Posted by: Prof. Kate | October 8, 2008

Leonhard Euler

On Tuesday evening there will be a public lecture sponsored by the Clay Mathematics Institute and hosted by Harvard’s math department.  It’s about Euler, a mathematician from the 1700s, who studied the sorts of things we study in Math 152.  Euler is the stuff of legend, dictating mathematical papers day and night after he lost his sight.  Today we are still reading the 25 000 pages of mathematics he produced.

One concept due to Euler, a precursor to a wealth of very rich and important ideas in mathematics, is the Euler characteristic of a general polyhedron (not necessarily just the regular ones we have been tossing around in class), which is to say, something made of vertices, edges and faces.

The Euler characteristic is easily defined as the number

vertices – edges + faces

For a simple polyhedron, that is one without holes (here are lots of examples), this number always comes out to two!  Try it on the five Platonic solids.  For example, a tetrahedron has 4 vertices, 6 edges and 4 faces:  4-6+4 = 2.  Actually, Euler only tried on simple polyhedra, and he proved that it was always equal to two.

But if you make a polyhedron like this one or this one (click for pictures), which has a hole in it, then you’ll find that the Euler characteristic comes out to something other than 2.  The Euler characteristic can be thought of as a way to detect holes in an object.  You can take any object you like, approximate it with a polyhedra full of holes (kind of like the animation in some video games), and the Euler characteristic will be related to the number of holes by the formula

euler characteristic = 2 x ( 1- the number of holes )

Try it!  Get out a pen, and grab your coffee mug or some other object with some holes, and “divide it up” into little faces, edges and vertices (make them small enough that you can imagine pushing them flat to make the object into a polyhedron).  Then count!


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  2. Apologize, I become to remember the euler formula exp(ipi)=-1. Maybe this formula can be used to answer how to prove one minus one equal zero or 1-1=0. Please correct me if I do a mistake

    Firstly , taking :

    1 + x = 0,

    to involve -1 then add -1 to the above both sides

    1-1 + x = -1

    Next, substitute x by exp(ipi)

    1-1 + (-1) = -1

    Finally we can prove that 1-1=0. But now I have a question, how to get 1-1 ? Thank for your help.

  3. @rohedi
    1-1 is not seem to be a simple question sir…..
    we are still confuse in understanding between “empty” and number of “null or 0” values.
    I guess more and more explanation from all netters here to discuss about it. thanks.

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