# Pythagorean Theorem intuition [closed]

I actually have a degree in Pure Mathematics, and this has always bugged me. Like a tiny, hardly-noticeable stone in your shoe you just can't get out.

Despite being at the core of geometry, arithmetic, and infinity, despite being able to prove it a dozen different ways, and despite it being the first "formula" anyone learns and one of the oldest known mathematical ideas - I can only see shadows and silhouettes of its truth.

Looked at from the point of view of geometry, I can see the statement plainly as a statement regarding the intrinsic notions of angle and area (using the proof of similar right-angle triangles) - however, the last step requires a correspondence between the notions of length and area to make complete (namely, the area of a shape increases quadratically as its scale increases linearly). This has always bothered me, since, the Theorem is an idea solely between angle and length.

There are, of course, simple proofs that don't depend on the notion of area (ex: Pythagorean Theorem Proof Without Words 6), however, sadly, they don't further my intuition, despite being clever/cute.

Staring at the statement blank in the face, I understand it far less intuitively then I do half of the ideas in algebra, analysis, number theory, etc.

## closed as unclear what you're asking by tilper, Matthew Towers, астон вілла олоф мэллбэрг, user91500, SharkosFeb 8 '17 at 10:38

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• Maybe if you see it as a theorem of algebra ? Like, if $E$ is a vector space with a scalar product which I will denote with $<.,.,>$, then any two vectors $x,y\in E$ such that $<x,y> = 0$ also satisfy $||x|| ^2 + ||y||^2 = ||x+y||^2$, with $||z|| := \sqrt{<z,z>}$. This equation is trivial, and it is the Pythagorean theorem. To me, this is the intuition behind it – Max Feb 6 '17 at 15:42
• What's bothersome about appealing to the notion of area for a theorem about angles and length? I'm surprised that would bother anyone with a pure math degree since such a person must have encountered several such proofs (i.e., proofs using ideas not found in the statement). Also, $a^2 + b^2 = c^2$ can be interpreted as a statement about areas since all three terms represent the areas of squares whose sides coincide with the sides of the triangle. Also, what are you asking exactly? For an intuitive way of thinking about the Pythagorean theorem? – tilper Feb 6 '17 at 15:46
• @tilper Please re-read and then kindly edit or remove your comment – Zach Feb 6 '17 at 15:51
• @Max, $\langle$ and $\rangle$ are \langle and \rangle. And $\cdot$ is \cdot. – tilper Feb 6 '17 at 16:33
• @Max : If you right-click on the notation, you get a menu in which the first item says "Show Math As", and then you choose "TeX Commands" and you'll see the code. And if you can click on "edit" (as in the case of a question or an answer, if not of a comment) then you can see the code as well. – Michael Hardy Feb 6 '17 at 16:58

## 2 Answers

I wonder if some way of looking at Einstein's proof of the Pythagorean theorem might help.

First recall that the areas of figures of the same shape are proportional to the squares of the distances involved. For example, if you have two regular pentagons, and the length of the side of one of them is $6$ times the length of the side of the other, then its area is $36$ times the area of the other.

Now look at the right triangle. Suppose the angles other than the right angle are $\theta^\circ$ and $(90-\theta)^\circ.$ Drop a perpendicular from the vertex of the right angle to the hypotenuse, splitting the triangle into two parts. The area of the original right triangle is the sum of the areas of those two parts. One of those two parts shares the angle of $\theta^\circ$ with the original triangle and has a right angle at the foot of the perpendicular that you dropped. It thus has two angles in common with the original triangle; therefore all three in common; hence it has the same shape as the original triangle, and its hypotenuse is one of the legs of the original triangle. Similarly the other part has the same shape as the original triangle and its hypotenuse is the other one of the legs of the original triangle. You've split the triangle into two triangles each having the same shape as the original triangle. And areas are proportional to the squares of distances.

• Yes, this is precisely the way I see the theorem; however, as you pointed out, it depends on the correspondence between the notions of length and area (at the very least, restricted to rectangular shapes). Actual work needs to be done in establishing that. Looking at it as a statement about adding areas, it is plain enough - but doesn't directly map into an understanding regarding length and angle. When viewed through algebraic geometry, the distinction never really occurs; but viewed purely from geometry, as the Greeks would have, it's not a given – Zach Feb 6 '17 at 17:25
• Though I don't have the historical knowledge to know if they had the sophistication at the time to establish the correspondence rigorously – Zach Feb 6 '17 at 17:28

The Pythagorean Theorem does seem to follow nicely from the similarity of right triangles formed within a right triangle by dropping a perpendicular from the right angle to the hypotenuse, and the fact that similar polygons are as the squares on corresponding sides (@Michael Hardy). These are Props. 8 and 20 in Book VI of Euclid's Elements. However he proves the PT in I,47 relying mainly on congruency of triangles (I,4) and the fact that a parallelogram is double a triangle on the same base and under the same height (I,41). The latter rests on I,37 (triangles on same base under same height are equal), which Newton said was the first thing in Euclid not obvious to him. Euclid I,47 makes no appeal to area increasing as the square of length, since his theory of proportion doesn't come until Book V, followed by its geometric application in VI. The "proof without words" (@Zach) seems to rest on Euclid VI,8, which does rest on words. Do you find even Euclid I,47 less than satisfying as proof of the Pythagorean Theorem?

• Thanks for the detailed reply! I have a copy of the Elements laying around somewhere (digitally ;) ). Let me find the relevant notions and get back to you! – Zach Feb 6 '17 at 19:04
• After reading Elements I,47 - I'm still left disappointed :(. Although a clever way to equate areas of the squares, it results in a statement equating areas, and requires VI,20 (trusting your reference) to make complete. My feeling (as of Max's comment above) is that the inspiration may come from the necessary constraints required of an inner product in order to introduce a notion of distance and angle through algebra. Perhaps the true essence is distilled and brought to light there. – Zach Feb 6 '17 at 19:19
• @Zach-How is VI,20 required? Each of the two smaller squares proves equal to one of two rectangles, and the two rectangles make the square on the hypotenuse. No proportion involved. – Edward Porcella Feb 6 '17 at 19:46
• I didn't lookup what V1,20 was. What I,47 shows is "the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides". You still need something to say the area of a square is proportional (equal) to the square of it's length. – Zach Feb 7 '17 at 14:27
• This may seem "obvious", but its quite subtle. The measure of length is distinct from the measure of area; two different notions we are able to perceive – Zach Feb 7 '17 at 14:31