Existence of a natural number X

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SUMMARY

The discussion centers on proving the existence of a natural number X such that for all natural numbers n greater than or equal to X, the inequality n² + n + 1 > M holds true, where M is also a natural number. The initial approach suggests setting X equal to M, leading to the conclusion that n² + n + 1 will indeed exceed M for n ≥ X. However, participants express concerns about the completeness of this proof and whether X should be defined more restrictively to ensure the validity of the argument.

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Homework Statement


Given [itex]M \in N[/itex], show that there exists an [itex]X \in N[/itex] such that for all [itex]n \geq X[/itex], [itex]n^2+n+1 \succ M[/itex]



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The Attempt at a Solution


Since both M and X are natural numbers and I am just trying to prove the existence of a certain natural number X, I thought that i could just set X = M.
Then, n^2+n+1 [itex]\geq[/itex] X^2+X+1 since n [itex]\geq[/itex] X.
And X^2+X+1=M^2+M+1[itex]\succ[/itex]M.
So, n^2+n+1[itex]\succ[/itex]M.
Is this a sufficient proof for the existence of X?
It just doesn't feel like a full proof, should X be more limited?
 
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If that is the correct statement of the problem, I don't see anything wrong with your argument. But I have an uneasy feeling like you do. Since it seems so trivial I wonder if the original problem is mis-copied or misunderstood.
 

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