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This is the second installment in my Elegant Math series.  Today we'll look at the proof that there are an infinite number of primes.

We discussed prime numbers in a previous post.  And we showed how they are the building blocks for all other counting numbers in Tuesday's proof.  So we understand how important prime numbers are, but now the big question:  How many prime numbers are there?

Oh, a cliff-hanger!  Click here to find out the answer!

 
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One of the biggest stumbling blocks for many math students is the requirement that they show their work and write in proper mathematical notation.

This is simultaneously the area in which other students thrive: If you can follow directions, you will succeed in math class.

Students are told that mathematics is a language like French or Spanish.  This mostly just frightens the students, or bores them if we're lucky.

What students are not often told is why we use a special language for math and why we need to follow special rules.

Click here to find out why!

 
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We've done a lot of fun, number-symbol free math so far.  Sometimes, taking away all those symbols makes patterns easier to see (check out Counting Like an Ancient Greek), but mathematicians use symbols for a reason.  Symbols often make it much, much easier to talk about ideas in math.


So I'm going to show you one of the most elegant proofs from mathematics.  Now, don't get scared!  You don't need to know much math to understand the beauty of this proof, and I'll walk you through the basic ideas you need to know.

The ancient Greeks wanted to think that all numbers could be expressed as fractions.  They wanted the world to be neat and tidy, and fractions are just that: neat and tidy.  Unfortunately, they stumbled upon some numbers that didn't always seem to be able to be expressed as fractions: square roots.  We call these neat, tidy fractions rational numbers.  Even whole numbers like 5 can be expressed as fractions: 5 = 5/1.  The messy numbers that cannot be expressed as fractions are, understandably, called irrational numbers.

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Right triangle where base = 1 and height = 1.
Pythagoras was looking at triangles, and he wanted to find the length of the third side of a right triangle where two sides had length one.  Using the famous theorem named after him, the length of the third side can be found:

A^2 + B^2 = C^2 or
1^2 + 1^2 = 2 (I'm using little carrots to show "squared" because this program doesn't have many fancy fonts. 1^2 is "one squared.")

To find the length of the third side, we just need to take the square root of two.  Easy!  It's uh.... what number do we square to get two?  1^2 is 1, 2^2 is 4.  What squares to make two?  It must be between 1 and 2.  Well, Pythagoras was stumped, because he couldn't find a fraction between 1 and 2 that squared to make 2.

Before we move on, I want to quickly talk about factors here.  The factors of a number are numbers that divide into that number.  For example, 2 is a factor of 6 because 6/2 = 3 (and 3 is a whole number).  In fact, 2 is a factor of every even number but not a factor of any odd numbers.  That's one way to define even and odd numbers.  

We say that two numbers have no common factors if, well, they have no factors in common!  When we talk about fractions, we have a reduced fraction when the numerator and the denominator have no common fractions.

6/2 is not reduced because 6 and 2 have a common factor: 2.  We could write 6/2 as (3*2)/(1*2) or (3/1)*(2/2).  2/2 = 1, so we can just get rid of it, and we see that 6/2 can be written more simply as 3/1 or 3.  

If all those numbers freaked you out, don't worry about it.  Just read on.  The proof is better:

Because of the limitations of my text editor, we'll let root2 mean "square root of two" and c^2 will mean "c squared" as before.  It doesn't really matter if we use traditional notation for the proof.  The elegance will still be there.

So let's say, for the sake of argument, that Pythagoras was wrong, and there is a way to represent root2 as a fraction.  Then:
  • root2 = a/b where a and b have no common factors
  • 2 = (a/b)^2 I just squared both sides because I don't like having square roots in my equations.
  • 2 = (a^2)/(b^2) You'll have to trust me on this one (or look it up); we're not going to prove this step here.
  • 2*(b^2) = a^2 Just some quick algebra: I multiplied both sides by b^2.
  • 2*b*b = a*a That looks a little nicer, doesn't it? It's an unnecessary step in the proof, but I want to make this easy to read!
  • This means that a is a multiple of two (if a^2 is a multiple of two, a must be also--another thing I'll let you figure out on your own), so we could say a=2*m.  That means a*a = 2*m*2*m = 4*m*m, so:
  • 2*b*b = 4*m*m Let's simplify by dividing both sides by two...
  • b*b = 2*m*m
  • So b is also a multiple of two!
  • But that's a contradiction: a and b have no common factors.  Our proof started by setting a and b as numbers without common factors and ended by saying they were both even (and therefore had 2 as a common factor).
  • This means that there are no numbers a and b for which root2 = a/b. Since two numbers can't both be even and have no common factors!
  • Therefore, root2 cannot be expressed as a fraction.  In other words, it is irrational.
The above proof is a bit long-winded because I wanted to make sure I explained each step.  To see the proof in a more elegant form (using an equation editor), you can check out this PDF I put together.
root_2_proof.pdf
File Size: 112 kb
File Type: pdf
Download File

I hope that you can see the simple beauty of this proof.  It uses very basic facts about numbers--all you need to know is basic arithmetic and a tiny bit of algebra, and we can show something as powerful as the existence of irrational numbers!