Making an open statement to satisfy two conditions help

[MtX]

Hello guys,

I have a question regarding mathematical logic that I'm stuck on. Here it is:

T = { (5,9), (4,9), (5,7), (6,5), (5,5), (6,3), (7,1), (6,1), (5,1), (4,1), (3,1), (2,1), (1,1), (0,1) }
F = { (5,4), (6,8 ), (2,11), (4,13), (8,1), (1,0) }

1) Make a simple open statement P(x,y) so (x,y) in T -> P(x,y) and (x,y) in F -> !P(x,y). Use only domain N, comparison operators (<, =, >), operations (+, -) and logical notation and don't use T or F in P.

2) Find an example (x,y) !in T so that P(x,y) and an example (x,y) !in F so that !P(x,y).

My thoughts:

1) I can't think of any general equation or formula so that T is true and F is false, but using cases I may be able to find something. Don't think I can use cases though because there's just too many... Next thing I did was look for patterns but I can't seem to find anything different from T and F. For T, 2x+y <= 19 and x+y <= 14 and -5 <= x-y <= 6 for all sets in T, but when we look at the sets in F, some of those sets satisfy the equations from T.. basically, NOT all of the sets in F are false, some are true.. what can i do to ensure all sets in T are true and all sets in F are false?

2) ?

 

[TenaliRaman]

My 2 cents :
1>
find the equation of the line that passes through all the points in T but not through any point in F.Let this function be f(x,y)
P(x,y) : (x,y) is a point in f(x,y)

-- AI

 

[BicycleTree]

Well, I don't know how simple your statement has to be but you could separate the points by regions (graphing them would help). You'd probably have to use five or six lines.

 

[MtX]

im not sure what you mean by graphing them.. i put all the coordinates on a x/y graph, connected them with a line..

 

[MtX]

after connecting all the sets of coordinates of T with a line, the line isn't even straight.. can't find the slope..