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.

Good work !!
Ronjan...



Keep it up.




Cheers
Crazy😁

Good Job!

Good job Ronjan ! 😁

chavat.mulga please let us know if you have more questions. 😀