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0613 vector problem 6th edition Round 1 p3

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May 3, 2021
vectoralgebra6th edition

Problem Statement

Introductory part We call an nn-tuple x=(x1,x2,...,xn)x = (x_1, x_2, ... , x_n), with xk∈Rx_k \in R (or respectively with all xk∈Zx_k \in Z) a real vector (or respectively an integer vector). The set of all real vectors (respectively all integer vectors) is usually denoted by RnR^n (respectively ZnZ^n). A vector xx is null if and only if xk=0x_k = 0, for all k∈{1,2,...,n}k \in \{1, 2,... , n\}. Also let UnU_n be the set of all real vectors x=(x1,x2,...,xn)x = (x_1, x_2, ... , x_n), such that x12+x22+...+xn2=1x^2_1 + x^2_2 + ...+ x^2_n = 1. For two vectors x=(x1,...,xn),y=(y1,...,yn)x = (x_1, ... , x_n), y = (y_1, ..., y_n) we define the scalar product as the real number x⋅y=x1y1+x2y2+...+xnynx\cdot y = x_1y_1 + x_2y_2 +...+ x_ny_n. We define the norm of the vector xx as ∣∣x∣∣=x12+x22+...+xn2||x|| =\sqrt{x^2_1 + x^2_2 + ...+ x^2_n}
The problem Let A(k,r)={x∈Un∣A(k, r) = \{x \in U_n | for all z∈Znz \in Z^n we have either ∣x⋅z∣≥k∣∣z∣∣r|x \cdot z| \ge \frac{k}{||z||^r} or zz is null }\}. Prove that if r>n−1r > n - 1 the we can find a positive number kk such that A(k,r)A(k, r) is not empty, and if r<n−1r < n - 1 we cannot find such a positive number kk.
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