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Jan 23, 2016 at 14:55 comment added Denis Pankratov Since $|y| \ge 1$, we have that $2^{|y|}$ is even and $3^{k_i - k_{i-1}}$ is odd. Their difference is odd, therefore at least 1 in absolute value.
Jan 23, 2016 at 9:04 comment added Anton Trunov Could you elaborate a little bit on why $|2^{|y|} - 3^{k_i - k_{i-1}}| \ge 1$ is true? I'm asking because this inequality alone could be used to reach a contradiction: $|2^{|y|} - 3^{k_i - k_{i-1}}| \ge 1$, multiplying both sides of it by $3^{k_{i-1}}$, we get $|3^{k_{i-1}} 2^{|y|} - 3^{k_i}| \ge 3^{k_{i-1}}$, thus, $|C (2^{|y|} - 1)| \ge 3^{k_{i-1}}$, which is a contradiction (by the reason provided in your proof).
Jan 22, 2016 at 17:24 history answered Denis Pankratov CC BY-SA 3.0