If $D$ is an ordered field and $c = x^2$, where $c \in D$. How many solutions are possible? Please provide a proof or an example.
I know that there could be no solutions if, for example, $D \in \mathbb{Q}$ and $c = 7$, also it could have two solutions when $D = \mathbb{R}$ and $c = 7$.
Are there any other possibilities for the number of solutions? Can there be exactly one, three or four solutions?
On
It's field.
Trivial case: c = 0. Then a = 0 is the only solution ($a \ne 0 => a^2 \ne 0.)
Non trivial case: c does not equal zero.
It's always possible that there is no solution. That's a possibility.
Suppose there were one solution $a$ so $a^2 = c$. Then $(-a)^2 = a^2 = c$ so c is also a solution. If $a = 0$ then $-a = a$ so if there can't be only one solution. If there is one, there are two.
Suppose $b$ is a third solution such that $b^2 = c$. Then $b^2 = a^2$ so $b^2 - a^2 =(b-a)(b+a) = 0$ so either b = a or b = -a.
Two hints:
First, suppose I tell you that $a$ is a solution to $x^2=c$. Then what do you know about the polynomials $x^2-c$ and $x-a$? What does this tell you about the possibility of having 3+ solutions?
Second, draw the graph of $y=x^2$ on the reals. Does this answer your question about the possibility of exactly one solution?