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, nd 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, we need to understand the concept of positive divisors.

A positive divisor of a positive integer x is a positive integer that divides x without leaving a remainder. For example, the positive divisors of 12 are 1, 2, 3, 4, 6, and 12.

To find the values of m, n, and k, we need to identify positive integers that have the required number of divisors.

First, let's find m such that f(m) = 3. We are looking for a positive integer with exactly three divisors. The only positive integers that satisfy this condition are perfect squares. For example, 4, 9, 16, and so on. So, we can choose m = 4.

Next, let's find n such that f(n) = 4. We are looking for a positive integer with exactly four divisors. One such example is 6, which has the divisors 1, 2, 3, and 6. Therefore, we can choose n = 6.

Finally, let's find k such that f(k) = 5. We are looking for a positive integer with exactly five divisors. One such example is 16, which has the divisors 1, 2, 4, 8, and 16. Therefore, we can choose k = 16.

So, m = 4, n = 6, and k = 16 satisfy the given conditions.

Now, let's determine if f is one-to-one (injective) and onto (surjective).

To prove that f is one-to-one, we need to show that different inputs map to different outputs. If there exist distinct positive integers a and b such that f(a) = f(b), then f cannot be one-to-one.

Suppose f(a) = f(b). This means that the numbers of divisors of a and b are equal. However, different positive integers can have the same number of divisors. For example, f(4) = 3 and f(6) = 4. Therefore, f is not one-to-one.

To prove that f is onto, we need to show that every positive integer y has a preimage under f, i.e., there exists an x in the domain such that f(x) = y. If there exists a positive integer y that does not have a preimage, then f cannot be onto.

Since f maps positive integers to the number of their divisors, every positive integer y has a preimage. For example, if y is a prime number p, then f(p^(p-1)) = p, as p^(p-1) has exactly p divisors. Therefore, f is onto.

Finally, let's prove or disprove if f(ab) = f(a) * f(b) for all positive odd integers a and b.

Let a and b be positive odd integers. We need to show that f(ab) = f(a) * f(b).

Since a and b are odd, their prime factorizations only include odd prime numbers. Let p1, p2, ..., pn be the distinct prime factors of a and q1, q2, ..., qm be the distinct prime factors of b.

The prime factorization of ab is given by (p1 * p2 * ... * pn) * (q1 * q2 * ... * qm). This implies that the total number of divisors of ab is equal to the product of the number of divisors of a and b.

Using f(ab) = f(a) * f(b), we can conclude that f(ab) = f(a) * f(b) for all positive odd integers a and b.

Therefore, the answer is yes, it is true that for all positive odd integers a and b, f(ab) = f(a) * f(b).