TOPIC: The Completeness Axiom

❯❯ Completeness Axiom Meaning And Definition

The completeness axiom is the final axiom that is needed in order to make a complete and continuous mapping between the real number system (analytic) and points on the real line (geometric).

Without this axiom, we cannot actually prove definitely the existence of the irrational numbers subset of the real number set.

Remember, that the proof showing that no rational number can satisfy the equation $x^2=2$, does not necessarily imply there exists a solution.

Thus, in order to establish the existence of the solution we utilize the completeness axiom, which we will now state:


Given the two sets $A \subseteq \R$ and $B \subseteq \R$, satisfying the conditions:

1) Both $A$ and $B$ are nonempty.

2) $\forall a \in A$, $\forall b \in B$, $a \leq b$.

Then, $\exists x \in \R $, such that:

1) $\forall a \in A$, $x \geq a$.

2) $\forall b \in B$, $x \leq b$.


Now, we can use this axiom to derive the existence of the solution for $x^2 =2$, and actually the existence of every positive real root of any natural order of a positive real number.

❯❯ Proof Of The Existence Of Positive Roots

The theorem we wish to establish is:

$\forall c \in \R$, $c \geq 0$, $\exists x \in \R$, $x \geq 0$, such that $ \forall n \in \N$, $x^n = c$.


Proof:


For the case when $c = 0$, the proof is trivial.

First we will need to use two lemmas in this proof, which we will state (without proving) now:

1) $\forall x, c \in \R$, $x > 0$, if $x^n < c$, then $\exists y \in \R$, $y > x$, such that $y^n < c$.

2) $\forall x, c \in \R$, $x > 0$, if $x^n > c$, then $\exists y \in \R$, $y < x$, such that $y^n > c$.

These two lemmas, in essence mean that, we can always increase\decrease the value of $x$ a little bit and still not reach the value of $c$.


Now, let's define the two sets:

$$A=\{x \in \R^+ \;|\; x^n \leq c, \; \forall n \in \N\} \\ B=\{x \in \R^+ \;|\; x^n \geq c, \; \forall n \in \N\}$$
Then, notice that both sets $A$ and $B$ are nonempty:

If $c \geq 1$, then $1 \in A$, $c \in B$.

If $1 > c > 0$, then $c \in A$, $1 \in B$.

Now, we will prove the following property (related to the order axioms):

Given $a,b \in \R$, $a \geq 0$, $b \geq 0$, then $ \forall n \in N$, $a \leq b$ iff $a^n \leq b^n$.


Proof:


If either $a$ or $b$ equal $0$ (or both), then the proof is trivial.

Thus, we will assume $a > 0$ and $b > 0$.

Left to right ($a \leq b$ $\rightarrow a^n \leq b^n$):

We can prove in two ways, either directly or by induction (direct proof utilizes the product of inequalities property repeated $n-1$ times). In essence they are the same.

By induction:

1) Base case ($n = 1$) is satisfied by assumption.

2) Assume true for some $k-1$.

3) Inductive step:

$0 < a^{k-1} \leq b^{k-1} \quad / \cdot 0 < a \leq b $

We use the product of inequalities a single time,

$a^k \leq b^k$.



Right to left ($a^n \leq b^n \rightarrow a \leq b$):

Now, we use proof by contradiction.
If false, then we assume that $a^n \leq b^n$ and $a > b$. Using product of inequalities $n$ times leads to $a^n > b^n$, and we get a contradiction! $\Box$


Now, observe that the sets $A$ and $B$ elements satisfy:

$\forall a \in A$, $\forall b \in B$, $a^n \leq c \leq b^n$.

Thus, $a \leq b$ (by the previously proved theorem).

These two observations (proved), namely, $A \neq \phi$, $B \neq \phi$ and $\forall a \in A$, $\forall b \in B$, $a \leq b$, mean that these two sets satisfy the completeness axiom, so $\exists x \in \R$, $\forall a \in A$, $\forall b \in B$, $a \leq x \leq b$.



Now, by the trichotomy property, there can be one of three possibilities:


1) $x^n < c$.

2) $x^n > c$.

3) $x^n = c$.



Assume (1) is true.

Then, using the first lemma we introduced as an auxiliary tool for the proof, we have:

$\exists y > x$, such that $y^n < c$. This implies $y \in A$, but $\forall a \in A$, $a \leq x$, so we get a contradiction.

Assume (2) is true.

Then, using the second lemma we introduced as an auxiliary tool for the proof, we have:

$\exists y < x$, such that $y^n > c$. This implies $y \in B$, but $\forall b \in B$, $b \geq x$, so we again get a contradiction.



From exhaustion, we deduce (3) must be true $\Box$.

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