Claim: The function f(x) = x + 1 from the set of integers to itself is onto.

Proof:

To prove that the function is onto, we need to show that for every integer y in the set of integers, there exists an integer x such that f(x) = y.

Let y be an arbitrary integer in the set of integers.

Existence of an integer x:
Since y is an integer, we can subtract 1 from y to obtain an integer x = y - 1.

Verification:
Now, let's verify whether f(x) = y.

f(x) = f(y - 1) = (y - 1) + 1 = y

Since f(x) = y, we have shown that for every integer y in the set of integers, there exists an integer x such that f(x) = y.

Therefore, the function f(x) = x + 1 is onto.

Explanation:

To determine whether a function is onto, we need to show that for every element in the codomain, there exists at least one element in the domain that maps to it.

In this case, the function f(x) = x + 1 maps each integer x to x + 1. So, for any given integer y in the codomain, we can find an integer x such that f(x) = y.

For example, if y = 5, we can choose x = 4. Then f(x) = f(4) = 4 + 1 = 5, which matches the value of y.

Similarly, for any other integer y, we can always find an integer x such that f(x) = y. This shows that the function is onto.

Conclusion:

The function f(x) = x + 1 from the set of integers to itself is onto, as we have shown that for every integer y in the set of integers, there exists an integer x such that f(x) = y. Therefore, the correct answer is option 'A' - True.