You can't assume the conclusion

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When we prove that \(\sqrt{2}\) is irrational, we need to show the following:
For any \(m, n \in \mathbb{Z}\), if \(m^2 = 2n^2\), then \(m\) is even.

The following "proof" is complete nonsense, but, believe it or not, I find many students doing something similar:

Suppose \(m\) is even. Then \(m = 2k\) for some \(k \in \mathbb{Z}\). Substituting this into \(m^2 = 2n^2\), we have \(4k^2 = 2n^2\). Comparing this with \(m^2 = 2n^2\), we have \(m^2 = 4k^2\), hence \(m = 2k\). This indicates that \(m\) is even. (Q.E.D.)

But what's exactly wrong? First of all, note that it starts with "Suppose \(m\) is even," which is the conclusion to be proved. In essence, what this "proof" says is that "if \(m\) is even, then \(m\) is even." Let \(Q\) denote the proposition "\(m\) is even." Then, this "proof" is essentially saying, "\(Q \implies Q\)." Let's see the truth table of this proposition: 

\begin{equation} \begin{array}{c|c} Q & Q \implies Q\\\hline \top & \top\\ \bot & \top\\ \hline \end{array} \end{equation}

That is, \(Q \implies Q\) is always true, whether or not \(Q\) is true. A proposition that is always true is called tautology. Tautology doesn't give us new information. In the present context, the proposition that "if \(m\) is even, then \(m\) is even" is true even if \(m\) is not even! 

By the way, there are other points that are wrong in the above proof. Try to spot them.

Here's the correct proof. We prove by contradiction:

Suppose \(m\) is odd. Then, \(m = 2k + 1\) for some \(k \in \mathbb{Z}\). Substituting this into \(m^2 = 2n^2\), we have \(4(k^2 + k) + 1= 2n^2\). The left-hand side of this equation is odd (\(4l + 1\) is odd where \(l = k^2+k\)), whereas the right-hand side is even. This is a contradiction. Therefore, \(m\) is not odd; that is, it is even. (Q.E.D.)

Let's examine this proof a little more formally.  Let \(P\) denote the proposition that \(m^2 = 2n^2\), and \(Q\) that "\(m\) is even." Then, we want to prove that \(P \implies Q\). But we know that this is equivalent to its contrapositive: \(\lnot Q \implies \lnot P\). Here \(\lnot Q\) means "not that \(m\) is even," that is, "\(m\) is odd." Similarly, \(\lnot P\) means \(m^2 \neq 2n^2\) (for any \(m, n\in\mathbb{Z}\)), from which it follows that \(m^2\) is not even. Indeed, in the above correct proof, we derived that \(m^2\) is odd by assuming that \(m\) is odd. Therefore, we have actually proved the contrapositive of the original proposition.




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