Janessa Olson

2022-07-07

Non-geometric Proof of Pythagorean Theorem

Is there a purely algebraic proof for the Pythagorean theorem that doesn't rely on a geometric representation? Just algebra/calculus. I want to TRULY understand the WHY of how it is true. I know it works and I know the geometric proofs.

Is there a purely algebraic proof for the Pythagorean theorem that doesn't rely on a geometric representation? Just algebra/calculus. I want to TRULY understand the WHY of how it is true. I know it works and I know the geometric proofs.

Jenna Farmer

Beginner2022-07-08Added 17 answers

The "modern" approach is this : first we define the field $\mathbb{R}$ (for instance, it's the only totally ordered field with the supremum property).

Then we define what a $\mathbb{R}$-vector space is : it's an abelian group with an external action of $\mathbb{R}$ satifying some axioms.

Then there is a notion of dimension : we can define a vector space of dimension 2.

The notion of Euclidean distance is obtained by defining what an inner product is : it's a symmetric bilinear form such that $\u27e8x,x\u27e9>0$ if $x\ne 0$. The distance is then $||x||=\sqrt{\u27e8x,x\u27e9}$. We also have the notion of orthogonality from this inner product.

Well, once you did that, then the Pythagorean theorem is a triviality : $||x-y|{|}^{2}=\u27e8x-y,x-y\u27e9=\u27e8x,x\u27e9-2\u27e8x,y\u27e9+\u27e8y,y\u27e9=||x|{|}^{2}+||y|{|}^{2}$ (assuming of course that $\u27e8x,y\u27e9=0$, the orthogonality hypothesis).

Of course all the work went into the definitions, which is contrary to the basic approach of geometry which deduces properties of distance from a set of axioms (generally ill-defined, but it can be made precise with a little work).

The interesting thing about this modern approach is that algebraic structures come before geometric content. This is powerful because algebraic structures have enough rigidity. For instance, if you start with a set of points and lines satisfying some incidence axioms, it's very hard to define what it means that it has a certain dimension. But if you have a vector space structure, then it's easy.

Of course it can be a little disappointing beacause it feels like we "cheated" : we made the theorem obvious by somewhat putting it in the definitions. But on the other hand, it's very clear and precise : can you properly define what distance or an angle is using "high school geometry" ? Not so easy. Even in Euclide's Elements, this is kind of put under the rug as "primitive notions". This approach makes everyting perfectly well-defined and easy to work with.

Then we define what a $\mathbb{R}$-vector space is : it's an abelian group with an external action of $\mathbb{R}$ satifying some axioms.

Then there is a notion of dimension : we can define a vector space of dimension 2.

The notion of Euclidean distance is obtained by defining what an inner product is : it's a symmetric bilinear form such that $\u27e8x,x\u27e9>0$ if $x\ne 0$. The distance is then $||x||=\sqrt{\u27e8x,x\u27e9}$. We also have the notion of orthogonality from this inner product.

Well, once you did that, then the Pythagorean theorem is a triviality : $||x-y|{|}^{2}=\u27e8x-y,x-y\u27e9=\u27e8x,x\u27e9-2\u27e8x,y\u27e9+\u27e8y,y\u27e9=||x|{|}^{2}+||y|{|}^{2}$ (assuming of course that $\u27e8x,y\u27e9=0$, the orthogonality hypothesis).

Of course all the work went into the definitions, which is contrary to the basic approach of geometry which deduces properties of distance from a set of axioms (generally ill-defined, but it can be made precise with a little work).

The interesting thing about this modern approach is that algebraic structures come before geometric content. This is powerful because algebraic structures have enough rigidity. For instance, if you start with a set of points and lines satisfying some incidence axioms, it's very hard to define what it means that it has a certain dimension. But if you have a vector space structure, then it's easy.

Of course it can be a little disappointing beacause it feels like we "cheated" : we made the theorem obvious by somewhat putting it in the definitions. But on the other hand, it's very clear and precise : can you properly define what distance or an angle is using "high school geometry" ? Not so easy. Even in Euclide's Elements, this is kind of put under the rug as "primitive notions". This approach makes everyting perfectly well-defined and easy to work with.

The distance between the centers of two circles C1 and C2 is equal to 10 cm. The circles have equal radii of 10 cm.

A part of circumference of a circle is called

A. Radius

B. Segment

C. Arc

D. SectorThe perimeter of a basketball court is 108 meters and the length is 6 meters longer than twice the width. What are the length and width?

What are the coordinates of the center and the length of the radius of the circle represented by the equation ${x}^{2}+{y}^{2}-4x+8y+11=0$?

Which of the following pairs of angles are supplementary?

128,62

113,47

154,36

108,72What is the surface area to volume ratio of a sphere?

An angle which measures 89 degrees is a/an _____.

right angle

acute angle

obtuse angle

straight angleHerman drew a 4 sided figure which had only one pair of parallel sides. What could this figure be?

Trapezium

Parallelogram

Square

RectangleWhich quadrilateral has: All sides equal, and opposite angles equal?

Trapezium

Rhombus

Kite

RectangleKaren says every equilateral triangle is acute. Is this true?

Find the number of lines of symmetry of a circle.

A. 0

B. 4

C. 2

D. InfiniteThe endpoints of a diameter of a circle are located at (5,9) and (11, 17). What is the equation of the circle?

What is the number of lines of symmetry in a scalene triangle?

A. 0

B. 1

C. 2

D. 3How many diagonals does a rectangle has?

A quadrilateral whose diagonals are unequal, perpendicular and bisect each other is called a.

A. rhombus

B. trapezium

C. parallelogram