You mention you're "okay with" (a) and (b):
(a) Just to double-check, you should have seen that it is not valid, as it asserts the converse of the first statement, $q\rightarrow p$, which is not equivalent to the first statement of the form $p \rightarrow q$.
- The inverse of $\;p\rightarrow q\;$ is $\;\lnot p \rightarrow \lnot q$, which is equivalent to (is the contrapositive of), the the converse of the implication. Neither the converse of an implication nor its inverse are equivalent to the original implication.
(b) is valid, as it states the equivalent in the form of the contrapositive of the first statement. That is, if we are given $p \rightarrow q$, then it is validly asserting $\lnot q \rightarrow \lnot p$.
c) If $n$ is a real number with $n>2$, then $n^2>4$. Suppose that $n≤2$. Then $n^2≤4$.
Let "$n$ is a real number with $n > 2$" be denoted by $p$.
Let "$n^2>4$" be denoted by $q$.
Then $n \leq 4 \equiv \lnot p$. The negation of "$n> 2$" is " n is not greater than 2, i.e., $n \leq 2$.
And $n^2 \leq 4 \equiv \lnot q$. Can you see why? The negation of "$n^2 > 4$ is "$n^2$ is not greater than 4", i.e., $n^2 \leq 4$.
The first statement in (c) is of the form $p\rightarrow q$. Then we have the assertion $\lnot p$, and the conclusion therefore $ \lnot q$.
$$[(p \rightarrow q) \land \lnot p] \not \rightarrow \lnot q$$
Hence, this represents the fallacy of denying the hypothesis:
- $p\rightarrow q$
- $\lnot p$
- $\therefore \lnot q$
which is not a valid inference.
The contrapositive of the first statement in (c) would be "If $n^2 \leq 4$, then $n \leq 2$, that is, if would be of the form $\lnot q \rightarrow \lnot p$.
Best Answer
It is called a fallacy because it is not a valid argument. What is a valid argument, then?
A valid argument is an argument where the conclusion can never be false when the premises are true.
Let's now build the truth table of your argument: \begin{array}{cc|ccc}p&q&p\to q&\neg p&\neg q\\\hline T&T&T&F&F\\T&F&F&F&T\\F&T&T&T&F\\F&F&T&T&T\end{array}
As you can see, there are two assignments of truth values to the sentences letters that make the premises true (last two lines of the table), yet only one of them also makes the conclusion true. Therefore, there is a case where the premises are true and the conclusion is false which, by definition, means that the argument is invalid.
Take a simple example, where $p$ means "it has rained" and $q$ means "the ground is wet". Assuming the first premise is true, i.e. "if it has rained, then the ground is wet", knowing $\neg p$, i.e. knowing that "it hasn't rained", isn't enough to conclude about whether "the ground is wet or not" (the ground might be wet for some other reason). This is the core idea of the fallacy.