Existence of a natural number X

  • Thread starter h.shin
  • Start date
  • #1
h.shin
7
0

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]



Homework Equations





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?
 

Answers and Replies

  • #2
LCKurtz
Science Advisor
Homework Helper
Insights Author
Gold Member
9,568
774
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.
 

Suggested for: Existence of a natural number X

Replies
9
Views
388
Replies
4
Views
11
Replies
29
Views
641
Replies
5
Views
436
Replies
4
Views
148
Replies
5
Views
352
Replies
4
Views
411
Replies
12
Views
385
Replies
5
Views
756
Top