SOLUTION: Sir/Mam, Respectfully,Kindly tell me how can i solve my maths problem.My question is given Below .Please try to solve my problem soon because my examination is near. The question

Algebra ->  Signed-numbers -> SOLUTION: Sir/Mam, Respectfully,Kindly tell me how can i solve my maths problem.My question is given Below .Please try to solve my problem soon because my examination is near. The question       Log On


   

Question 618052: Sir/Mam,
Respectfully,Kindly tell me how can i solve my maths problem.My question is given Below .Please try to solve my problem soon because my examination is near. The question are-" If p and q consecutive natural numbers (in increasing order), then which of the following is true?
The options are-
(i) q^2 < p
(ii) 2p > p
(iii) (q+1)^2 > p^2 Thanking you.
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
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.