Proof that the cube root is unique

Question

I was wondering if there was anything wrong with the following proof. I sometimes get tripped up on proving blindingly obvious facts, like that the cube root is unique.

Proof (Uniquess of Cube Root): Suppose that $\forall x \in \mathbb{R}$, $\exists z,y \in \mathbb{R}$, $z \neq y$ and WLOG $z>y$ such that $z^3=x$ and $y^3=x$. Then either $x^3 < y^3$ or $x^3 > y^3$. But $y^3=z^3$, which is a contradiction. Hence, $y=z$.

Answer 1

Let $x^3=y^3$. Thus, $$0=x^3-y^3=(x-y)(x^2+xy+y^2),$$

which gives $x=y$ or $x^2+xy+y^2=0$, which gives $$0=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4},$$ which gives $x=y=0$ and we got $x=y$ again.

Answer 2

I think there is a typo with "either $x^3 < y^3$ or $x^3 > y^3$" as mentioned by Stinking Bishop and it should be $z$ instead of $x$, but the otherwise it's correct. Here are a few comments:

• Since you assumed WLOG that $z>y$ then you don't need to say $z\neq y$
• For "either $z^3 < y^3$ or $z^3 > y^3$", since we have $z>y$, you only need that $z^3>y^3$. But $z^3=x=y^3$, which is a contradiction.

Answer 3

$$\forall x\in\mathbb R:\exists\, y<z\text{ (WLOG)}:x=y^3=z^3$$ but $$y<z\implies y^3<z^3,$$ a contradiction.