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!

  • $\begingroup$ 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 $x>0\implies 0=x+(-x)>0+(-x)=-x\implies x>0>-x\implies x\ne -x.$ $\endgroup$ – DanielWainfleet Jun 30 at 7:01

You got the basic idea, but I would write it somewhat differently.
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.

  • $\begingroup$ Thanks for the feedback. On checking the question again it seems to want me to prove that there are at most two roots, for which there is only one root only for the case where the element is 0. I've made some changes taking into consideration your comment, so I hope you would do me the pleasure of reviewing my answer again. $\endgroup$ – Aden Jun 30 at 6:31

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