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: a^b for \( a^{b} ; Q \) for \( \mathbb{Q} ; a / b \) for \( \frac{a}{b} ; a \backslash \) neq \( b \) for \( a \neq b \); \neg for \( \neg \), wedge for \( \wedge \), vee for \( \vee, \backslash \) rightarrow for \( \rightarrow, \backslash i m p l i e s \) for \( \Longrightarrow \), lequiv for \( \equiv, \backslash \) subset for \( \subset, \backslash \) subseteq for \( \subseteq, \backslash \) times for \( \times \), \( \backslash \) leftrightarrow for \( \leftrightarrow \); \exists for \( \exists \) and \( \backslash \) forall for \( \forall, \backslash \) 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 defining an odd integer. An odd integer can be expressed in the form \( x = 2k + 1 \), where \( k \) is an integer. Now substituting \( x \) into the expression \( 9x + 5 \): \[ 9x + 5 = 9(2k + 1) + 5 = 18k + 9 + 5 = 18k + 14 \] We notice that \( 18k + 14 \) can be factored as: \[ 18k + 14 = 2(9k + 7) \] Since \( 9k + 7 \) is an integer (as both \( 9k \) and \( 7 \) are integers), it follows that \( 2(9k + 7) \) is an even integer. Thus, \( 9x + 5 \) is even when \( x \) is an odd integer. Hence, we conclude that if \( x \) is an odd integer, then \( 9x + 5 \) is indeed even, as required. Therefore, we have shown the relationship: \[ \text{If } x \text{ is odd } \Rightarrow 9x + 5 \text{ is even.} \]
