A set with property p satisfies that there is no 0 and reciprocal in the set.

Therefore, the set {- 1, 2,3} satisfies the property p.

Then:

When a∈A, b∈A, (a+b)∈A, a=- 1, b=3 or a=3, b=- 1.

S={(- 1，3)，(3，- 1)}

When a∈A, b∈A, (a-b)∈A, a=2, b=- 1 or a=2, b=-3.

T={(2，- 1)，(2，3)}

The second question:

Let there be k elements in set A, then there are at most k 2 elements in set T, that is, n ≤ k 2 (when any two elements A and B in a can satisfy A ∈ a, b ∈ A, (A-B) ∈ A, n = k 2).

When set a satisfies property p:

ai≠0，ai+aj≠0

Therefore:

Set t does not contain (ai, ai) (ai-ai = 0,0? A), such a number pair is * * * K.

The number pair (ai, aj) and the number pair (aj, ai) in the set T do not exist at the same time (if they exist at the same time, (ai-aj)∈A and (aj-ai)∈A, (ai-aj)+(aj-ai)=0, then the set A does not satisfy the property P.

So:

n≤(k^2-k)/2

Namely:

n≤k(k- 1)/2

The third question:

( 1)

When (a, b)∈S, (b, a)∈S, (a+b, b)∈T, (a+b, a)∈T((a+b)∈A).

When pairs (a, b) and (c, d) belong to S, a=c and b=d are not established at the same time, so a+b=c+d and b=d are not established at the same time. Therefore, when pairs (a, b), (c, d) all belong to S, pairs (b, a), (a).

At this time, m≤n (at this time, it is only proved that any ordered number pair in S can be found in T)

(2)

When (a, b)∈T, (a-b, a)∈S((a-b)∈A)

When number pairs (a, b) and (c, d) all belong to T, a=c and b=d are not established at the same time, so a-b=c-d and a=c are not established at the same time, so when number pairs (a, b) and (c, d) all belong to S, number pairs (a-b, a)

At this time, n≤m (at this time, it is only proved that any ordered number pair in T can be found in S).

From (1) and (2):

m=n

note:

1. Parentheses are comments, not a problem-solving process.

2. It is the same as the solution of test paper analysis, but the process is more detailed. If there is anything you don't understand, please ask.