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Prove that for all positive reals $k,j$ with $2k>3j$, there exists positive integers $s,t$ such that

$k > |\sqrt{s}-\sqrt{t}| > j$

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What have you tried? –  Jonas Meyer Oct 25 '12 at 6:38
    
Two months later, you unaccepted my answer and then accepted your own? Why the change of heart? –  Benjamin Dickman Dec 30 '12 at 5:56
    
Lol... I just want to see what happens when your own answer gets accepted. Sorry if that offends or anything. –  Christmas Bunny Dec 31 '12 at 17:44

3 Answers 3

up vote 0 down vote accepted

Let's show that there are positive integers $s, t$ such that:

$3j/2 > |\sqrt{s} - \sqrt{t}| > j$.

Note that $2k > 3j$ iff $k > 3j/2$, so that if we can find $s, t$ as above we'll be done.

Observe $\sqrt{n+1} - \sqrt{n} = 1/(\sqrt{n+1} + \sqrt{n}) \longrightarrow 0$ as $n$ tends to infinity. Pick $n$ large enough for this difference to be less than $j/2$. Then choose $a \in \mathbb{Z}^+$ such that $3j/2 > a(\sqrt{n+1} - \sqrt{n}) > j$.

(This will be possible, since increasing our prospective $a$ by $1$ amounts to taking "steps" of size less than $j/2$. Thus, you don't run the risk of totally "overstepping" the $j/2$ gap between $3j/2$ and $j$.)

Pick $s = a^2 (n+1)$ and $t = a^2 n$.

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For every positive real numbers $u\lt v$, there exists some positive integers $t\lt s$ such that $u\lt\sqrt{s}-\sqrt{t}\lt v$.

Proof: Pick any integer $t\geqslant1/[4(v-u)^2]$ and consider $x_n=\sqrt{t+n}-\sqrt{t+n-1}$ and $s_n=\sum\limits_{k=1}^nx_k=\sqrt{t+n}-\sqrt{t}$.

For every $n\geqslant1$, $x_n=1/(\sqrt{t+n}+\sqrt{t+n-1})\lt1/(2\sqrt{t})\lt v-u$ and $s_n\to+\infty$ hence there exists $k\geqslant1$ such that $s_{k-1}\lt u\lt s_k$, namely, $k=\inf\left\{i\geqslant1\,\mid\,s_i\gt u\right\}$. Then, $u\lt s_k\lt v$. Choose $s=t+k$.

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This is like a generalized version of my proof below. Make the difference between $\sqrt{n+1} - \sqrt{n} := \epsilon$ less than $v - u$, then move along in an integer number of $\epsilon$ increments until you fall between $u$ and $v$. If this first occurs after $a$ moves, use $s = a^2 (n+1)$ and $t = a^2 n$. –  Benjamin Dickman Oct 29 '12 at 0:31

Another solution:

Select a rational $b/a$ with $a,b$ positive integers such that $2k>b/a>3j$. By multiplying the numerator and denominator by a constant integer, if necessary, we could ensure $3a^2>b$. Now we claim $s=a^2+b,t=a^2$ satisfy the condition. Proof: $k>\frac{b}{2a}>\sqrt{a^2+b}-\sqrt{a^2}=\frac{b}{\sqrt{a^2+b}+a}>\frac{b}{3a}>j$

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For a similar problem, see Putnam 2011 B1. (You can find it, e.g., here: math.northwestern.edu/~mlerma/problem_solving/putnam/…) –  Benjamin Dickman Oct 26 '12 at 6:44

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