How to complete the proof: If $x\geq 0$ and $x^2<2$, then there is $y>x$ with $y^2<2$

Question

How to complete the proof: If $x\geq 0$ and $x^2<2$, then there is $y>x$ with $y^2<2$

My attempted proof:

Let $y:=x+\varepsilon$ and $\varepsilon >0$. So, we observe $(x+\varepsilon)^2\overset{!}{<}2 \Leftrightarrow x^2+2\varepsilon x+\varepsilon ^2<2 \Leftrightarrow 2\varepsilon x+\varepsilon^2 < 2-x^2$.

We know that $x^2<2 \Longleftrightarrow 0<2-x^2$. Therefore $ 2\varepsilon x+\varepsilon^2 $ has to be greater or equal to zero. What do I have to do now? Can you please help?

Answer 1

$ 0\le x$ and $x^2<2$ implies that $$0\le x<\sqrt 2$$

Let $$y=\frac {x+\sqrt 2}{2}$$ then you have $$x<y<\sqrt 2 $$ Thus $$ x^2 <y^2 <2$$

Answer 2

Just choose $$y=\sqrt{\frac{x^2+2}2}$$ then clearly we have $x^2\lt y^2\lt2$ and hence $y\gt x$ as the positive square root preserves order.