Textbook's corollary of $S := \{y \in \mathbb{R} : 0 \leq y^2 \leq 2 \}$

1. Jul 28, 2012

operationsres

I can't understand a statement in a proof in a textbook.

I'm going to terminate the proof at the line that I don't understand.

1. The problem statement, all variables and given/known data
Prove that there exists an $x \in \mathbb{R}$ such that x2=2.

2. Their proof until line I don't understand
For this, we define $$S:= \{y \in \mathbb{R} : 0 \leq y^2 \leq 2\}$$ Since $0 \in S$, S is a nonempty. Further, S does not contain real numbers y≥2 since the last inequality implies that $y^2 - 2 = (y-2)(y+2)+2 \geq 2$ ............

3. Reason why I don't understand
In the definition of S, they say that $0 \leq y^2 \leq 2$. Subtracting 2 from everything, $-2 \leq y^2 - 2 \leq 0$, which leads to $y^2 - 2 \geq -2$ ... But they've claimed that $y^2 - 2 \geq 2$ - why?

2. Jul 28, 2012

Re: Textbook's corollary of $S := \{y \in \mathbb{R} : 0 \leq y^2 \leq 2 \}[/ite It is most likely a typo. If y^2 - 2 >= 2, then y^2 >= 4, which is clearly wrong. 3. Jul 28, 2012 Robert1986 Re: Textbook's corollary of [itex]S := \{y \in \mathbb{R} : 0 \leq y^2 \leq 2 \}[/ite I think it means this: If [itex]y \in S$ then $y^2 \leq 2$ and so $y^2 - 2 \leq 0$.

However, if $y \geq 2$ then $y-2 \geq 0$ and so $y^2 - 2 = (y-2)(y+2)+2 \geq 2$. That is $y^2 - 2 > 0$, which is a contradiction.