Question

Let f:N×N→N be the function given by: for every pair of natural numbers (n,m)∈N×N, f(n,m)=2ⁿ⁻¹ (2m−1). What can you safely conclude?

Choose all that apply.
a. f(3,4)=28
b. f is injective.
c. f fails to be fully-defined.
d. If f(n,m) is odd, then necessarily n=1.

Answer 1

The function f(n,m)=2^n-1(2m-1) is fully-defined and injective. Statement a, f(3,4)=28, is true and so is statement d, if f(n,m) is odd, then n=1 must be true for that to happen.

Step-by-step explanation:

The function f:N × N → N is defined as f(n,m) = 2n−1(2m−1) for every pair of natural numbers (n,m). Using this definition, let's consider the statements:

• a. f(3,4)=28: By plugging in the values, f(3,4)=23-1(2×4-1)=4×7=28, so this statement is true.
• b. f is injective: To determine if f is injective, we must check if every two distinct pairs (n,m) produce distinct outcomes. Since increasing n or m will produce a greater value of f(n, m), the function is injective.
• c. f fails to be fully-defined: The function is given for all pairs of natural numbers, so it is fully-defined.
• d. If f(n,m) is odd, then necessarily n=1: Since 2n-1 is a power of 2, it will always be even unless n=1, as 20 is 1 which is odd. Therefore, this statement is true.