The question is, "Determine whether each of these arguments is valid. If an
argument is correct, what rule of inference is being used? If it is not, what logical error occurs?
a) If $n$ is a real number such that $n>1$, then $n^2>1$.
Suppose that $n^2>1$. Then $n>1$.
b)If $n$ is a real number with $n>3$, then $n^2>9$. Suppose that $n^2≤9$. Then $n ≤3$.
c) If $n$ is a real number with $n>2$, then $n^2>4$. Suppose that $n≤2$. Then $n^2≤4$.
The only one I am having difficulty with is part c). The answer to part c), is that it is a fallacy of denying the hypothesis. I thought, "perhaps by taking the contrapositive of the conditional statement, it might make things a little more lucid." However, I was wrong on that idea. If someone could explain part c), I'd be so quite grateful. Thank you!
Best Answer
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 $p \rightarrow q$.
(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 $\lnot q \rightarrow \lnot p$.
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$.
Hence, this represents the fallacy of denying the hypothesis:
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 $\lnot q \rightarrow \lnot p$.