# Should my 8th graders see a proof of the Pythagorean Theorem?

I've been teaching the Pythagorean Theorem in my 8th grade class, and I noticed something odd. In the book I'm using, the sequence goes something like this:

• Motivate the idea of distances on a grid
• Talk about areas on a lattice grid
• Work on finding the areas of irregular shapes -- including tilted squares
• Talk about the relationship between squares and side lengths, introduce irrationals and things like that.
• Look at squares coming off of a triangle. By now, students can find the area of all 3 squares coming off the right triangle, and derive the equation a^2+b^2=c^2 just by observation from a few example problems.

While I think this is very cool, it's not really a proof. They see a few example cases, and then the sum of the squares coming off of the legs of the triangle are exactly the same as the area of the square off the hypotenuse! And if we introduce labels a, b, and c instead of numbers, they can derive the Pythagorean Theorem this way. I noticed that a lot of teachers teach the Pythagorean Theorem this way. There are a lot of visual gifs and videos showing that the areas of the squares coming off the triangle add up, but is that technically a proof? At this point, the students are actually very close to seeing a more algebraic proof by the time they get to that last bullet point!

All they would need to do is take the tilted square (c^2) and find the area with respect to labels a and b.

There's two ways to do this. The "inside" way and the "outside way". The inside way is adding up all the inside areas using labels a and b. The outside way is taking the outer box and subtracting off the corner triangles. Using real numbers, this isn't very difficult, but using a, b, and c as the numbers is a huge "cognitive leap" for these students. So at the end of the day, my question is: Should I be leading students through an actual proof of the Pythagorean Theorem? Or is an exploration of deriving the formula good enough?

I hope this all made sense! If not, I'll do my best to clear things up.

• Related, but with no satisfactory answers yet: matheducators.stackexchange.com/q/10540 – Tommi Brander Mar 16 '16 at 14:27
• The wikipedia on Pythagorean theorem has a possibly neater picture of your proof. – Simply Beautiful Art Mar 18 '16 at 14:17
• You talk about doing "a more algebraic proof." Have these 8th-grade students even learned algebra yet? E.g., California recently gave up on its disastrous attempt to force all kids in public schools to take algebra in 8th grade. A proof can be general without being algebraic, e.g., Euclid never knew algebra. I assume a non-algebraic, general proof is what the authors of the common core standard have in mind. – Ben Crowell Jan 1 '17 at 19:57

If you are in the United States, at a public school, then you should explain a proof because this is one of the common core state standards: http://www.corestandards.org/Math/Content/8/G/B/6/

I would suggest using the following proof: use a line perpendicular to the hypotenuse and passing through the vertex of the right angle. This divides the triangle into two similar right triangles. Using this similarity, one can prove the theorem.

This proof is somewhat preferable to others because it connects to the larger story of similarity, instead of being a one off proof.

Even if common core were not an issue, I think that mathematics is about making sense of patterns. Surely you should investigate, and notice the pattern first. Just noticing the pattern is not enough: math is not an experimental science. We should seek to understand why the pattern holds. So yes, we should try and find an explanation (a proof).

• Oh, very interesting. I'll have to look at a proof done this way. I guess I'm surprised my book doesn't go through a more formal proof if it's a common core standard... And I think you're right. The only good counter-argument I've heard is that time is an issue, and they'll see a proof later on. To go through a proof would take more time away from other standards, and these students would ideally go through a formal proof when they take Geometry in high school. – Wmol Mar 16 '16 at 14:59
• @Wmol Here is a video of this proof: khanacademy.org/math/basic-geo/basic-geo-pythagorean-topic/… – Steven Gubkin Mar 16 '16 at 19:34
• @Wmol Perhaps even more interesting, however, is that the decomposition alone proves the theorem without any algebra. Since each triangle is similar, each is in the same proportion to the square of their hypoteni. Since the areas of the two smaller sum to the area of the original, it must also be the case that the sum of the areas of the squares of the legs is equal to the square of the hypotenuse. – Steven Gubkin Mar 16 '16 at 19:37
• Wow, I gotta say I'm surprised that I haven't seen that proof before...and I love geometry! Loved it in high school and loved it again when I took non-euclidean/euclidean geometry in my math undergrad. I'm a little confused by how the decomposition proves the theorem though. I definitely believe it, but I'm confused as to why the sums of the squares off the two smaller hypoteni are equivalent to the square off the original hypotenuse. I mean it's clearly true, but I can't figure out why via this proportional reasoning. I blame my exhaustion! – Wmol Mar 16 '16 at 21:20
• @StevenGubkin I love this proof, but it's not clear to me that students in 8th grade have a sufficient grip on 'area scaling as a square of the expansion factor' for it to be convincing / instructive. Have you taught it to students of this age? – NiloCK Jan 1 '17 at 21:34

I think an important aspect of this question -- one that I don't think has been mentioned yet in the other answers -- is the verb "see", as in "Should my 8th grader see a proof". While "Explain a proof of the Pythagorean Theorem and its converse" is indeed one of the Common Core Standards (and thanks to Steven Gubkin for providing the link in his answer) it's important to notice that the Standards describe what students should be able to do, not what students should see the teacher do.

