# Real Analysis using the Archimedean Principle

Let $x$ be a real number. Then there exists a natural number $n$ such that $3^n > x$.

proof:

Let $x$ be a real number.
Suppose that $x$ is an upperbound of the natural numbers.
We know $1$ is a natural number, so the natural numbers are non empty.
Then the natural number have a supremum, say $B = sup(N)$ where $N$ is the natural numbers.
This implies $\frac{1}{3} B < B$.
So $\frac{1}{3} B$ is not an upperbound of $N$.
Then there exists a natural number $n$ such that $\frac{1}{3} B < n$.
This implies $B < 3n$.
Then the natural numbers are closed under multiplication.
So $3n$ is a natural number.
This is a contradiction of $B = sup(N)$.
Therefore, $x$ is not an upperbound of the natural numbers.
So there exists a natural number $m$ with $x < m$.

Let $S = \{3^n|n \in \mathbb{R}\}$.

Now I need to use this to show that there is $n$ with $3^n > x$. But I am confused on where to go from here and how to apply the first part of the proof.

• Can you find an $n$ with $n>x$? – Slade Oct 13 '14 at 1:38

If you know $\exists n \in \mathbb{N}$ such that $n > x$, then since $3^{n} > n$ for all $n \in \mathbb{N}$, it follows that $3^{n} > x$ for the $n$ satisfying $n > x$.

Proof by induction that $3^{n} > n$:

Base case: For $n = 1$, we do have $3^{1} > 1$.

Inductive assumption: Assume for $n-1$ that $3^{n - 1} > n - 1$

Now we need to show $3^{n} > n$. But $$3^{n} = 3^{n-1}3 > (n-1)3 = 3n - 3 > n$$ if $n \geq 2$. So we proved if $n \geq 2$, then $3^{n} > n$ (and the base case shows this inequality holds of $n = 1$...).

You might decide that you don't believe that $3n - 3 > n$ if $n \geq 2$, so we will have to prove this claim, too, by induction.

Base case: For $n = 2$, $3(2) - 3 > 2$ holds.

Inductive assumption: Suppose for $n$ that $3(n-1) - 3 > n - 1$.

We must prove $3(n) - 3 > n$.

But by assumption we have $$3(n - 1) - 3 > n - 1$$ and $$3(n - 1) - 3 = 3n - 3 - 3$$ so $$3n - 3 - 3 > n -1$$ or $$3n - 3 > n + 2 > n$$ so we proved $3n - 3 > n$ (for $n \geq 2$).