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 i am not able to find time to solve this proofs...if any one can solve 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 cube-n) 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..


  • ronjan
    Might 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(x-1) > 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^2-x-1>0 => (x-1/2)^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^3-n = 0, which is trivial to the proof..

    Now, n^3-n = n(n^2-1)=n(n+1)(n-1) Now, put n = 3k-1/3k/3k+1 u will get one 3k out of the equation, which means n^3-n is divisible by 3 for all n. Proved.

    d) x^3-x^2 >0 =>x^2(x-1)>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 = 3k-1 , n^3+2n = (3k-1)(9k^2-6k+3) = 3(3k-1)(3k^2-2k+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 Jain
    Anil Jain
    Good work !!

    Keep it up.

  • crook
    Good Job!

    Good job Ronjan ! 😁

    chavat.mulga please let us know if you have more questions. πŸ˜€

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