MHB Show that a quantified statement is true:

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The discussion revolves around proving the quantified statement involving the inequality 2n + 100 ≤ λn for sufficiently large n. It highlights the concept of big-O notation, indicating that while 2n + 100 is greater than n for all n > 0, there exists a constant λ (like 3) that satisfies the inequality eventually. The key point is finding a positive integer m such that the inequality holds for all n ≥ m. The conclusion reached is that m can be determined as 100, thus validating the statement. Overall, the discussion clarifies the relationship between the expressions and the conditions under which the inequality holds.
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for example this question:
∃λ∈R+, ∃m∈Z+,∀n∈m..+∞,2n+100≤λn

To be honest I'm really struggling to understand this math, and I'm actually not even totally sure what this question is called. If anyone could explain this to me or point me to some good tutorials I'd really appreciate it.
 
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This seems like the statement saying that $2n+100$ is $O(n)$. If you don't know what the big-O notation is, please ignore this.

The statement says that even though $2n+100>n$ for all $n>0$, we can find a positive constant $\lambda$ such that $2n+100\le\lambda n$. For example, $\lambda=3$ looks promising. The second subtlety is that $2n+100\le3n$ does not hold for all $n>0$, but only eventually, i.e., from some point $m$ on. Can you find such $m$ if $\lambda=3$?
 
So that would mean m = 100, which proves the statement to be true.

This doesn't seem that bad, thanks for the help.
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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