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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.

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    The set $\{y+1 \colon y \in N\}$ is the set of all numbers in $N$ that are _successors_ of the natural numbers in $N$. Let $y_0$ denote the smallest natural number. Then, the question asks whether there is an $x \in N$ such that $2x \leq y_0 + 1$2012-09-20

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