Existence of Square Roots

We use induction on $n$, a direct algebraic technique. Base case $n=1$: $a < b$ is the hypothesis. Inductive step: suppose $a^n < b^n$ for some $n \ge 1$; we show $a^{n+1} < b^{n+1}$. Since $0 < a < b$, multiplying $a^n < b^n$ by $a>0$ (order is preserved under multiplication by a positive element) gives

Multiplying $a < b$ by $b^n > 0$ gives

Chaining these (transitivity of $<$) gives $a^{n+1} < b^{n+1}$. By induction the claim holds for all positive integers $n$. $\blacksquare$

Induction on $n$. For $n=1$, $(1+b)^1 = 1+b = 1 + 1\cdot b$, so equality holds. Assume $(1+b)^n \ge 1+nb$. Multiply both sides by $1+b>0$ (order preserved):

since $nb^2 \ge 0$. This proves the inductive step, and the lemma follows by induction. $\blacksquare$

We prove existence and then uniqueness. This is the classical least-upper-bound (supremum) construction, a direct application of the Completeness Axiom followed by a trichotomy elimination argument.

Step 1: Define the candidate set and show it is nonempty and bounded above. Let

Nonempty: Let $t_0 = \dfrac{x}{x+1}$. Then $0 < t_0 < 1$ (since $x>0$ forces $x < x+1$, and dividing by $x+1>0$ gives $t_0<1$; also $t_0>0$ since $x>0$). By Monotonicity of powers applied with $a = t_0 < 1 = b$... more directly, since $0<t_0<1$, repeated multiplication by $t_0$ only decreases the value (each factor is less than $1$ and positive), so $t_0^n \le t_0 < 1 \le x+1$. We want $t_0^n < x$ specifically; here is a clean way to see it. Since $0<t_0<1$ we have $t_0^n \le t_0$ (immediate from the Monotonicity Lemma's inductive proof: multiplying $t_0<1$ repeatedly by $t_0>0$ shows $t_0^{k+1}=t_0\cdot t_0^k \le t_0^k$ for all $k\ge 1$, using $t_0\le 1$, so the sequence $t_0, t_0^2,\dots$ is non-increasing). Hence $t_0^n \le t_0 = \frac{x}{x+1} < x$ (the last inequality holds because $\frac{x}{x+1}<x \iff x < x(x+1) \iff 1 < x+1 \iff 0<x$, true). So $t_0 \in E$, and $E \ne \varnothing$. Bounded above: Let $M = x+1$. We claim every $t\in E$ satisfies $t < M$. Suppose instead $t \ge M = x+1 > 1$. By Monotonicity of powers (with $a=1\le$ ... more directly, since $t\ge x+1>1$, monotonicity gives $t^n \ge (x+1)^n \ge x+1 > x$ where $(x+1)^n\ge x+1$ follows from the Bernoulli-type Lemma with $b=x$: $(1+x)^n \ge 1+nx \ge 1+x$ since $n\ge1$. This contradicts $t \in E$, which requires $t^n < x$. Hence every $t \in E$ satisfies $t< M$, so $E$ is bounded above by $M = x+1$.

Step 2: Apply completeness. Since $E$ is nonempty and bounded above, the Completeness Axiom guarantees that

exists as a real number, and $y > 0$ (because $t_0 \in E$ and $t_0>0$, so $y \ge t_0 > 0$). We now show $y^n = x$ by eliminating the other two trichotomy possibilities: $y^n < x$ and $y^n > x$.

Step 3: Rule out $y^n < x$. Suppose $y^n < x$. We will produce $h>0$ small enough that $(y+h)^n < x$, exhibiting an element of $E$ larger than $y$, contradicting that $y$ is an upper bound for $E$. Choose $0 < h < 1$ to be fixed shortly. Using the binomial expansion (a finite algebraic identity valid in any ordered field) and bounding each binomial coefficient term crudely, one obtains the standard estimate: for $0<h<1$,

To justify this, write $(y+h)^n = \sum_{k=0}^n \binom{n}{k} y^{n-k}h^k$. Since $0<h<1$ we have $h^k \le h$ for $k\ge 1$ (as $h^k = h\cdot h^{k-1} \le h\cdot 1^{k-1}=h$, using $h<1$ and induction on $k$), so

where the last equality is again the binomial expansion applied to $(y+1)^n = \sum_{k=0}^n \binom{n}{k}y^{n-k}$. Let $C = (y+1)^n - y^n > 0$ (positive because $(y+1)^n > y^n$ by Monotonicity of powers, since $y+1>y>0$). We now choose

which is a valid positive real number since both quantities in the minimum are positive ($x-y^n>0$ by our current supposition, and $C>0$). This $h$ satisfies $0<h<1$, so the estimate above applies:

the last inequality because $y^n < x$ implies $\frac{y^n+x}{2}<x$. Thus $y+h \in E$, yet $y+h > y = \sup E$, contradicting the definition of supremum as an upper bound. Hence $y^n < x$ is impossible.

Step 4: Rule out $y^n > x$. Suppose $y^n > x$. We produce $k$ with $0<k<y$ such that $(y-k)^n > x$; this will show $y-k$ is a smaller upper bound for $E$ than $y$, contradicting that $y$ is the least upper bound. First we check that $y-k$ is indeed an upper bound for $E$ once $(y-k)^n>x$ is established: if $t\in E$ then $t^n < x < (y-k)^n$, and since $t>0$ and $y-k>0$, Monotonicity of powers (contrapositive form: if $t^n<(y-k)^n$ with both positive then $t<y-k$, for otherwise $t\ge y-k$ would force $t^n\ge (y-k)^n$ by the Lemma) gives $t< y-k$. So $y-k$ is an upper bound for $E$. Now we find such a $k$. As in Step 3, write for $0<k<y$ (so that $y-k>0$):

It is cleaner to apply the Step 3 estimate to the pair $(y-k, k)$ in place of $(y,h)$: since $0<k<1$ can also be arranged, and $(y-k)+k = y$, Step 3's inequality (with $y$ replaced by $y-k$ and $h$ replaced by $k$) gives

using $(y-k)+1 \le y+1$ and monotonicity to bound $(y-k+1)^n - (y-k)^n \le (y+1)^n$ crudely (a nonnegative quantity bounded by a positive one suffices for our purposes). Let $D = (y+1)^n>0$. Then

so

Choose

again valid since $y^n - x>0$ by supposition and $D>0$. Then $0<k<y$ (so $y-k>0$, and the estimate above applies), and

the last inequality because $y^n>x$ implies $\frac{y^n+x}{2}>x$. So $(y-k)^n > x$, confirming (via the argument two paragraphs above) that $y-k$ is an upper bound for $E$ strictly smaller than $y$. This contradicts that $y=\sup E$ is the least upper bound. Hence $y^n>x$ is impossible.

Step 5: Conclude existence. Steps 3 and 4 eliminate $y^n<x$ and $y^n>x$. By the trichotomy law for the order on $\mathbb{R}$, exactly one of $y^n<x$, $y^n=x$, $y^n>x$ holds; the only surviving possibility is

Since $y>0$, this proves existence of a positive $n$th root of $x$.

Step 6: Uniqueness. Suppose $y_1, y_2 > 0$ both satisfy $y_1^n = x = y_2^n$. If $y_1 \ne y_2$, then without loss of generality $y_1 < y_2$ (trichotomy). By Monotonicity of powers, $y_1 < y_2$ implies $y_1^n < y_2^n$, i.e. $x < x$, a contradiction. Hence $y_1 = y_2$, proving uniqueness. $\blacksquare$