#### Answer

PART A
Let (a,b) be a point on the graph. If the graph is origin-symmetrical, then (-a,-b) must also be a point on the graph.
1. By x-axis symmetry, (a, -b) is also a point on the graph.
2. If (a, -b) is a point on the graph, then by y-axis symmetry, (-a,-b) is a point on the graph.
3. If (-a,-b) is a point on the graph, then the graph is symmetrical with respect to the origin (by definition).
* the converse of the original statement is the statement, "If a graph is symmetrical about the origin, then it is symmetrical about the x- and y-axis".
counter-example. f(x) = y = x^3.
1. Take any point (x,y) on the graph.
2. f(-x) = (-x)^3 = -(x^3) = -y . . . (-x,-y) is a point on the graph
3. If (x,y) and (-x,-y) are both on the graph, then it is symmetrical about the origin.
4. But if you insert x = 2 and y = 8, you see that (-2,8) and (2,-8) are not on the graph, i.e. there is no x-axis or y-axis symmetry, repsectively, even though (2,8) is a point on the graph.
PART B
1. If the graph has origin-symmetry, then for point (a,b) on the graph, (-a,-b) is also on the graph
2. Take point (-a,-b): if there is x-axis symmetry, then (-a,b) is also on the graph.
3. If (-a,b) is on the graph, then there is symmetry about the y-axis (by defintion)
4. In step 2, you could also first assume y-axis symmetry. If (-a,-b) is on the graph, then by y-axis symmetry, so too is (a, -b). If (a, -b) is on the graph, then there is is x-axis symmetry.

#### Work Step by Step

The proof given above was formulated using the appropriate axioms and deductions.