Prove or Disprove: ∃x ∈ N such that ∀y ∈ N, 2x ≤y + 1 The first thing that I tried to do is: let y be an arbitrary natural number. I then tried to choose a value for x, but I cannot think of a value in which 2x ≤ y + 1.. 
So I then tried to prove the negation: ∀x ∈ N, ∃y ∈ N, 2x > y + 1
So I then let x be an arbitrary natural number and tried to set a value for y .... but the smallest value that y can be is x, correct? Because anything less than x and y would not be a natural number all of the time. So I do not know how to solve this, because I can't think of anything that works.

Translate the statement into simple English:

Is there a natural number (call it $x$), such that for every other natural number $y$, the number $y + 1$ is at least twice as large as $x$?

And try rephrasing/simplifying that statement:

Is there a natural number $x$ which is less than or equal to $(y+1)/2$, for every natural number $y$?

(The translation into simple English is meant to help you with the quantifiers, here — for the algebra of dividing by two, using math rather than English will probably be less cumbersome.)
Because we're talking about a number which is smaller than or equal to some expression $f(y)$, where $f(y)$ grows with $y$, your best approach is to consider the smallest allowed $y$, and see if any values of $x$ work.
