On page 120 of Terence Tao's Analysis I, 3rd Ed., there's one step I cannot understand in showing the Proposition 5.5.12 There exists a positive real number $x$ such that
$x^2 = 2$:
T. Tao started with arguing by contradiction and considered two assumptions $x^2 < 2$ and $x^2 > 2$, which both should lead to contradictions.
Taking the first assumption, given that $x^2 < 2$ (from hypothesis) and $x \le 2$ (due to 2 being an upper bound), why can we get some $0 < \epsilon < 1$ such that $x^2 + 5\epsilon < 2$ holds? There is a similar trick for the case where $x^2 > 2$ (see the below link) but I believe understanding the aforementioned one should be enough for explaining both.
This link is the complete proof taken from the book, in which you can see I underlined the step confuses me
Best Answer
Let $\epsilon := \frac{2- x^2}{10} > 0$. Then $$x^2 + 5 \epsilon = x^2 + \frac{2 - x^2}{2} = \frac{x^2}{2} + 1 < \frac{2}{2} + 1 = 2,$$ since not only $x^2 \leq 2$, $x^2 < 2.$