Sections 13-3 and 13-4: Ruling Out and Indirect Proof (Day 105)

Section 13-3 of the U of Chicago text is on Ruling Out Possibilities, and Section 13-4 of the text is on Indirect Proof.

Here are a few things that I want to point out. First of all, some texts refer to the Law of Ruling Out Possibilities in Section 13-3 by another Latin name, modus tollens. Here is a link to the Metamath reference to modus tollens.

http://us.metamath.org/mpegif/mt4.html

As we can observe in the proof at the above link, modus tollens is essentially modus ponens (The Law of Detachment) applied to the contrapositive (Law of the Contrapositive, or contraposition.)

Section 13-3 is another section that lends itself to an activity, since many of its questions are actually logic problems, like the ones that often appear in puzzle books. Therefore, most of my worksheet depends on Section 13-4, although I did sneak in one logic problem as a preview for the upcoming activity day on Friday.

But Section 13-4 is the big one. This section is on indirect proof. I've delayed indirect proofs long enough -- now is the time for me to cover them. Actually, indirect proofs aren't emphasized in the Common Core Standards, but they were in the old California State Standards, where they were known as "proofs by contradiction."

What, exactly, is an indirect proof or proof by contradiction, anyway? The classic example in geometry is to prove that a triangle has at most one right angle. How do we know that a triangle can't have more than one right angle? It's because if a triangle were to have two right angles, the third angle would have to have 0 degrees -- since the angles of a triangle add up to 180 degrees -- and we can't have a zero angle in a triangle. Therefore a triangle has at most one right angle.

And voila -- that was an indirect proof! Notice what we did here -- we assumed that a triangle could have two angles -- the opposite (negation) of what we wanted to prove. Then we saw that this assumption would lead to a contradiction -- a triangle containing a zero angle. Therefore the original assumption must be false, and so the statement that we wanted to prove must be true. QED

Indirect proofs are often difficult for students to understand. One way I have my students think about it is to imagine that they are having a dream. Normally, when one is dreaming, one can't tell that they are having a dream, unless something impossible happens, such as a pig flying in the background, or the dreamer is suddenly a young child again. I recently had a dream where I was suddenly younger again, and I was flying off the ground! Naturally, as soon as those impossible events happened, I knew that I was in a dream.

And so a proof by contradiction works the same way. We begin by assuming that there is a triangle with two right angles, and then we see our flying pig -- a triangle with a zero angle. And as soon as we see that flying pig, we know that we were only dreaming that there was a triangle with two right angles, because there's no such thing! And so all triangles really have at most one right angle. So an indirect proof is really just a dream.

We saw how an indirect proof was needed when we were trying to prove that there exists a circle through any three noncollinear points A, B, and C. The proof that such a circle exists requires an indirect proof to show that the perpendicular bisectors m of AB and n of BC actually intersect. The indirect proof goes as follows: assume that they don't intersect -- that is, that they are parallel. Then because, m is perpendicular to AB and parallel to n, by our version of the Fifth Postulate, AB must be perpendicular to n. Then, now that n is perpendicular to both AB and BC, by the Two Perpendiculars Theorem, AB and BC are parallel. But B is on both lines, so we must have, by our definition of parallel, that a line is parallel to itself -- that is, AB and BC are on the same line. But this contradicts the assumption that A, B, and C are noncollinear. Therefore the perpendicular bisectors m and n aren't parallel -- so that they actually exist.

On the worksheet, notice that ABCD being both a trapezoid and a square is not a contradiction, since by our definition of trapezoid, a parallelogram (and hence a square) is still a trapezoid.