Let f:ℤ+ → ℤ+ be the function defined by: for each x ∈ ℤ+, f(x) is the number of positive divisors of x.

- find integers m, n, and k so tha f(m)=3, f(n) = 4 anf f(k) = 5
-is f one-to one? Explain, is f unto? prove
is it true that all positive odd integers a and b, f(ab) = f(a) f(b)? prove answer

To find integers m, n, and k such that f(m) = 3, f(n) = 4, and f(k) = 5:

1. Start with the given function f(x):ℤ+ → ℤ+, where f(x) represents the number of positive divisors of x.
2. For f(m) = 3, we need to find an integer m whose number of positive divisors is 3.
- One possible solution is m = 2^2 = 4, as the positive divisors of 4 are 1, 2, and 4.
- Another solution is m = 3^2 = 9, as the positive divisors of 9 are 1, 3, and 9.

3. For f(n) = 4, we need to find an integer n whose number of positive divisors is 4.
- One possible solution is n = 2^3 = 8, as the positive divisors of 8 are 1, 2, 4, and 8.
- Another solution is n = 3^1 = 3, as the positive divisors of 3 are 1 and 3.

4. For f(k) = 5, we need to find an integer k whose number of positive divisors is 5.
- One possible solution is k = 2^4 = 16, as the positive divisors of 16 are 1, 2, 4, 8, and 16.
- Another solution is k = 5^1 = 5, as the positive divisors of 5 are 1 and 5.

So, one possible set of integers that satisfy f(m) = 3, f(n) = 4, and f(k) = 5 is m = 4, n = 8, and k = 16.

To determine if the function f(x) is one-to-one and onto:

1. One-to-one: A function is one-to-one if every input maps to a unique output.
- To check if f(x) is one-to-one, we need to find different integers x and y such that f(x) = f(y).
- If we find such integers, then f(x) is not one-to-one.

2. Onto: A function is onto if every element in the codomain has a corresponding preimage in the domain.
- To prove f(x) is onto, we need to show that for every integer y in the codomain, there exists an integer x in the domain such that f(x) = y.

To determine if all positive odd integers a and b satisfy f(ab) = f(a)f(b):

1. Let's take two positive odd integers, a and b.
2. The prime factorization of any odd number will only contain odd prime numbers.
- Let's assume a = p1^a1 * p2^a2 * ... * pn^an and b = q1^b1 * q2^b2 * ... * qm^bm, where p and q are odd primes.
3. The number of positive divisors of ab can be calculated by multiplying the exponents of each prime factor of a and b incremented by 1.
- f(ab) = (a1 + 1)(a2 + 1)...(an + 1) * (b1 + 1)(b2 + 1)...(bm + 1)
4. The number of positive divisors of a can be calculated as f(a) = (a1 + 1)(a2 + 1)...(an + 1).
5. Similarly, the number of positive divisors of b can be calculated as f(b) = (b1 + 1)(b2 + 1)...(bm + 1).

Therefore, f(ab) = (a1 + 1)(a2 + 1)...(an + 1) * (b1 + 1)(b2 + 1)...(bm + 1) = f(a) * f(b).

Hence, it is true that for all positive odd integers a and b, f(ab) = f(a)f(b).