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Can someone please help me with this proof:

Let $f, g : \mathbb N → \mathbb Z^{+}.$ We say that $g$ is eventually dominated by $f$ if and only if there exists $n_{0} \in \mathbb Z^{+}$ such that every natural number $n$ greater than or equal to $n_{0}$ satisfies $g(n) ≤ f(n).$ We can express this in a predicate $P(f,g)$: “$g$ is eventually dominated by $f$" where $f, g : \mathbb N → \mathbb Z^{+}$ as $$P(f, g): \exists n_{0} \in \mathbb Z^{+}, \forall n \in \mathbb N, n ≥ n_{0} \implies g(n) ≤ f(n) $$

(a) Let $f(n) = n^2$ and $g(n) = n + 165$. Prove that $g$ is eventually dominated by $f$.

Now I know that the order of quantifiers is important for this part and that we start by introducing $n_{0}$ by giving it a concrete value. I have no idea where to go from there.

(b) Prove the following statement, which is a generalization of the previous part: $ \forall a, b \in \mathbb Z^{+}, g(n) = an + b $ is eventually dominated by $ f(n) = n^2$.

I'm assuming here we need to pick a suitable $n_{0}$ that other variables can easily depend on.

1 Answers 1

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You just need to find a $n_0 \in \mathbb N$ such that $f(n) \ge g(n), \forall n \ge n_0$................................................1)

In part b), let us find $n_0$. Let us assume such an $n_0$ satisfying 1) does NOT exist. Then $\forall n \in \mathbb N, f(n)\lt g(n)$

or, $n^2 < an + b$............................2)

or, $n^2 - an - b < 0$ [ since for 2) to be correct the RHS of the inequality must be positive, we can switch it to the other side without changing the inequality sign].

or, $n(n-a) - b < 0$ .............................3)

Now, Note that $(a+b) \ge 1$ [since $g(1) \in \mathbb Z^+$]

choose n = (a+b).

Then , from 3):

$(a+b).b - b<0$ , or $b(a+b - 1)<0$ . Clearly, the last one is impossible since both the factors are positive.Contradiction.

Hence, there always exists an $n_0 \in \mathbb N$ such that $\forall n \ge n_0, f(n) \ge g(n)$