CSE390D Notes for Wednesday, 10/9/24
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open problems
took a long time to prove Fermat's Last Theorem
x^n + y^n = z^n has no nontrivial solutions for n > 2
hailstone conjecture is still unproven
given n, even n -> n/2, odd n -> 3n + 1, do you get to 1?
are there true statements that can't be proven true? completeness
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number theory, chapter 4, many applications (cryptography, hashing)
definition:
If a and b are integers with a != 0, we say that a divides b if
there is an integer c such that b = ac (or equivalentely if b/a is
an integer). When a divides b we say that a is a factor or
divisor of b, and that b is a multiple of a. The notation "a | b"
denotes that a divides b.
division algorithm:
for a and d positive integers, there are unique integers q and r
such that:
a = dq + r with 0 <= r < d
a is dividend, d is divisor, q is quotient, r is remainder
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we looked at mod 7
definition:
If a and b are integers and m is a positive integer, then a is
congruent to b modulo m if m divides a-b. We use the notation
a == b (mod m) to indicate that a is congruent to b modulo m. We
say that a == b (mod m) is a congruence and that m is its modulus.
If a and b are not congruent, we write a !== b (mod m)
Prove: given positive integer m, if a==b (mod m) and (c == d) (mod m):
a + c == b + d (mod m) and a * c == b * d (mod m)
proof:
a == b (mod m), so exists k such that a - b = km
c == d (mod m), so exists j such that c - d = jm
a + c = (b + km) + (d + jm) = (b + d) + m(k + j)
(a + c) - (b + d) = m(k + j)
a * c = (b + km) * (d + jm) = bd + dkm + bjm + kjm^2 =
= bd + m(dk + bj + kjm)
a * c - b * d = bd + m(dk + bj + kjm) - bd = m(dk + bj + kjm)
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therefore when working in a modulus, we get congruence classes where
any representative of class will do in computing a result. Examples:
7008 ^ 312 % 7 = 1
14006 ^ 175 % 7 = 6 (-1)
1000 ^ 3491 % 1001 = 1
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applications:
hashing
simple ciphers
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positive n is prime if n > 1 and only factors are p and 1.
Other are called composite.
Fundamental Theorem of Arithmetic:
Every integer greater than 1 can be written uniquely as a prime or
as the product of two or more primes, where the prime factors are
written in nondecreasing size.
prove: infinitely many primes
assume not, then primes = p1, p2, ..., pn
consider q = p1 * p2 * ... * pn + 1
bigger than any p, therefore not prime
therefore it has prime factors
let pi be a prime factor, pi | q
but (q - p1 * p2 * ... * pn) = 1
so pi | 1
contradiction
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Stuart Reges
Last modified: Wed Oct 9 13:56:42 PDT 2024