Here's the question in mind.
Let $$A = \left\{r : r \quad \text{is a rational number and} \quad r^2 < 2\right\}$$Prove that $A$ has no largest number.
(Hint: if $r^2 < 2$, and $r > 0$, choose a rational number $δ$ such that $0< δ < 1$ and $δ < (2-r^2)/(2r+1)$ show that $(r + δ)^2< 2$)
I was wondering if I could answer this question by assuming that $r$, which is an element of $A$, is the largest element. Can I then construct a new element $p$, where $p = r+δ$ and $p^2 < 2$. If that is the case, how would I go about proving that $p^2 $is in fact less than $2$? Thanks.
Following the hint in the given question....
Consider a rational number $r$ such that $r^2\lt 2$. Now, consider $p=(r+\delta)$ where $\delta$ is a positive real less than $1$ satisfying $\delta\lt\dfrac{2-r^2}{2r+1}$. So, obviously we have $p\gt r$. Then, we have,
$$\delta\lt\frac{2-r^2}{2r+1}\implies r^2+2r\delta\lt 2-\delta\implies (r+\delta)^2\lt 2-\delta+\delta^2\tag1$$
Since $0\lt\delta\lt 1$, we have $\delta^2\lt\delta\implies -\delta+\delta^2\lt 0\implies 2-\delta+\delta^2\lt 2\tag2$
Using $(1)$ and $(2)$, we get,
$$(r+\delta)^2\lt 2\implies p^2\lt 2$$
Hence, the set $\{r\in\Bbb Q\colon r^2\lt 2\}$ has no maximum.