Question 618052
p, q are consecutive numbers ---> write as p, p+1


i) q^2 < p
(p+1)^2 < p
p^2 + 2p + 1 < p
p^2 + p + 1 < 0
There are no solutions to this inequality.


ii) 2p > p
2 > 1
This is always true.


iii) (q+1)^2 > p^2
(p+2)^2 > p^2
p^2 + 4p + 4 > p^2
4p + 4 > 0
4p > -4
Natural numbers are positive, so this is always true.