22

Here is a simple construction. Adjust to taste. 1. Draw a line $l$ passing through a point O. 2. Construct circles of radius 3 and 4 with centre O. Call them $C_3$ and $C_4$. Let the intersection of $C_4$ and $l$ be A. (Note that OA is 4 units.) 3. Construct an interval of length 5 from A to $C_3$. Call it B. (Note that OB is 3 units, and AB is 5 units.) ...


21

The Chinese came up with the following a long time ago. Probably something better, but this is the gist of it. Let's start with a right triangle with height b=4 and base a=3. We know it has some hypotenuse, c, but we don't know it's length because that's what we're trying to prove. Now we're going to make 3 copies, and rotate them each 90°. Next we're ...


19

The fact that there is a 3-4-5 triangle that is a right triangle is unique to the Euclidean plane. There is no such triangle in the spherical or hyperbolic planes. Since the Pythagorean theorem is equivalent to the parallel postulate, any proof that a 3-4-5 triangle is a right triangle will somehow depend on the Pythagorean theorem/parallel postulate.


18

Graph paper (or square floor tiling) to the rescue! Proof by picture for a 3 4 5 triangle: Because the drawing is on the grid and not the skew tiling of the square on the hypotenuse, determining the area is not inscrutable. The blue square (it is a square by adding angles of the triangle) is area 25 by adding the four blue triangles (obviously 6 each) and ...


16

The best explanation I know comes from this answer by Emanuele Paolini (the only thing I did was to redo the pictures, please go and upvote Emanuele's post). The point is that squares in the usual picture doesn't have to be squares. The only thing is that we need to use the area (so that is scales with a square of the scale), it could have been pentagons. ...


13

Diagram shows that there exists a 345 triangle that is right-angled. It is clear by inspection that an angle greater than 90 between 3 and 4 leads to a hypotenuse longer than 5. Similarly an angle less than 90 leads to a hypotenuse of less than 5


11

A complete different approach reasoning with the area $A$ of the triangle 3-4-5. Use Heron's Formula to show that $A =6$. Conclude that the height to the side with length 4 must be 3 since $A = ah/2$. Thus, the length of the side (3) equals the length of the height, which finishes the proof, since the height is perpendicular by definition. The crucial step ...


10

A number of proofs (43!) can be found at cut-the-knot.org. Some of these are described below. I don't know how hands on you want the students to get with the visual aspect. I remember doing these proofs around that age. I recall this progression and it made sense to me why Pythagoras's Theorem was true. Geometrical demonstration: Pythagoras's Theorem can ...


10

First copy your rectangle like this to make a big square of side-length m with a square of side-length d drawn inside it. The big square minus the small square leaves four half-rectangles (coloured pink in this picture). Half of the pink area is two half-rectangles, and so is the area of one whole rectangle.


6

I don't think I can do better than Giles answer, but here is an answer which gets the converse of PT without proving PT first: Let $AB=3$, $BC=4$ and $AC=5$. Draw a point $D$ on the line segment $AC$ with $AD=9/5$ and $DC=16/5$. Then triangles $ABC$ and $ADB$ are similar, since they have the same angle at $A$ and proportional sides. So $\angle ABC = \angle ...


6

It seems the Stanford Encyclopedia of Philosophy has a useful article that at least tangentially addresses your interesting question: "Can visual thinking lead to discovery of an idea for a proof in more advanced contexts? Yes. Carter (2010) gives an example from free probability theory. [...] Reflection on a diagram such as Figure 9 does the work." ...


4

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4

You might find the essay by Marjorie Senechal, "Visualization and visual thinking." Geometry's Future (1991): 15-21, published by the Consortium for Mathematics (COMAP), of interest. In particular she addresses the distinction between "visual thinking" and "visualization." (You can also find articles that cite this work by searching on "Google Scholar"; ...


4

First, I am a great fan of Visual Complex Analysis. Nevertheless, I disagree with the statement that Mathematics today [...] is mostly built on abstract symbolic manipulation rather than on [...] visual (intuition) Even the areas of mathematics that do not easily lend themselves to visualization—say, abstract algebra, or mathematical logic—...


4

If you have Pick's formula at your disposal, you can draw your favourite right triangle on grid paper and count. Actually, you can do the counting for any triangle with grid point vertices, but of course (by what we know) we get equality iff the triangle is a right one.


4

According to the Wikipedia entry for Pythagorean theorem, a proof of the converse of the Pythagorean theorem without assuming the Pythagorean theorem can be found in Stephen Casey, "The converse of the theorem of Pythagoras," The Mathematical Gazette, Vol. 92, No. 524 (July 2008), pp. 309-313.


4

Not an answer; rather an observation. If you can convince that the blue circle of radius $3$, and the black circle of radius $4$, meet the red circle of radius $\frac{5}{2}$ at the same point, then the right angle follows from Thales' Theorem.           But I have to admit, I don't see how to intersect the circles without using ...


4

Check out the book "Proofs Without Words" for a bunch of nice visual proofs of the Pythagorean Theorem. http://www.maa.org/publications/books/proofs-without-words


3

Maybe you can do President Garfield's proof and combine it with a history lesson. page 161 of the New-England Journal of Education, April 1, 1876 (image from Google Books) Note: M. C. = Member of Congress.


3

Perhaps this YouTube video on Visualizing Group Theory, by Nathan Carter (Bentley Univ), may help. Here is a snapshot illustrating $C_3 \times C_4$:           Blue represents the cyclic group $C_4$ horizontally, and red the cyclic group $C_3$ vertically. Nathan is the author of the wonderful book, Carter, Nathan. Visual group ...


3

Here is one of my favorite visualizations of the Pythagorean Theorem: There is also a fun activity here. Have the kids draw two adjacent squares of any size of their choosing, and use the technique shown above to have them cut the squares into 5 pieces that can be arranged to form one larger square. The fun part about this is that the kids get do ...


3

Here's a link to a Google+ posting about exactly this. In addition to additional links, it has a GIF using water to demonstrate the equalness of squares.


2

For complex variables, there is the famous "visual" book Tristan Needham, Visual Complex Analysis


2

Cut out four identical 345 triangles and put the biggest angle from each triangle together at a point. Then observe that all 4 identical angles fit to make a revolution, which implies each of the angles is a right angle. A picture is provided below. More work would be required to prove this construction, but hopefully this is the level of proof you were ...


2

Here is a Java applet showing Euclid's proof: http://neil-strickland.staff.shef.ac.uk/courses/MAS100/pclock.html


2

to illustrate $A \times B,$ draw two orthogonal axis and mark the points of $A$ on one and the points of $B$ on the other. the elements of $A \times B$ are the points where lines through points on the axes meet. or is this too obvious and adequate for the purposes.


2

The slides of Dror Bar-Natan are excellent examples of such mathematical "infographics".


1

In addition to Needham's famous book mentioned in Gerald Edgar's answer, there is Visual Complex Functions: An Introduction with Phase Portraits by Elias Wegert. The book is richly illustrated with colour diagrams of complex functions, especially the "phase portraits" mentioned in the title, as can be seen on the author's website for the book.


1

When I was originally taught groups, one concrete example was the symmetry transformations of an object, like an equilateral triangle ($S_3$) or a rectangle ($V_4$). Usually you label or colour-code the corners so you can tell the difference between them. One easy way to form a Cartesian product would be to have transformations of both object simultaneously....


1

Usual bubbles can be freezed. See Bubbles freezing at -26°C and Frozen Bubbles 01-24-11. According to WikiHow no special bubble mix is required. Obvious idea: use very cold water to minimize the required time.


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