I was looking for examples of Proof by contradiction to teach my A-Level students and chanced upon this website:http://www.personal.kent.edu/~rmuhamma/Philosophy/Logic/ProofTheory/proof_by_contradictionExamples.htm
I don't believe that Example 4 (appended below) is proof by contradiction and I hope that someone is able to concur. I think the method used is actually Proof by Contrapositive where the author shows that (not Q) implies (not P) and therefore P implies Q. Proof by Contradiction should entail supposing that the contrary is true at the beginning and then hope to find a contradiction along the way. The contrary would be if $n^2$ = odd then n is even.
Going with this idea, if we let $n^2$ = 2p + 1 and let n = 2q where p, q are integers, there must exist some p and q such that $$(2q)^2=2p+1$$ $$4q^2=2p+1$$ $$p=2q^2-1/2$$ The right hand side expression is not an integer (given the 1/2) and therefore p is not an integer. We arrive at a contradiction since we said at the beginning that p and q are integers. The contrary is shown to be false and therefore the original premise must be true ie if $n^2$ is odd then n is odd.
Is my argument correct?
Example 4: Prove the following statement by contradiction: For all integers n, if $n^2$ is odd, then n is odd.
Proof:
Suppose not. [We take the negation of the given statement and suppose it to be true.] Assume, to the contrary, that ∃ an integer n such that $n^2$ is odd and n is even. [We must deduce the contradiction.] By definition of even, we have
n = 2k for some integer k.
So, by substitution we have
n . n = (2k) . (2k)
= 2 (2.k.k)
Now (2.k.k) is an integer because products of integers are integer; and 2 and k are integers. Hence,
n . n = 2 . (some integer)
or $n^2$ = 2. (some integer)
and so by definition of $n^2$ even, is even.
So the conclusion is since n is even, $n^2$, which is the product of n with itself, is also even. This contradicts the supposition that $n^2$ is odd. [Hence, the supposition is false and the proposition is true.]
Your proof by contradiction is fine (except that it contained a typo that I fixed in an edit).
But also the proof appended below can be classified as a proof by contradiction.
The other starts with "suppose not" which means that an integer $n$ must exist with: $$n^2\text{ is odd and }n\text{ is even }\tag1$$
Then he reaches the conclusion that:$$n^2\text{ is even}\tag2$$
It is evident that $(1)$ and $(2)$ cannot be true together so a contradiction is found.
The conclusion is that $(1)$ is false.
Actually both a proof by contradiction and a proof by contrapositive rest on the Boolean assumption that a statement is true or false and there is no third possibility (tertium non datur).