Does this simple proof-by-contradiction, also require contrapositive?

Question

Simple exercise 6.2 in Hammack's Book of Proof. "Use proof by contradiction to prove"

"Suppose $n$ is an integer. If $n^2$ is odd, then $n$ is odd"

So my approach was:

Suppose instead, IF $n^2$ is odd THEN $n$ is even

Alternatively, then you have the contrapositive, IF $n$ is not even ($n$ is odd), then $n^2$ is not odd ($n^2$ is even).

$n = 2k+1$ where $k$ is an integer. (definition of odd)

$n^2 = (2k+1)^2$

$n^2 = 4k^2 + 4k + 1$

$n^2 = 2(2k^2 + 2k) + 1$

$n^2 = 2q + 1$ where $q = 2k^2 + 2k$

therefore $n^2$ is odd by definition of odd.

Therefore we have a contradiction. Contradictory contrapositive proposition said $n^2$ is not odd, but the derivation says $n^2$ is odd. Therefore the contradictory contrapositive is false, therefore the original proposition is true.

Not sure if this was the efficient/correct way to prove this using Proof-By-Contradiction.

Answer 1

The contrapositive of the original statement i.e. "If $n$ is even the $n^2$ is even" is easy to prove.

Let $n=2k$, $k\in\mathbb Z$ then $n^2=(2k)^2=2(2k^2)$ is even.

Answer 2

To prove $$ n^2\text{ is odd}\implies n\text{ is odd}\tag{1} $$ by contradiction, you need to prove that $$ n^2\text{ is odd}\wedge n\text{ is even}\tag{2} $$ is false. That is, you need to suppose that $n^2$ is odd and that $n$ is even and obtain a contradiction from those two statements.

This method of proof becomes clearer when the implication $$ n^2\text{ is odd}\implies n\text{ is odd} $$ is written in a logically equivalent way as $$ \neg((n^2\text{ is odd})\wedge\neg(n\text{ is odd}))\tag{3} $$ The proof by contradiction assumes the negation of the statement and obtains a known contradiction from it. In this case, you see that the negation of $(3)$ is $(2)$.

You propose to show $$ n^2\text{ is odd}\implies n\text{ is even}\tag{4} $$ is false in order to show $(1)$. That is incorrect.

For example, one could prove that $$ x>0\implies\sin(x)\geq0 $$ is false and yet $$ x>0\implies\sin(x)<0 $$ is also false.

In fact, what you did is show the converse of $(1)$. That is, you showed $$ n\text{ is odd}\implies n^2\text{ is odd} $$

In this case, in order to prove $(1)$, a proof of its contrapositive is the simplest way to go. Indeed, if $n=2k$ is even, then $n^2=(2k)^2=2(2k^2)$ is even. Here there is no real difference between the proof by contradiction and the proof by contrapositive: the hypothesis that $n^2$ is odd in $(2)$ doesn't need to be used.

Answer 3

Last night I read this as a perfectly acceptable claim, but, as has been pointed out, your negation was not simply a "harder case" but instead the converse. Apologies! Your proof happened to work here because the stronger relationship (i.e. $\iff$)

A couple of things to note though. For a proof by contradiction would simply need to state that: there exists some $n^2$ which is odd, which has an even $n$ (perhaps phrased better: Suppose that you have $n^2$ which is odd for a corresponding $n$ even). With this in mind, consider the following direct proof by contradiction.

Assume that for some $n^2$ which is odd, we have $n = 2k$.

\begin{align} n^2 &= n\cdot n\\ &= (2k)(2k)\\ &= 2(2k^2) \\ 2k^2 &\in \mathbb{Z} \ \ \ \ \text{call it q} \\ n^2 &= 2q \implies 2 | n^2 \end{align}

and so we have reached a contradiction, so our assumption must be incorrect!

I do want to stress that, as with most propositions, there are multiple ways in which to prove this statement, as has been pointed out using the contrapositive is the most efficient way of proving it, but as the exercise asked to use a contradiction method, the above would work!

Answer 4

Your proof looks correct to me, but I would like to share with you my strategy for proof by contradiction.

Consider the if/then statement $p\Rightarrow q$. In your case, $p$ represents "$n^{2}$ is odd" and $q$ represents "$n$ is odd". To achieve the proof by contradiction, we want to show that when $p$ is true, then it is impossible for $\neg q$, that is to say $n$ is even, to also be true. So we will assume that $p$ and $\neg q$ are both true (i.e., at the same time).

Applying this to your particular problem, we have $n^{2}$ is odd and $n$ is even. By definition, $n=2k$ for some integer $k$. Then $$n^{2}=(2k)^{2}=4k^{2}=2\cdot(2k^{2}).$$ Since the integers are multiplicatively closed, we have that $n^{2}$ is $2$ times an integer. Hence $n^{2}$ is even, which contradicts the assumption that $n^{2}$ is odd.

This tells us two things:

1. The logical statement "$n^{2}$ is odd and $n$ is even" is a false statement.
2. The logical statement "$n$ is even implies $n^{2}$ is even" is a true statement. This is the contrapositive ($\neg q\Rightarrow\neg p$) of the original statement, which always has the same truth value as the original statement.