means that gof is indeed the identity function on N.
Now, let's prove the claim that for no function h : N -> N, does it hold that fog is the identity function on N.
Suppose h : N -> N is a function such that fog is the identity function on N. This implies that h(f(n)) = n for all n in N.
Consider the case when n = 0. We have:
h(f(0)) = 0
h(0 + 1) = 0
h(1) = 0
Now, consider the case when n = 1. We have:
h(f(1)) = 1
h(1 + 1) = 1
h(2) = 1
However, we already know from the previous case that h(1) = 0. This leads to a contradiction, since a function cannot assign two different values to the same input.
Therefore, there is no function h : N -> N that satisfies the condition fog = identity function on N.
Hence, we have found a function f : N -> N and a function g : N -> N such that fog is the identity function on N, but there is no function h : N -> N that satisfies the condition fog = identity function on N for all n in N.
Let's illustrate the example of a function that meets the given conditions. Consider the function f: N -> N defined as f(x) = x + 1. This function takes a natural number x and returns the next natural number.
Now, suppose there exists a function g: N -> N such that (g o f)(x) = x for all x in N, i.e., the composition of g and f is the identity function on N.
To prove that such a g function does not exist, let's assume that there exists a function g: N -> N with (g o f)(x) = x. We will reach a contradiction.
Let's consider any natural number n. Since (g o f)(x) = x for all x in N, it must hold true for n as well. Therefore, (g o f)(n) = n.
Now, plugging in the definitions of f and g: (g o f)(n) = g(f(n)) = g(n + 1).
Since g(f(n)) = n holds for all n in N, it means that g(n + 1) = n.
However, this implies that for every natural number n, g(n + 1) is equal to the previous natural number n, which is clearly contradictory as it violates the injectivity (one-to-one property) of g.
Hence, there is no function g: N -> N such that (g o f)(x) = x for all x in N.
To summarize:
- The function f: N -> N is defined as f(x) = x + 1.
- The claim states that there exists a function g: N -> N such that (g o f)(x) = x for all x in N.
- However, it has been proven that no function g: N -> N satisfies this condition, as it leads to a contradiction.
Therefore, no such function g exists for the given function f.