Vinogradov's Theorem/Lemma 1

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Lemma



For sufficiently large $N$ the major arcs are pairwise disjoint, and the minor arcs are non-empty.


Proof

Suppose that for some admissible $\dfrac {a_1} {q_1} \ne \dfrac {a_2} {q_2}$ we have:

$\map \MM {q_1, a_1} \cap \map \MM {q_2, a_2} \ne \O$

Then using the definition of the major arcs, for $\alpha$ in the intersection we have:

\(\ds \size {\frac {a_1} {q_1} - \frac {a_2} {q_2} }\) \(=\) \(\ds \size {\frac {a_1} {q_1} - \alpha + \alpha - \frac {a_2} {q_2} }\)
\(\ds \) \(\le\) \(\ds \size {\alpha - \frac {a_1} {q_1} } + \size {\alpha - \frac {a_2} {q_2} }\) Triangle Inequality
\(\ds \) \(\le\) \(\ds 2 \frac Q N\)

and

\(\ds \size {\frac {a_1} {q_1} - \frac {a_2} {q_2} }\) \(=\) \(\ds \size {\frac {a_1 q_2 - a_2 q_1} {q_1 q_2} }\)
\(\ds \) \(\ge\) \(\ds Q^{-2}\)

This shows that:

$N \le 2 Q^3 = 2 \paren {\log N}^3$

But by Power Dominates Logarithm, this is not the case for sufficiently large $N$.

Therefore the major arcs must be disjoint.

We have that the major arcs are pairwise disjoint closed intervals.

So by Cover of Interval By Closed Intervals is not Pairwise Disjoint it is not possible that $\MM = \closedint 0 1$.

So it follows that:

$\MM \ne \O$

$\Box$