Problem A. 436. (October 2007)

A. 436. Prove that

$\big| \big\{n\sqrt2\,\big\} - \big\{n\sqrt3\,\big\}\big| > \frac1{20n^3}$

for every positive integer n.

(5 pont)

Deadline expired on November 15, 2007.

Solution. Let $t=\big\{n\sqrt2\,\big\}-\big\{n\sqrt3\,\big\}$ and $k=\big[n\sqrt3\big]-\big[n\sqrt2\big]$ . Then $t=k-\big(\sqrt3-\sqrt2\big)n\ne0$ , since $\sqrt3-\sqrt2=\sqrt{5-2\sqrt6}$ is irrational.

Write

$t = \frac{ \big(k-(\sqrt3-\sqrt2)n\big)\big(k-(\sqrt3+\sqrt2)n\big) \big(k+(\sqrt3-\sqrt2)n\big)\big(k+(\sqrt3+\sqrt2)n\big)}{ \big(k-(\sqrt3+\sqrt2)n\big)\big(k+(\sqrt3-\sqrt2)n\big) \big(k+(\sqrt3+\sqrt2)n\big)} =$

$= \frac{k^4-10k^2n^2+n^4}{ \big(t-2\sqrt2n\big)\big(t+2(\sqrt3-\sqrt2)n\big)\big(t+2\sqrt3n\big)}.$

The numerator is an integer which is not 0. In the denominator we have

$\big|t-2\sqrt2n\big| \le 2\sqrt2n+\frac1{20} \le \left(2\sqrt2+\frac1{20}\right)n,$

$\big|t+2(\sqrt3-\sqrt2)n\big| \le 2(\sqrt3-\sqrt2)n+\frac1{20} \le \left(2\sqrt3-2\sqrt2+\frac1{20}\right)n$

and

$\big|t+2\sqrt3n\big| \le 2\sqrt3n+\frac1{20} \le \left(2\sqrt3+\frac1{20}\right)n,$

therefore

$|t| \ge \frac1{ \big(2\sqrt2+\frac1{20}\big) \big(2\sqrt3+2\sqrt2-\frac1{20}\big) \big(2\sqrt3+\frac1{20}\big)n^3} > \frac1{7n^3}.$

We obtained that

$|t| \ge \min\left(\frac1{20},\frac1{7n^3}\right) \ge \frac1{20n^3}.$

Statistics:

7 students sent a solution.

5 points: Korándi Dániel, Lovász László Miklós, Nagy 235 János, Tomon István.

3 points: 1 student.

1 point: 1 student.

Unfair, not evaluated: 1 solutions.

Problems in Mathematics of KöMaL, October 2007

7 students sent a solution.
5 points:	Korándi Dániel, Lovász László Miklós, Nagy 235 János, Tomon István.
3 points:	1 student.
1 point:	1 student.
Unfair, not evaluated:	1 solutions.

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