Question

1. Consider the function h:Z+ +Z+ defined by h(n) = l{k e Z+ : k|n}l. The bars around the set mean that we are taking the size of the set. Thus h(n) is the number of positive divisors of n. (a) Make a table of values for h(n) for 1 sn < 10. Write one or two sentences describing how you found the values in the table. (b) Find the value of h(90). Explain how you found your answer. (c) Find the value of h(2). The answer will be a simple formula involving m. (d) Is h a one-to-one function? Why or why not? (e) Is h an onto function? Why or why not? (f) Describe the set h+(2). Explain your reasoning.

Answer 1

(a) Obviously $h(1)=1$ and for a prime number ,
f(p) = 2
since its only divisors are and itself.

1	2	3	4	5	6	7	8	9	10
1	2	2	3	2	4	2	4	3	4

For the other values one checks it manually,

(b) First note that
90 = 2 x 32 x 5
. All the divisors of are of the form $2^{\varepsilon_2} \times 3^{\varepsilon_3} \times 5^{\varepsilon_5}$ for $\varepsilon_2 \in \{ 0,1\}$ , $\varepsilon_3 \in \{ 0,1,2\}$ and $\varepsilon_5 \in \{ 0,1\}$ . So, the positive divisors of are in correspondence with how many combinations of exponents we can choose. In this case, $h(90) =(1+1)\times (2+1)\times (1+1) = 12$ .

(c) All the divisors of $2^m$ are $2^0 , 2^1 ,\dots, 2^m$ . Therefore, .

(d) The function is not one-to-one. As we saw in (a), . That is, two different values have the same output.

(e) The function is onto. Indeed, for $n \in \mathbb Z^+$ we have as we saw in (c).

(f) The set $h^{\leftarrow} (2)$ is the set of prime numbers. Note that for we have that $h(n)\geqslant 2$ since and are always divisors. If it means that does not have any other divisor, hence it is prime.