Thread: Proof of checkerboard pieces

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Originally Posted by chillerbros17

discrete2.doc

Problem is in attachment.

B) It is imposible as part b has two white squares adjacent, which
never occurs on a real chess board.

RonL

Originally Posted by chillerbros17

discrete2.doc

Problem is in attachment.

6. Prove the following theorems. State the method of proof that you are using in parts (a) and (c).





(a):

Thm: The sum of two prime numbers, each greater than 2, is never a prime. Restatement: For all integers p, q, if p is a prime greater than 2 and q is a prime greater than 2, then p + q is not prime.

Proof by Contradiction (Reductio ad absurdum)

If p and q are primes greater than 2, then p and q are both odd, and may
be written p=2a+1, q=2b+1, for some positive integers a, and b.

Now suppose p+q is prime, then as it is greater than 2, it must be odd.
But:

p+q=2a+1+2b+1=2(a+b+1).

Thus p+q is divisible by 2, and hence not prime (as p+q !=2), which
is a contradiction, hence our assumption is false, and p+q is not prime.

RonL

discrete2.doc

Here's is b with the right pattern. Sorry for the mistake.

Originally Posted by chillerbros17

discrete2.doc

Here's is b with the right pattern. Sorry for the mistake.

See the attachment. This shows how to fit two of each shape together to make a 4x4 section of the chess board pattern. Four of these fitted together
will make the entire board.

RonL