<s>[INST] Prove that $\exists !c \in \mathbb{R} \exists ! x \in \mathbb{R} (x^2 + 3x + c = 0)$ This is an exercise from Velleman's "How To Prove It". I am struggling with how to finish the final part of the uniqueness proof, so any hints would be appreciated!


  
*a. Prove that there is a unique real number $c$ such that there is a unique real number $x$ such that $x^2 + 3x + c = 0$.
  

Proof: Let $c = \frac{9}{4}$.  Let $x = -\frac{3}{2}$. It follows that $x^2 + 3x + c = \frac{9}{4} - \frac{9}{2} + \frac{9}{4} = 0$. To show that $x$ is unique, let $y \in \mathbb{R}$ be arbitrary such that $y^2 + 3y + c = 0$. So $y^2 + 3y + \frac{9}{4} = 0$, and $(y+\frac{3}{2})^2 = 0$. It immediately follows that $y = -\frac{3}{2} = x$.
Now to show that $c$ is unique, let $d, e \in \mathbb{R}$ be arbitrary such that $\exists ! x \in \mathbb{R} (x^2 + 3x + d = 0)$ and $\exists ! x \in \mathbb{R} (x^2 + 3x + e = 0)$. This means that $\exists x \in \mathbb{R}(x^2 + 3x + d = 0)$, $\exists x \in \mathbb{R}(x^2 + 3x + e = 0)$,  $\forall y \in \mathbb{R} \forall z \in \mathbb{R} ((y^2 + 3y +d = 0 \wedge z^2 + 3z + d = 0 )\rightarrow y =z)$, and $\forall y \in \mathbb{R} \forall z \in \mathbb{R} ((y^2 + 3y +e = 0 \wedge z^2 + 3z + e = 0 )\rightarrow y =z)$. (How can we show that $d = e$ to finish this?)
 [/INST] Your approach is complicated and I do not recommend it. Instead, consider the quadratic formula applied here:
$$
x = \frac{-3 \pm \sqrt{9 - 4c}}{2}
$$
Let $d$ be another number such that there exists a unique $x$ in which $x^2 + 3x + d = 0$. We note that $d \not> \frac{9}{4}$, as otherwise $x$ it not real (so no such real $x$ exists). We also see that $d \not< \frac{9}{4}$, as otherwise this $x$ is not unique. Thus, $d = \frac{9}{4} = c$ necessarily.
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