Prove 100,000,000 is a member of the set of functions defined by Big Theta of g(n), where g(n) = 1
You haven't provided a definition, but here is a standard one:


  Let $f,g$ be functions from $\mathbb{N}$ to $\mathbb{R}$. We say that $f(n) \in \Theta(g(n))$ (usually written $f(n) = \Theta(g(n))$) if there exist a natural number $N$ and positive real numbers $0 &lt; a &lt; b$ such that for all natural $n \geq N$ it holds that $$ ag(n) \leq f(n) \leq bg(n). $$


Roughly speaking, this says that the two functions $f,g$ have the same order of growth – they differ by a constant multiple. The definition is reflexive ($f(n) = \Theta(f(n))$), symmetric ($f(n) = \Theta(g(n))$ implies $g(n) = \Theta(f(n))$) and transitive ($f(n) = \Theta(g(n))$ and $g(n) = \Theta(h(n))$ imply $f(n) = \Theta(h(n))$), and so forms an equivalence relation. Each equivalence class is an "order of growth".

In your question, $f(n) = 10^8$ and $g(n) = 1$. So you have to find natural $N$ and positive reals $0 &lt; a &lt; b$ such that the following holds for all $n \geq N$:
$$ ag(n) \leq f(n) \leq bg(n) \Longleftrightarrow a \leq 10^8 \leq b.$$ 
I'm sure you can find such $N,a,b$.