Prove that if \( x \) is an odd integer, then \( 9 x+5 \) is even. Write your answer explaining every step and your reasoning in the box provided below. Symbol HINTS: If necessay, use the following commands: \( \mathrm{a}^{\wedge} \mathrm{b} \) for \( a^{b} ; \mathrm{Q} \) for \( \mathbb{Q} ; \mathrm{a} / \mathrm{b} \) for \( \frac{a}{i} ; \mathrm{a} \) \neq b for \( \mathrm{a} \neq b \); ineg for 7 , Iwedge for \( \wedge \), Ivee for \rightarrow for \( \rightarrow \), \implies for \( \longrightarrow \), \equiv for \( \equiv \), \subset for \( C \), \subseteq for Itimes for \( \times \), \leftrightarrow for \( \leftrightarrow \); \exists for \( \exists \) and \forall for \( \forall \), \therefore for \( : \therefore \) in for \( \in \) on the keyboard.
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To prove that if \( x \) is an odd integer, then \( 9x + 5 \) is even, we start by expressing an odd integer. An odd integer can be written in the form: \[ x = 2k + 1 \] for some integer \( k \). Now, we will substitute this expression into \( 9x + 5 \): \[ 9x + 5 = 9(2k + 1) + 5 \] Distributing the \( 9 \) gives us: \[ = 18k + 9 + 5 = 18k + 14 \] Next, notice that \( 18k \) is an even number (since it’s a multiple of \( 2 \)), and \( 14 \) is also an even number. The sum of two even numbers is even, so: \[ 18k + 14 \text{ is even} \] Thus, we have shown that if \( x \) is an odd integer, then \( 9x + 5 \) is indeed even. Therefore, we conclude: \[ \therefore 9x + 5 \text{ is even when } x \text{ is odd} \]
