Discrete-math 3

\begin{matrix} N_{o d d} = {2 n - 1 ∣ n \in N} \\ N_{e v e n} = {2 n ∣ n \in N} \\ Let a \in N_{e v e n} \\ There exists n, such as a = 2 n \\ a is a product of 2 natural numbers, so a \in N \\ Let b \in N_{o d d} \\ There exists n, such as b = 2 n - 1 \\ b is a difference of two natural numbers, so b \in Z \\ n \in N ⟹ n \geq 1 ⟹ 2 n - 1 \geq 1 ⟹ b \in N \\ (N_{e v e n} \cup N_{o d d}) \subseteq N \\ Let n \in N, if n is odd, there exists k \in N : n = 2 k - 1 \in N_{o d d} \\ if n is even, there exists k \in N : n = 2 k \in N_{e v e n} \\ ⟹ n \in N ⟹ n \in N_{o d d} \lor n \in N_{e v e n} ⟺ \\ ⟺ N \subseteq (N_{e v e n} \cup N_{o d d}) \\ ⟹ (N_{o d d} \cup N_{e v e n}) = N \end{matrix}

\begin{matrix} A_{n} = {\frac{m}{n} | m \in N}, n \in N \\ Prove: ⋃_{n \in N} A_{n} = Q^{+} \\ Let x \in ⋃_{n \in N} A_{n} \\ Then there exists n \in N, such that x \in A_{n} \\ Therefore exists m \in N, such that x = \frac{m}{n} \in Q^{+} \\ ⟹ ⋃_{n \in N} A_{n} \subseteq Q^{+} \\ Let x \in Q^{+} \\ Then exists m \in Z, n \in N, such that x = \frac{m}{n}, x > 0 \\ Since x > 0, m > 0 is also true, thus m \in N \\ Therefore x \in A_{n} ⟹ x \in ⋃_{n \in N} A_{n} \\ ⟹ Q^{+} \subseteq ⋃_{n \in N} A_{n} \\ ⟹ ⋃_{n \in N} A_{n} = Q^{+} \end{matrix}