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If a polygon has exactly three sides, then it is a triangle. Jeri drew a polygon with exactly three sides. Therefore, Jeri drew a triangle.

A. [tex]p \rightarrow q \\
q \\
\therefore p[/tex]

B. [tex]p \rightarrow q \\
\begin{array}{l}
\sim q \\
\therefore \sim p
\end{array}[/tex]

C. [tex]p \rightarrow q \\
\begin{array}{l}
\sim p \\
\therefore \sim q
\end{array}[/tex]

D. [tex]p \rightarrow q \\
\begin{array}{c}
p \\
\therefore q
\end{array}[/tex]


Sagot :

Given the premises, let's choose the correct logical argument and explain it:

- Premise 1: If a polygon has exactly three sides, then it is a triangle.
- Premise 2: Jeri drew a polygon with exactly three sides.
- Conclusion: Therefore, Jeri drew a triangle.

Let's translate these statements into logical terms:
- Let [tex]\( p \)[/tex] be "Jeri drew a polygon with exactly three sides."
- Let [tex]\( q \)[/tex] be "Jeri drew a triangle."

Now the premises can be written as:
1. [tex]\( p \rightarrow q \)[/tex] (If a polygon has exactly three sides, then it is a triangle.)
2. [tex]\( p \)[/tex] (Jeri drew a polygon with exactly three sides.)

The logical argument we need to conclude is [tex]\( q \)[/tex] (Therefore, Jeri drew a triangle.).

Given the above, let's examine each option:

Option A:
[tex]\( p \rightarrow q \)[/tex]
[tex]\( q \)[/tex]
[tex]\(\therefore p \)[/tex]

This argument form is not applicable to our premises, because it starts from [tex]\( q \)[/tex] and concludes [tex]\( p \)[/tex], reversing the logic we need.

Option B:
[tex]\( p \rightarrow q \)[/tex]
[tex]\(\sim q \)[/tex]
[tex]\(\therefore \sim p \)[/tex]

This is the contrapositive argument. This states that if [tex]\( q \)[/tex] is not true, then [tex]\( p \)[/tex] is not true. This argument also doesn’t align with our premises.

Option C:
[tex]\( p \rightarrow q \)[/tex]
[tex]\(\sim p \)[/tex]
[tex]\(\therefore \sim q \)[/tex]

This is the converse fallacy and is not logically valid for our situation, as it suggests that not having a polygon with exactly three sides implies not having a triangle, which isn't what we need.

Option D:
[tex]\( p \rightarrow q \)[/tex]
[tex]\( p \)[/tex]
[tex]\(\therefore q \)[/tex]

This is the direct application of modus ponens, a valid logical argument form. Since our premises match this form:

1. If Jeri drew a polygon with exactly three sides ([tex]\( p \)[/tex]), then Jeri drew a triangle ([tex]\( q \)[/tex]) - [tex]\( p \rightarrow q \)[/tex].
2. Jeri drew a polygon with exactly three sides ([tex]\( p \)[/tex]).

Therefore, we can conclude:
3. Jeri drew a triangle ([tex]\( q \)[/tex]).

Hence, the correct answer is:

Option D:
[tex]\( p \rightarrow q \)[/tex]
[tex]\( p \)[/tex]
[tex]\(\therefore q \)[/tex]
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