Floor function equation and proving a polynomial root exist

[matrix_204]

I am having some difficulty in understanding this problem.
1. Prove that there is a real x, s.t x-|_ x _|=2 or that there isn't.
(|_ x _|= m (element of) Z and m<=x<m+1.)
Read as, "the floor function of x is equal to m, element of Z(integers), and m is less than or equal to x, which is less than m+1.

How am i suppose to solve this problem, and what kind of definitions or facts can i use to solve this problem? Could someone tell me the steps that are required to solve this problem?

And this is the second problem.
2. Show that there is a real x s.t. f(x)=x^5+x+1=0. f(x) is continuous because it is a polynomial. Let s be a real s.t. f(-s)<0<f(s). Then by I.V.T.(intermediate value theorem) there is at least one x in (0-s,0+s) s.t. f(x)=0 if f(-s)<0<f(s).

Could someone just clarify this for me, because i think i am able to do it, just don't understand on what to do first. Thanks in advance.

 

[vsage]

LaTeX helps so much with these sorts of problems.

1. Is part of the text proposing solutions as well? subtracting m from each part of the inequality you have fetches 0<= x-m < 1 which shows that x - floor(x) = 2 doesn't have a solution.

 

[matrix_204]

Well not really, no solutions are given. I just don't understand how the floor functions work. And like you said for 1. it doesn't have a solution, then what do i have to say in the proof to prove that it has no solution.

 

[matrix_204]

Anyone know how to do the second problem, urgent help needed.

 

[Inquisitive_Mind]

Hint for Q2: what's f(1) and f(-1)?