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.

Hint: Let $x$ be the smallest natural number ($0$ or $1$, depending on the convention being used).
