>>10892095Cases
1. p > 0 & k > 0 then
p + k > p + k, which isn't possible
2. p> 0 & k < 0
p - k > p + k
0 > 2k
k < 0 so p > k
3. p < 0 & k> 0
p+k > -p + k
2p > 0 which contradicts case
4. p < 0 & k < 0
p-k > -p+k
2p > 2k
p > k
5. k = 0 & p > 0
p > p which isn't possible
6. k = 0 & p < 0
p> -p
2p > 0 which contradicts case
7. p=0 & k > 0
k> k which is false
8. p = 0& k < 0
-k > k
0 > 2k
k < 0 so p > k
