Proof of Existence and Uniqueness of the Nonnegative Square Root of a Nonnegative Real Number

Axiom numbers below refer to Field, and conditions 1 and 2 are the two order-compatibility conditions of Ordered Field. Claims of Elementary Arithmetic in an Ordered Field and of Zero Products and Elementary Identities in a Field are cited by number. The order $\le$ is a total order, so it is reflexive, antisymmetric, transitive and total, and these four properties are used by name. Rearrangements of sums are made freely using axioms 1, 2, 3 and 4.

Preliminary (halving). Put $2=1+1$. Claim 1 of Elementary Arithmetic in an Ordered Field gives $0\le1$, so claim 2 gives $0\le2$; and $2-1=1$, so claim 3 gives $1\le2$. If $2=0$, then $1\le0$, and antisymmetry with $0\le1$ would give $1=0$, contradicting axiom 6. Hence $2\ne0$, and claim 4 of Elementary Arithmetic in an Ordered Field gives $2^{-1}\ne0$ and $0\le2^{-1}$. Moreover, for every $u\in\mathbb{R}$, axioms 6, 9, 5 and 7 give

Uniqueness. Let $r,s\in\mathbb{R}$ satisfy $0\le r$, $0\le s$, $r^{2}=a$ and $s^{2}=a$. Claim 4 of Zero Products and Elementary Identities in a Field gives

so claim 3 of Zero Products and Elementary Identities in a Field gives $s-r=0$ or $s+r=0$.

If $s-r=0$, then adding $r$ to both sides gives $s=r$.

If $s+r=0$, then adding $r$ to both sides of $0\le s$ and using condition 1 gives $r\le s+r=0$; with $0\le r$, antisymmetry gives $r=0$, and then $s=s+0=s+r=0=r$.

In both cases $s=r$, so there is at most one such $r$.

Existence. Put

Step 1 ($S$ is nonempty). Reflexivity gives $0\le0$, and $0^{2}=0$ by claim 1 of Zero Products and Elementary Identities in a Field, so $0^{2}\le a$ by hypothesis. Hence $0\in S$.

Step 2 ($a+1$ is an upper bound for $S$). Let $x\in S$. By totality either $x\le a+1$, and there is nothing to prove, or $a+1\le x$; assume the latter. Since $0\le x$, claim 5 of Elementary Arithmetic in an Ordered Field gives $(a+1)x\le xx=x^{2}$, and $x^{2}\le a$, so $(a+1)x\le a$ by transitivity. Since $(a+1)-1=a$ and $0\le a$, claim 3 gives $1\le a+1$, and claim 5 with $0\le x$ gives $1x\le(a+1)x$; as $1x=x$ by axioms 6 and 8, transitivity yields $x\le a$. Combined with $a+1\le x$ this gives $a+1\le a$ by transitivity. Adding $-a$ to both sides and using condition 1 gives $1\le0$, and antisymmetry with $0\le1$ gives $1=0$, contradicting axiom 6. Hence $x\le a+1$.

Step 3 (the least upper bound). By Steps 1 and 2 the set $S$ is a nonempty subset of $\mathbb{R}$ that is bounded above in the sense of Upper Bound and Least Upper Bound, so by The Real Numbers it has a least upper bound $r\in\mathbb{R}$. Since $0\in S$ and $r$ is an upper bound for $S$, we get $0\le r$.

We show $r^{2}=a$. Suppose $r^{2}\ne a$. By totality either $r^{2}\le a$ or $a\le r^{2}$, and we derive a contradiction in each case.

Case A: $r^{2}\le a$ and $r^{2}\ne a$. Put $d=a-r^{2}$. Claim 3 of Elementary Arithmetic in an Ordered Field gives $0\le d$, and $d\ne0$, since $d=0$ would give $a=r^{2}$ on adding $r^{2}$.

Put $c=(r+r)+1$. Claim 2 gives $0\le r+r$ and then $0\le c$; and $c-1=r+r$, so claim 3 gives $1\le c$. If $c=0$ then $1\le0$, and antisymmetry with $0\le1$ would give $1=0$, contradicting axiom 6; hence $c\ne0$, and claim 4 gives $c^{-1}\ne0$ and $0\le c^{-1}$.

Put $e=dc^{-1}$. Condition 2 gives $0\le e$, and claim 3 of Zero Products and Elementary Identities in a Field gives $e\ne0$, since $d\ne0$ and $c^{-1}\ne0$.

