Prove that $y-x < \delta$

Question

In Hardy's Pure Mathematics it says if $x^2<2$, $ \ \ y^2>2$, $ \ \ 2-x^2 < \delta$ and $y^2 - 2 < \delta$, then $y-x<\delta$. I added the last two inequalities to get $(y+x)(y-x)<2\delta$.

How do I proceed from here?

Answer 1

With the information given it is easy to show that $0<y^2-x^2<2\delta$, forcing $\delta$ to be positive. Then $$x^2<2+\delta \tag 1$$ and we can also see from $y^2>2$ that $$-y^2<-2\tag 2$$ The sum of $(1)$ and $(2)$ gets us $$x^2-y^2 =(x-y)(x+y)<\delta \tag 3$$ It is also clear from what was given that $x^2-y^2<0$, making exactly one of the quantities $(x-y),(x+y)$ negative and exactly one positive. If $(x-y)$ is positive then $(y-x)$ is negative so trivially $y-x<\delta$. Clearly we need only consider the case that $(x-y)$ is negative. This gets us that $y-x$ and $y+x$ are positive quantities. Can you proceed from here, either by contradiction or directly, to show that $y-x<\delta$? There is more to unpack from what is given that may be useful. For example, $y>0$ when $(y+x),(y-x)$ are both greater than zero.