# Proving if an element of an ordered field has a square root and is not 0, then it has exactly two square roots, one positive and one negative

Assume $$b$$ is a square root of an element $$n$$, and let $$a$$ be an unknown element which is also a square root of $$n$$.

If $$n$$ has only one root, then $$a = b$$.

We have,

$$a^2 - b^2 = n - n = 0$$ so $$(a + b)(a - b) = 0$$

which implies

$$a = b$$ or $$a = -b$$

hence $$a$$ can take the value of $$b$$ and $$-b$$, however this contradicts our initial assumption of $$n$$ having one root, so it has at least two roots.

Since the field is ordered, we have $$b ≠ -b$$ so they are two different square roots of $$n$$. From the axiom of an ordered field, we have $$b > 0$$, $$b < 0$$ or $$b = 0$$. Now $$b ≠ 0$$ as if $$b = 0$$, then $$b^2 = n = 0$$, which is against our assumption that $$n ≠ 0$$. Hence $$b > 0$$ or $$b < 0$$, so $$n$$ has one positive and negative square root, which happen to be additive inverses of each other.

Assume on the other hand that n has three square roots, $$a$$, $$b$$ and $$c$$, where $$a$$ and $$c$$ are unknown elements, then

$$a^2 = b^2 = c^2 = n$$

$$(a - b)(a + b) = 0$$, $$(b - c)(b + c) = 0$$, $$(a - c)(a + c) = 0$$

so $$a = b$$ or $$a = -b$$, $$b = c$$ or $$b = -c$$, $$a = c$$ or $$a = -c$$

Now if $$a = c$$, we get

$$c = b$$ or $$c = -b$$

The same holds for $$a = -c$$

So we are left with exactly two square roots and not three, by which this argument can be extended if we assume any number of square roots above two.

Therefore if a non-zero element of an ordered field has a square root, then it must have two square roots where one is positive and one is negative.

Any feedback is welcomed!

• In any field a polynomial of degree $d$ has at most $d$ zeroes. So $x^2-n=0$ for at most $2$ values of $x.$ In any field we have $x^2=(-x)^2$ for all $x$. In an ordered field $x>0\implies x+y>y$ for all $y.$ In particular, with $y=-x,$ we have $x>0\implies 0=x+(-x)>0+(-x)=-x\implies x>0>-x\implies x\ne -x.$ – DanielWainfleet Jun 30 at 7:01

You started off by assuming $$a$$ and $$b$$ are square roots of an element $$n$$.
That's helpful to figure out that $$b$$ and $$-b$$ are both square roots of $$n$$, but not really part of the proof of the statement in the title. Also, when you said $$b$$ and $$-b$$ are two different square roots of $$n$$, you should have pointed out (as you did later) that holds if $$b\ne0$$.
To prove the statement in the title, let $$n$$ be a non-zero element of an ordered field, and assume $$n$$ has a square root, say $$b$$; i.e., there is an element $$b$$ of the ordered field such that $$b^2=n$$. Then $$-b$$ is also a square root of $$n$$, because $$(-b)(-b)=b^2$$ follows from the axioms of an ordered field. Furthermore, for an ordered field, $$b>0$$, $$b=0$$, or $$b<0$$. If $$b=0$$ then $$b^2=0$$, which contradicts the assumption $$n\ne0$$. If $$b>0$$ then $$-b<0$$, and if $$b<0$$ then $$-b>0$$. Thus, $$n$$ has two square roots, one positive and one negative.
It's not clear whether you need to prove $$n$$ has two square roots and not more, or just that $$n$$ has (at least) two square roots, but your approach for proving the former looks valid.