Some proofs to be solved

Replies

• 

  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

  Good work !!
  Ronjan...
  
  
  
  Keep it up.
  
  
  
  
  Cheers
  Crazy😁
• 

  crook

  Good Job!
  
  Good job Ronjan ! 😁
  
  chavat.mulga please let us know if you have more questions. 😀