Lesson 13-4: Indirect Proof (Day 134)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

If the area of the larger circle is 80pi sq. units, the area of the inner circle is ____pi square units.

[Here's the additional given info: the radius of the larger circle is the diameter of the smaller circle.]

To solve this problem, we begin by writing:

pi R^2 = 80pi

where R is the larger radius. We notice that R = d = 2r, where r is the smaller radius. So we substitute this in:

pi(2r)^2 = 80pi
4pi r^2 = 80pi

Dividing both sides by 4:

pi r^2 = 20pi

And since pi r^2 is the desired area of the smaller circle, the area is 20pi. The question already supplies us with the pi, so all we need to fill in is 20 -- and of course, today's date is the twentieth.

Solving the problem this way enables us to avoid finding either R or r and the nasty square root of 20.

Lesson 13-4 of the U of Chicago text is called "Indirect Proof." In the modern Third Edition of the text, indirect proof appears in Lesson 11-3.

This is what I wrote last year about today's lesson:

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.

[2019 update: The eighth graders from yesterday's post had to answer the question true or false: "A triangle can contain two right angles." This was Question 12 on WS #85. Of course, the eighth graders didn't have to write a formal indirect proof.]

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.

Returning to 2019, let me post the second side of the worksheet that I posted yesterday. This lesson naturally leads itself to activity, but today isn't an activity day. So I omit the extra activity worksheets from last year.

This post is relatively short and sweet. Let me close this post with something interesting. Today, just before 3 PM, Pacific Time, was the spring equinox. Just after 3 PM, there was thunder, lightning, rain, and hail where I live. In Southern California, hail is relatively rare. But hail is how spring greets us today!