Quick Math Proofs

Meta- Stuff

I'm not claiming any original material here, just things that I've gathered over the years that I find pretty interesting.

The Square Root of 2 is Irrational

This is a rather famous proof, so I'm not going to elaborate too much. However, it's a proof by contradiction, so we're going to start by assuming the converse is true and trying to reach a logical contradiction. Let's get started.

Suppose √(2) is rational. Let's say √(2) = a / b where a and b are "natural numbers"– that is, positive, non-zero integers (1, 2, 3, and so on). Let's also suppose that a and b are relatively prime– that is, the fraction is in its lowest possible terms. a shares no divisors with b.

With these suppositions made, here's our proof:

  1. Start with: ∃ a, bN where a and b are coprime and a / b = √(2).
  2. Get rid of the radical on the right by squaring both sides: 2 = ( a / b )2 = a2 / b2
  3. Multiply both sides by b2: 2 b2 = a2. So, a2 is even. This implies a is even.
  4. Since a is even, it can be written as 2 m, giving us: 2 b2 = ( 2 m )2
  5. Multiply out the squared term on the right-hand side: 2 b2 = 4 m2
  6. Divide both sides by two: b2 = 2 m2. Here's our contradiction: b2 is even, so b is even.
  7. If a and b are both even, they could be divided by two, contradicting (1). Q.E.D.

I want to end by saying that this proof originally disappointed me. You could seemingly replace 2 by any other number and have the proof still work. This is kinda touchy, but the simplicity of the odd and even characteristics really make it work well for √(2). After further consideration, it grew on me.

The Square Root of Any Prime Number is Irrational

I originally found this next one by searching for further truths about irrational numbers after my √(2) conundrum. I found this proof on Everything2, but the English was pretty bad and it didn't properly explain things.

This proof goes a little beyond my title. It actually proves all numbers that are not perfect squares are irrational. This is another proof by contradiction. We start by assuming p is prime and that there exists some a and b so that √(p) = a / b. Let's do some math.

  1. Multiply both sides by b: b √(p) = a.
  2. Let c = ⌊√p⌋ ... If the font doesn't display this properly, that's c = floor[ √(p) ], where the floor function rounds down its argument.
  3. Let's set up a new equation and examine d = b( √(p) - c ). d is an integer since we're multiplying b times √(p) minus its decimal part c.
  4. √(p) - c gives you a small integer close to the actual square root. d = b( √(p) - c ) = b√(p) - bc (by distribution).
  5. Let's multiply both sides by √(p): d√(p) = bp - √(p) bc
  6. Let's rewrite d once more to factor out the b: d√(p) = b( p - c√(p) )
  7. Since c < 1 and √(p) < p, then p - c√(p) > 1 and d < b.
  8. Here's our contradiction: According to (1), b was supposed to be the smallest number that when multiplied by √(p) gave us an integer, but we've proven ∃ dZ, d < b: d √(p) ∈ Z... that is to say, there exists an integer d that, when multiplied by our square root of a prime number (or non-perfect-square number), yields an integer.

This is a really beautiful proof that makes more sense the more you look and play with it. It's difficult to describe in plain English and mathematical notation becomes almost a necessity. (8) is kinda chunky, but it makes lots of sense. Things like the proof of √(2) might lead you to believe that these proofs will make things like √(25) drift off into irrationality, but it doesn't. In fact, if you plug √(25) or any other perfect-square into the above proof, c is equal to √(25) and d turns into zero. Then, of course d is less than b and d times √(25) is also 0. So, everything falls apart and rationality is restored.

Proof of the Nonexistence of a Largest Prime Number

I found this one in a 1960s textbook on number theory. It's breaking the trend of "prove-this-is-irrational" from the previous two. It's a proof by contradiction that assumes there is a largest prime number.

  1. Let p be the largest prime number.
  2. Let q = ( p × p - 1 × p - 2 × . . . × 3 × 2 × 1 ) + 1
  3. Let r = the largest prime number that divides q.
  4. r > p. This is in contradiction with (1).

So, q - 1 is a hugely composite number. Every number including p divides it. q, however, can't be prime since (1) says that p is the largest prime, and every number less than or equal to p divides q with a remainder of one. Therefore, there can't be a largest prime number.

From here, you can either go back to writings, or go back home.