# Why is this type of reductio ad absurdum not taught more?

I recently read in a book about a proof that Archimedes did. I don't remember the exact details and I don't have the book on me right now, but it involved proving an equality. So let's say proving that $A=B$. In order to do this, he first assumed that $A>B$ and found a contradiction. He then assumed that $A<B$ and once again, found a contradiction. Therefore, he concluded, it must be that $A=B$.

I had never thought of this proof method or encountered it, so I thought it was very neat. Of course you're doing twice as much proving, if you like, but I feel like proving an inequality can often be easier than an equality because you can 'neglect' certain terms which don't effect the inequality.

I don't know if it is just the education I have had, but I am an undergraduate now and, like I said, I've never encountered this proof. I feel this is a shame because it would be a handy tool to know for any student. So my question is, why is this type of proof not used more often and taught?

The book was Journey through genius by William Dunham

• I'm not convinced that this is especially rare among methods of proof, but here is a thought: If the ultimate result is that $A = B$, then there may be some sort of symmetry in the proposition that can be exploited to avoid, as you describe it, "doing twice as much proving." Dec 27, 2016 at 0:53
• There is another form of proof which I think you might find cool: to prove that $A=B$, prove that both $A \le B$ and $A \ge B$ are true.
– Ovi
Dec 30, 2016 at 14:21
• @Ovi I guess this is similar to proving $A=B$ when talking about sets.
– user7607
Dec 30, 2016 at 14:53
• @Leonhard Yeah, nice connection!
– Ovi
Dec 30, 2016 at 15:43

This form of proof is quite common, especially in real analysis. Any time you are proving something about ordered fields -- and therefore have access to a law of trichotomy -- this form of argument is likely to come up.

Just to take one example, consider the set $S = \{x \in \mathbb{Q}^+ | x^2 \le 2 \}$. One proves that this set is bounded, but has no least upper bound; this demonstrates that $\mathbb{Q}$ is not complete. Then we move to $\mathbb{R}$, which has the completeness property, so the same set $S$ has a least upper bound $r \in \mathbb{R}$. We want to prove that $r^2 = 2$. So, first we assume that $r^2 > 2$, and derive a contradiction, which eliminates that possibility; then we assume $r^2 < 2$, and derive a contradiction, which eliminates that possibility. The only option left is that $r^2 = 2$.

Direct proofs are generally favored over proof by contradiction, or at least that's what I've been taught more than once. I think it's because it's usually simpler to do a direct proof, but of course each problem is different.

I don't think I've ever seen a proof by contradiction as you described to show that $A=B$ but I have seen several times the very similar approach of showing $A\le B$ and $A\ge B$ directly.

The argument I think you're referring to is a common "low level" argument — for practical purposes its applications are usually subsumed by "higher level" idea.

For example, one might use that sort of argument to prove the theorem

If $\small f(x) \leq g(x)$ for all $x$, then $\small \sup_x f(x) \leq \sup_x g(x)$. If the limits exist, then $\small\lim_{x \to a} f(x) \leq \lim_{x \to a} g(x)$

and this theorem is convenient to use for typical applications.

Another example of applying this idea is the approach to the fundamentals of integration by upper and lower Riemann sums. Then once you have the theory of integration, a wide class of problems where one might have used that argument form one instead sets up an integral.