Proposition: The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.

a

^{2}+ b

^{2}= c

^{2}

Proof: Suppose you have a square with sides (a + b) that has been cut into five pieces: a smaller inscribed square with sides c and four right triangles with sides a, b, and c. The area of the larger square is given by (a + b)

^{2}= a

^{2}+ 2ab + b

^{2}. The area of the larger square can also be given by the sum of the interior square and the four right triangles: c

^{2}+ 4(ab/2) = c

^{2}+ 2ab. Since the two areas are equal, we can set them equal to each other and subtract what they have in common.

a

^{2}+ 2ab + b

^{2}= c

^{2}+ 2ab Subtract 2ab from both sides

a

^{2}+ b

^{2}= c

^{2}Pythagorean Theorem

2. Prime Numbers

(a) Proposition: There are infinitely many prime numbers.

Proof: Suppose not. Then, there exists only finitely many prime numbers. We can label these finitely many primes *p _{1}, p_{2}, …, p_{n}*. Call the product of all these primes plus 1

*N*.

*N*=

*p*+ 1

_{1}* p_{2}* …* p_{n}*N*is thus larger than any number in the set

*p*. Since

_{1}, p_{2}, …, p_{n}*N*>

*p*, and

_{n}*p*is the largest prime number,

_{n}*N*is not prime. Therefore,

*N*must be divisible by a positive integer other than itself and 1. Based on the fundamental theorem of arithmetic, it follows that

*N*has some prime factorization. This means one of the numbers in the above list of primes must divide

*N*. Thus, there exists an integer,

*i*with 1 ≤

*i*≤

*n*so that

*p*divides

_{i}*N*. As

*p*divides

_{i}*N*, and thus the product of all primes, it must also divide

*N – (p*= 1. Since

_{1}* p_{2}* …* p_{n})*p*> 2, it is impossible for

_{i}*p*to divide one. This contradicts the idea that

_{i}*N*is not prime so there cannot be only finitely many primes.

(b) Algorithm: First try n + 1, if the returned number is prime, stop there.

If returned number is not prime, plug into the following formula: (2n + 1)

If the returned number is still not prime, add 2 until you reach a prime number

Any number returned by the algorithm satisfies the condition 0 < n < A. As we previously proved there is an infinite number of primes, there will always be an output that is larger than n. A problem with this formula is that it as you begin to input larger numbers, the possibility of encountering a prime number gets larger and larger so you will have to add 2 at an increaisng rate.

n | n + 1 | prime? | 2n + 1 | prime? | ~ + 2 | prime? |
---|---|---|---|---|---|---|

0 | 1 | No | 1 | No | 3 | Yes |

1 | 2 | Yes | N/A | N/A | N/A | N/A |

2 | 3 | Yes | N/A | N/A | N/A | N/A |

6 | 7 | Yes | N/A | N/A | N/A | N/A |

10 | 11 | Yes | N/A | N/A | N/A | N/A |

17 | 18 | No | 35 | No | 37 | Yes |

21 | 22 | No | 43 | Yes | N/A | N/A |

26 | 27 | No | 53 | Yes | N/A | N/A |