A strategies to find proofs.

We've talked about proof strategies, (direct, indirect proofs, proofs of implications, and proofs by contradictions.). How do you find which of these proof strategies might work? One trick is the "try it out" example, or the "failing to make a counter-example might suggest a proof" method. For instance, consider the following problem:

Prove or disprove that 3x² -8y = 1 has no integer solutions.

Lets first try to make a counter example. We can change the problem a little bit and say:

3x² -8y = 1 has integer solutions <==> 3x² = 8y+1 has integer solutions.

So let's consider possible integer solution to this equation by
plugging in values integer values for x, y, and see if we ever get the two sides to be equal:

 3x2    8y + 1 
-----   ------
0         1
3         9
12        17
27        25
48        33
          41
          49

Ok, so we didn't any case where the two sides are equal.  This is
not a proof that there are no integer solutions (!), but it
might suggest to you that the theorem is true, and encourage you to
look for a proof.  Even more, it might suggest how to find the proof.

So, prove  3x2 =  8y + 1  has no integer solutions.

Proof:
Suppose, for contradiction, that
3x2 =  8y + 1  has integer solutions, where x = a, and y = b.
  (good style tip, if you instantiate a variable, use a different letter than the problem statement).

3a2 =  8b + 1,  a,b are integers
a = sqrt( (8b+1) / 3 ).
.
.
.


now maybe you could prove that sqrt( (8b+1) / 3 ) is irrational and
therefore not an integer, although you would have to do this proof in
general (for any possible b).  It isn't clear that we have good tools
for this.  This is a good chance to go back and look at our (failed) attempt to
find a solution.  What patterns do you see?

(left side is odd and even, right side is always odd)

This suggests a proof strategy.  Let's break the proof into two
parts, when a is odd and when a is even.  The odd case is easy, but we still have to prove it.  The even case will be a little harder

Proof:
Suppose, for contradiction, that
3x2 =  8y + 1  has integer solutions, where x = a, and y = b.

3a2 =  8b + 1  has integer solutions, where x = a, and y = b.
We seperate in two cases, a is even and a is odd.  

  (Requirement!.  your cases *must* cover all possibilities.  since a
   is either even or odd, a will always fit into one of the above
   cases).

Case 1: a is even
  3a2 =  8b + 1.
  a = 2k                      for some integer k, defn. of even.
  3(2k)2 =  8b + 1.
  3*4k2 =  8b + 1.
  12k2 =  8b + 1.
  2*6k2 =  8b + 1.
  2*6k2 =  8b + 1.
  2*6k2 =  2*(4b) + 1.
  2*6k2  is even
  2*(4b) + 1 is odd
  2*6k2 !=  2*(4b) + 1, contradiction


Case 2: a is odd
  3a2 =  8b + 1.
  a = 2k+1                      for some integer k, defn. of even.
  3(2k+1)2 =  8b + 1.
  3*(4k2+4k+1) =  8b + 1.
  4*(3k2+3k)+3 =  8b + 1.
  4*(3k2+3k)+3 =  4(2b) + 1.

  4*(3k2+3k)+3 mod 4 = 3
  4(2b) + 1 mod 4 = 1, so

  4*(3k2+3k)+3 !=  4(2b) + 1, contradiction


math fact:
If a mod c != b mod c then a != b

What is the "abstract argument structure of this proof"?
We are trying to prove a theorem P

We failed to directly prove P, so we tried a proof by contradiction:

~P --> F
--------
T

that failed, so we did the following:

~P --> (a v b)
a --> F
b --> F
--------
P

this is called proof by cases.  Note that a, b do not have to be
mutually exclusive, but the statment ~P --> (a v b) must be true
(there can't be a case where neither a nor b is true.




Ok, now on to the last kind of proof.  This is an argument strategy
that mixes quantified and unquantified statements.  This is a strategy
for proving statements of the form:

(forall integers n>0) P(n),

One way to prove these statements is:

P(1),
(forall integers n>0) P(n)-->P(n+1).
------------------------------------
(forall integers n>0) P(n)

This proof strategy is called "induction"..

Prove, for int. n > 0, 3 | n³ - n

  (first, always always good to try a few examples).
n = 1,2,3,4:   0,6,

Direct proof:

  n3 - n
= n(  n2 - 1)
= n (n + 1) (n - 1)
= (n - 1) n (n + 1)

split into cases.  n = 3k  OR n = 3k + 1 OR n = 3k + 2.

Case 1:  n = 3k
 ...proof.
Case 2:  n = 3k+1
 ...proof.
Case 3:  n = 3k+2
 ...proof.

(proving divisible by 3, "Show me the Three!").

Now, an induction proof of the same thing...


Four pieces to induction proof.

Base case.
Induction Hypothesis.
Induction Step.
Conclusion.

Prove that L pieces on a 2^k x 2^k checkerboard can always cover all squares but one.