Consider the relation R on P dened by:R = (x; y) | y = 2^i * x for some i 2 N:(a) Prove that R is a partial order on N.(b) Identify the minimal elements of R.(c) Let A = {1; 2; 3; 5; 6; 12}. Draw the Hasse diagram for the poset dened by this relation

(a) To prove that R is a partial order on N, we need to show that it is reflexive, antisymmetric, and transitive.

Reflexive: For all x in N, we can choose i=0 (since 2^0 = 1), so x = 2^0 * x. Therefore, (x, x) is in R for all x in N.

Antisymmetric: Suppose (x, y) and (y, x) are in R. Then by definition of R, y = 2^i * x and x = 2^j * y for some i, j in N. Substituting the first equation into the second, we get x = 2^j * 2^i * x = 2^(i+j) * x. Since x ≠ 0, we can divide both sides by x to get 1 = 2^(i+j). The only way this can happen is if i + j = 0, which implies i = j = 0. Therefore, x = y.

Transitive: Suppose (x, y) and (y, z) are in R. Then by definition of R, y = 2^i * x and z = 2^j * y for some i, j in N. Substituting the first equation into the second, we get z = 2^j * 2^i * x = 2^(i+j) * x. Therefore, (x, z) is in R.

(b) The minimal elements of R are the elements x such that there is no y ≠ x with (y, x) in R. In other words, these are the elements x such that there is no y ≠ x with x = 2^i * y for some i in N. The only such elements are the powers of 2, since any other natural number can be written as 2^i * y with y ≠ 1.

(c) The Hasse diagram for the poset defined by this relation on the set A = {1, 2, 3, 5, 6, 12} is a diagram where each element of A is represented by a point, and there is a line from x to y if and only if (x, y) is in R and there is no z such that (x, z) and (z, y) are in R. The points are arranged so that if (x, y) is in R, then y is higher than x in the diagram. The diagram will look like this:

12

/
6 2 / \ /
5 3 1

This diagram shows that 1 is related to 2, 2 is related to 12, 3 is related to 6, and 6 is related to 12, with 5 being unrelated to any other elements.