Define $h=e$ if $e\le1$, and $h=1$ otherwise; in the second case totality gives $1\le e$. In either case, using reflexivity and $1\ne0$ from axiom 6,

Since $h\le1$ and $0\le h$, claim 5 gives $hh\le h1$, that is $h^{2}\le h$ by axiom 6. Claim 5 of Zero Products and Elementary Identities in a Field gives

Adding $r^{2}+(rh+rh)$ to both sides of $h^{2}\le h$ and using condition 1 gives

the equality because, by axioms 8, 9 and 6,

Since $h\le e$ and $0\le c$, claim 5 gives $ch\le ce$, and by axioms 5, 8, 7 and 6,

Hence $hc\le d$ by axiom 8, and adding $r^{2}$ and using condition 1 and transitivity gives

Claim 2 gives $0\le r+h$, so $r+h\in S$, and therefore $r+h\le r$ because $r$ is an upper bound for $S$. Adding $-r$ to both sides and using condition 1 gives $h\le0$; with $0\le h$, antisymmetry gives $h=0$, contradicting $h\ne0$.

Case B: $a\le r^{2}$ and $r^{2}\ne a$. Put $d=r^{2}-a$; as in Case A, $0\le d$ and $d\ne0$.

First, $r\ne0$: if $r=0$ then $r^{2}=0$ by claim 1 of Zero Products and Elementary Identities in a Field, so $a\le0$, and antisymmetry with $0\le a$ gives $a=0=r^{2}$, contradicting $r^{2}\ne a$.

Put $c=r+r$. Claim 2 gives $0\le c$. By axioms 6, 9 and 8, $c=r1+r1=r(1+1)=r2$, and $2\ne0$ by the preliminary, so $c=0$ would give $r=0$ by claim 3 of Zero Products and Elementary Identities in a Field; hence $c\ne0$, and claim 4 gives $c^{-1}\ne0$ and $0\le c^{-1}$.

Put $h=\bigl(dc^{-1}\bigr)2^{-1}$. Two applications of condition 2 give $0\le h$, and two applications of claim 3 of Zero Products and Elementary Identities in a Field give $h\ne0$. By axioms 5, 8, 7 and 6,

We claim that $r-h$ is an upper bound for $S$. Let $x\in S$, and suppose for contradiction that $x\le r-h$ fails; by totality $r-h\le x$. Since $r$ is an upper bound for $S$, also $x\le r$.

Claim 3 of Elementary Arithmetic in an Ordered Field applied to $x\le r$ gives $0\le r-x$, and applied to $r-h\le x$ gives $0\le x-(r-h)$; since $x-(r-h)=h-(r-x)$, claim 3 read in the other direction gives $r-x\le h$. Claim 2 gives $0\le r+x$, and adding $r$ to both sides of $x\le r$ using condition 1 gives $r+x\le r+r=c$.

Claim 4 of Zero Products and Elementary Identities in a Field gives $r^{2}-x^{2}=(r-x)(r+x)$. Claim 5 of Elementary Arithmetic in an Ordered Field, applied to $r-x\le h$ with $0\le r+x$, gives $(r+x)(r-x)\le(r+x)h$; applied to $r+x\le c$ with $0\le h$, it gives $h(r+x)\le hc$. By axiom 8 and transitivity,

Adding $x^{2}-d2^{-1}$ to both sides and using condition 1 gives $r^{2}-d2^{-1}\le x^{2}$. Since $x\in S$ we have $x^{2}\le a$, so transitivity gives $r^{2}-d2^{-1}\le a$, and $a=r^{2}-d$ by the definition of $d$. Adding $d-r^{2}$ to both sides and using condition 1 gives $d-d2^{-1}\le0$; by the preliminary $d=d2^{-1}+d2^{-1}$, so $d-d2^{-1}=d2^{-1}$ and hence $d2^{-1}\le0$. Condition 2 gives $0\le d2^{-1}$, so antisymmetry gives $d2^{-1}=0$, and then $d=d2^{-1}+d2^{-1}=0$, contradicting $d\ne0$.

Therefore $x\le r-h$ for every $x\in S$, so $r-h$ is an upper bound for $S$. Since $r$ is a least upper bound, $r\le r-h$. Adding $h$ to both sides and using condition 1 gives $r+h\le r$, and adding $-r$ gives $h\le0$; with $0\le h$, antisymmetry gives $h=0$, contradicting $h\ne0$.

Both cases are impossible, so $r^{2}=a$. Together with $0\le r$ from Step 3 this proves existence, and uniqueness was proved above.