Some proofs to be solved
Hey Ppl,
I got some very tricky problems to be solved...i need to submit it on friday to my prof..but i already have 2 tests this week..so i am not able to find time to solve this proofs...if any one can solve them..it wud be gr8.
here we go..
Prove/Disprove the following proposition
1. x is real, (x sqr)>(x)
2. x is real, (xsqr)>(x+1)
3.n is natural no, (n cuben) is divisible by 3.
4. x is real , (x cube) > (x sqr)
5.x,y, real, x sqr+ y sqr=9.
6.n is natural no, (n cube +2n) is divisible by 3.
7. N is natural, 2(n+2) (less or equal to)(n+2) sqr.
do reply if u know the proof..
I got some very tricky problems to be solved...i need to submit it on friday to my prof..but i already have 2 tests this week..so i am not able to find time to solve this proofs...if any one can solve them..it wud be gr8.
here we go..
Prove/Disprove the following proposition
1. x is real, (x sqr)>(x)
2. x is real, (xsqr)>(x+1)
3.n is natural no, (n cuben) is divisible by 3.
4. x is real , (x cube) > (x sqr)
5.x,y, real, x sqr+ y sqr=9.
6.n is natural no, (n cube +2n) is divisible by 3.
7. N is natural, 2(n+2) (less or equal to)(n+2) sqr.
do reply if u know the proof..
Replies

ronjanMight sound dumb...but...
Two questions :
1. (x) means fractional part of x, or simply x ?
2. If x means simply x,
a)x^2  x >0 => x(x1) > 0, means x >0 and x>1 =>x>1 or x<0 and x<1 => x<0
so for 0<=x<=1, x^2 >x doesn't hold
b)x^2x1>0 => (x1/2)^2 3/2 >0 disprove it as done in part a.
c) Any n>1 can be written as 3k  1 or 3k or 3k+1.
For n = 1, n^3n = 0, which is trivial to the proof..
Now, n^3n = n(n^21)=n(n+1)(n1) Now, put n = 3k1/3k/3k+1 u will get one 3k out of the equation, which means n^3n is divisible by 3 for all n. Proved.
d) x^3x^2 >0 =>x^2(x1)>0 Now, x^2>0 for all x<>0. So, x^3>x^2 for all x>1. Hence, disproved.
e) Put x>9, y>9 obv. x^2+y^2>9 , hence disproved.
f) Again for n = 1, it is 3 hence divisible,
for n >3, n^3+2n = n(n^2+2) . hence if n = 3k it is divisble by 3,
if n = 3k1 , n^3+2n = (3k1)(9k^26k+3) = 3(3k1)(3k^22k+1) divisible by 3...prove similarly for 3k+1, hence proved.
g) 2(n+2)<=(n+2)^2
=> [(n+2)1]^2>=1
=>[n+1]^2>=1 which is true for all n>=1, hence proved. 
Anil JainGood work !!
Ronjan...
Keep it up.
Cheers
Crazyπ 
crookGood Job!
Good job Ronjan ! π
chavat.mulga please let us know if you have more questions. π
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