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 $$ \frac{a}{i} ; \mathrm{a} $$
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Question

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 $$ \frac{a}{i} ; \mathrm{a} $$ 
eq b for $$ \mathrm{a} 
eq b $$; ineg for 7 , Iwedge for $$ \wedge $$, Ivee for ightarrow for $$ ightarrow $$, \implies for $$ \longrightarrow $$, \equiv for $$ \equiv $$, \subset for $$ C $$, \subseteq for Itimes for $$ 	imes $$, \leftrightarrow for $$ \leftrightarrow $$; \exists for $$ \exists $$ and \forall for $$ \forall $$, 	herefore for $$ : 	herefore $$ in for $$ \in $$ on the keyboard.

UpStudy AI · Accepted Answer

Let $$ x $$ be an odd integer. By definition of an odd integer, there exists an integer $$ k $$ such that

$$
x = 2k + 1.
$$

Now, consider the expression $$ 9x + 5 $$. Substitute $$ x = 2k + 1 $$ into the expression:

$$
9x + 5 = 9(2k + 1) + 5.
$$

Distribute the $$ 9 $$ across the terms in the parentheses:

$$
9(2k + 1) = 18k + 9.
$$

Thus, the expression becomes:

$$
9x + 5 = 18k + 9 + 5.
$$

Combine the constant terms:

$$
18k + 9 + 5 = 18k + 14.
$$

Notice that $$ 18k $$ is even because it is equal to $$ 2(9k) $$. Also, $$ 14 $$ is even as it can be written as $$ 2 \times 7 $$. Therefore, we can factor $$ 2 $$ out of the expression:

$$
18k + 14 = 2(9k + 7).
$$

Since $$ 9k + 7 $$ is an integer (as it is a sum of products and constants), this shows that $$ 9x + 5 $$ is of the form $$ 2 \times \text{(integer)} $$, which by definition means it is even.

Hence, we have proven that if $$ x $$ is an odd integer, then $$ 9x + 5 $$ is even.
If $$ x $$ is an odd integer, then $$ 9x + 5 $$ is even.

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