Mathematical and Physical Journal
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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.


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

= \frac{k^4-10k^2n^2+n^4}{

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 


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


 |t| \ge \frac1{
\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}.


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 solution.

Problems in Mathematics of KöMaL, October 2007