I mention this because it is common for teachers (at all levels!) to think that if we show student a proof, they now know the proof. I think that is a fallacy and a dramatic oversimplification of how students actually learn the skills of proof and reasoning. If you want your students to learn to prove the Pythagorean Theorem, you have to engage them in the activity of proving so that they construct the argument themselves. That requires scaffolding and careful planning that goes beyond the question of "What proof do I show?"

I think the usual visual proof of PT uses something like:

(From Math Is Fun website)

Make cardboard cutouts of the outer and inner squares and the 4 identical triangles. Fit them together as shown.

Comparing areas, you can obtain:

(a+b)^2 = c^2 + 4(ab/2)

Getting rid of the brackets and cancelling 2ab from bothsides, you obtain as required:

a^2 + b^2 = c^2

Maybe one student in the class will "get it?" Might be worth it, though, to light a fire under that one student. Make it quick though. I would think that most students at that level won't get it no matter how long you take, but it may help them remember the formula if your presentation is sufficiently animated.

• Ahh yeah this is the proof I was talking about. Specifically the outside method. I think you're right, maybe a few students would get it. It could be worth it! I just don't know when to throw in things like this. A lot of my classes are 35 students each, so teaching for 1-3 students wastes the time of the other 32 students. It's a tough balance. Now that I think about it, maybe this would be an appropriate topic for Algebra. It might be tough for pre-algebra students, but it could be appropriate and more meaningful in algebra 1. – Wmol Mar 16 '16 at 18:22

A broader answer regarding the teaching of the Pythagorean Theorem (including the proof):

http://www.magicalmaths.org/could-this-be-the-best-pythagoras-theorem-lesson-ever/

And an image from the proof -

• Could you please expand this post to give the gist of the linked source. – quid Mar 28 '16 at 23:08
• I find the original proof by rearrangement simpler than the one that you posted, also this one is not proven on the source website, and the numerical values for the sides may confuse students into thinking that this works only for specific side lengths. One of the pillars of geometry course is that it is taught with nothing more than a ruler (without markings) and a pair of compasses. As much as I like using squared paper for almost everything, in this case squared paper is not appropriate. – Rusty Core May 10 at 16:16

Actually the most 'intuitive' proof of the Pythagoras theorem involves cutting the big square into only 3 pieces and reassembling them into the two smaller squares (using merely translations). Absolutely no subtraction is involved! (See the last section below for why subtraction of area is a nontrivial issue.)

For students

(Cut the green square along the red line and reassemble into the yellow and blue squares.)

Note that the proof is still not that simple. I think the easiest way is as follows. Start with the yellow and blue squares, then construct the point at the top and the 4th point at the bottom by translations of the appropriate points. Now use translations to prove that the green shape has equal sides, and use the four congruent triangles to prove that it has equal angles, and hence it is a square. The four congruent triangles also show that the area of the green square is the same as the total area of the yellow and blue squares.

For teachers

All that is used here is scissors-congruency, which for polygons turns out to be equivalent to having the same area. (It is not equivalent for polytopes in general.) But what is area in the first place? In modern mathematics we could just use the Lebesgue measure, but arguably there is a more natural definition that extends the idea of scissors-congruency, which turns out to yield the Jordan measure and yet is not hard for a bright student to grasp.

Specifically we say that $A \precsim B$ iff $A$ can be cut into finitely many pieces along simple curves and reassembled by rigid motions (translation or rotation or reflection) to fit within $B$ such that none of the pieces overlap (except perhaps at boundaries). Now for each non-negative real $x$ choose some fixed rectangles $R_x$ with length $x$ and width $1$. Finally we define that $A$ has area $x$ iff ( both $R_y \precsim A$ for any non-negative real $y$ such that $y < x$, and $A \precsim R_z$ for any non-negative real $z$ such that $x < z$ ).

(Remark: If we naively try to define that $A$ has area $x$ iff $R_x \precsim A \precsim R_x$, then it will non-trivially work for polygons but will even more non-trivially fail for the circle! That is why we have to relax the condition via the notion of limits.)

It is now easy to establish all the standard properties of area, except for one, namely that $R_x$ has area $y$ iff $x=y$. This property is essentially the same as the fact that if two shapes have the same area and we remove a region of the same shape from each then the two resulting shapes still have the same area. Since many proofs of the Pythagoras theorem use this fact, there is thus a legitimate reason to prefer a proof that does not use it.

Once we have this last property, we can easily show that any region that has area has a unique area. To see why, take any region $A$ that has areas $x,y$ where $x < y$. Let $z = \frac12(x+y)$. Then trivially $R_t \precsim R_z$ for any non-negative real $t < x$. Also $R_z \precsim A \precsim R_t$ for any non-negative real $t > x$, the first inequality because $z > y$. Thus $R_x$ has area $z$ and hence a contradiction.

For completeness, here is a sketch of a proof of that last property. Take any measurable shape $A$ that has area $x$. Then $x \le |A|$, by additivity and monotonicity of the Lebesgue measure since there is a finite partition of $R_x$ into measurable regions that fit within $A$ without any overlap. Hence $R_x$ cannot have area $y$ if $x < y$. Maybe this is cheating a bit, but I do not see an easier way.