Let $w$ be a positive rational number and $w^2>2$. Prove that there exists a positive rational number $x$ such that $x^2>2$ and $x<w$. A condition is you can not use the property of real numbers such as the density property of real numbers (don't say that $w$ is a real number).
Asked
Active
Viewed 93 times
-2
-
@Michael: you need a rational number between $w$ and $\sqrt{2}$, so just taking the average won't work. – Qiaochu Yuan Nov 17 '15 at 00:17
4 Answers
1
The usual answer is to use Newton's method, which preserves rational numbers: $$ x=\frac12\left(w+\frac2w\right) $$
Rudin's book contains a slightly different expression. See here.
1
Hint: If $p-q\sqrt{2}>0$ for $p,q$ positive integers, multiply by $0<3-2\sqrt{2}<1$.
Thomas Andrews
- 177,126
1
Let $w=\frac{p} {q}$ the irreducible fraction. Let $v_0= \inf_n \{ v=\frac{p'} {q'}, \text{ such that } p'=10^n p - 1, ~q' = 10^n q, ~ v^2 > 2\}$ . Such a value exists because the inequality is strict. Then you have $v_0 <w$, $v_0 ^2>2$ and $v_0 $ is rational.
user288227
- 308
0
You don't really need to do anything fancy here; continuity of the squaring function will do the trick. Consider $$w-\epsilon$$ as $\epsilon \to 0^+$, with $\epsilon>0$ rational (if you like, take a sequence $\epsilon_1, \epsilon_2, \ldots$ of rational numbers decreasing to zero).
Then $(w-\epsilon)^2$ approaches $w^2>2$. Pick $\epsilon$ small enough so that $(w-\epsilon)^2>2$. No reference to real numbers necessary, even though I used continuity.
Dustan Levenstein
- 12,941