Prove that $y-x < \delta$ 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?
 A: 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.
A: This is exercise 4, p.11 from Hardy's "A course of Pure Mathematics". Some thoughts of the author related to the problem are supplied in a remark by the end of p.8 of the same book. 
The exercise's exact statement is: 

"If $x,y$ are approximations to $\sqrt{2}$, by defect and by excess respectively, and $2-x^2<\delta$, $y^2-2<\delta$, then $y-x<\delta$. "

If we consider the $x,y$ as values $x_n,y_n$ of some positive sequences,  approximating $\sqrt{2}$ by defect and by excess respectively, and $0<\delta<2$, then the initial inequalities $2-x^2<\delta$, $y^2-2<\delta$ can be solved to give:
$$
\sqrt{2}<y<\sqrt{2+\delta} \ \textrm{   and  } \ \sqrt{2-\delta}<x<\sqrt{2}
$$
Now, if we sum the above we get:
$$
x+y>\sqrt{2}+\sqrt{2-\delta}>\sqrt{2-\delta}+\sqrt{2-\delta}=2\sqrt{2-\delta}
$$
and since:
$$
(y+x)(y-x) < 2\delta
$$
we have that:
$$
2\sqrt{2-\delta}(y-x) < 2\delta \Rightarrow  \\ \\ \sqrt{2-\delta}(y-x) < \delta 
$$
Now, confining $0<\delta<1$ we have that $\sqrt{2-\delta}>1$, thus the last inequality produces:
$$
y-x<\sqrt{2-\delta}(y-x)< \delta \Rightarrow \\ \\ \\ 
 y-x< \delta
$$
which is what the exercise asks for. 
