commit c93abd2e46ffc7d5c55eb9360bc0baef0714902f
Author: Andrew <andrewlaack1@gmail.com>
Date: Tue, 9 Apr 2024 08:12:10 -0500
Init commit of abstract math
Diffstat:
6 files changed, 101 insertions(+), 0 deletions(-)
diff --git a/FibonacciNumbers.md b/FibonacciNumbers.md
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+# Fibonacci Numbers
+
+Abstract Math 10.5.
+
+## Notes:
+
+**Definition:** The set of numbers in the form $F_n = F_{n-1} + F_{n-2}$ starting from 1 as the first value.
+
+
diff --git a/FundamentalTheoremOfArithmetic.md b/FundamentalTheoremOfArithmetic.md
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+# The Fundamental Theorem of Arithmetic
+
+Abstract Math 10.4. Can be proven through [[StrongInduction.md]]
+
+## Notes:
+
+**Definition:** Any integer greater than 1 has a unique prime factorization.
+
+This means any number can be given in the form a * b * ... * z where all numbers multiplied together are prime numbers.
+
+The proof of this is quite interesting. Basically it states, if a number is not the prime factorization of any other numbers, then it can't be made because all numbers can be made by multiplying primes together thus this number must be prime meaning it factorizes itself.
+
+
diff --git a/Graphs.md b/Graphs.md
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+# Graphs
+
+Abstract Math 10.2.
+
+## Notes
+
+**Definition:** A graph is a configuration consisting of vertices and edges.
+
+**Cycle:** A cycle in a graph is a set of vertices such that traversal can be done back to itself.
+
+**Tree:** Trees are defined simply as graphs without cycles. There is no implication about split numbers or anything of the sort, but something interesting is that in all cases it must be true that the number of edges is one less than the number of vertices. This can be proved through [[StrongInduction.md]].
+
+
+
+
diff --git a/Induction.md b/Induction.md
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+# Induction
+
+Proof by induction from W11 abstract algebra. Induction is used to prove a statement relating to infinite sets of elements. This is not to be confused with inductive reasoning which is assumptions based on past data.
+
+## Notes
+
+**Definition:** This type of proof is done by proving that the first is true and how that subsequently means the rest are true (think dominoes).
+
+Steps to prove:
+1. Prove first statement is true
+2. Prove that the statement $S_k \implies S_{k+1}$ is true.
+
+**Basis Step:** The first step above is called the basis step because $S_1$ is generally quite simple.
+
+**Inductive Step:** The second step above is called the inductive step. This is most commonly done via direct proof ie. assuming $S_1$ is true.
+
+**Inductive Hypothesis:** This is the assumption that $S_k$ is true. Based on this assumption, we must then prove that it follows $S_K \implies S_{k+1}$.
+
+To prove this, it seems, we need to first prove that the 0th or first (generally) term of the sequence is true. This will generally be 0 or 1 indexed so such proofs are generally straightforward. Then we find an elements written out form, and then we find it's form + 1 in the index. Using this technique we should be able to show that the +1 index is in some way related to the +0 index.
+
+I am not sure if you need to prove the first element or if the logic to backpropogate is sufficient.
+
+Use if then notation for the current so that if the first then the second. Since we know the first we then need to prove the then follows logically in all cases across the domain of the function.
+
+If the set of numbers is bidirectional like $\Z$, then we need to prove that both $S_n \implies S_{n+1}$ and that $S_n \implies S_{n-1}$ for all numbers in the domain.
+
+When using induction the common form is $S_k \implies S_{k+1}$, but it is equally valid to prove that $S_{k-1} \implies S_k$ if that is easier to show.
+
+[[StrongInduction.md]] Is another type of induction.
diff --git a/SmallestCounterExample.md b/SmallestCounterExample.md
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+# Smallest Counterexample
+
+Abstract Math 10.3. This is similar to [[Induction.md]] and [[StrongInduction.md]]
+
+## Notes:
+
+**Definition:** Assume that the first element of a series is true and that not all other elements of the series are also true. We find the first element that is untrue denoted as $S_k$ and show that $S_{k-1}$ being true and $S_k$ being untrue is contradictory.
+
+**Steps:**
+1. Check that first statement $S_1$ is true
+2. Suppose not every statement $S_n$ is true
+3. Let k > 1 be the first instance where $S_k$ is false
+4. Show that $S_{k-1}$ being true and $S_k$ being false are contradictory
+
+
diff --git a/StrongInduction.md b/StrongInduction.md
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+# Strong Induction
+
+Abstract Math 10.2. Weak induction is the normal form of induction discussed in [[Induction.md]]
+
+## Notes
+
+**Definition:** Strong induction is the process of proving one or more prior true statements implies a later one much like weak induction, but with strong induction we can prove in the form of $S_{k-5} \implies S_{k+1}$ so long as k-5 is in the domain and that every value between k-5 and k+1 has been shown to be true.
+
+
+Steps:
+
+
+1. Prove the first statement $S_1$ or more if needed.
+2. Given any integer k$\geq$ 1, prove $(S_1 \wedge S_2 \wedge S_3 \wedge ... \wedge S_k) \implies S_{k+1}$
+
+A good example of this is an equation that does not factor nicely. If I know that $S_1$ is true, but I can't factor $S_2$ in a satisfactory way to prove that for each n+1 the statement is true, then proving a few until finding an instance of something factoring well can solve this issue.
+
+
+
+
