update solution 1

answers/zk_solution_1.py

@@ -1,3 +1,76 @@
-version https://git-lfs.github.com/spec/v1
-oid sha256:6d10f0baf016782cb934c3f7f8580a7fcb8c956536d004393ef2d66e73d618f4
-size 2016
+# Zero Knowledge Bootcam
+from scipy.optimize import fsolve
+import numpy as np
+import matplotlib.pyplot as plt
+
+
+def question_one():
+ """
+ 1. Working with the following set of integers s = {0,1,2,3,4,5,6}
+ What is
+ a) 4 + 4
+ b) 3 x 5
+ c) what is the inverse of 3?
+ """
+ # opereations are caried out by modulo 7
+ s = {0, 1, 2, 3, 4, 5, 6}
+ p = 7
+
+ a = (4+4) % p
+ b = (3*5) % p
+ c = pow(3, -1, p) # pyhton internal inverse modular function pow()
+
+ # For c we can use Fermat's little theorem where a^-1 = a^(p-2) mod p
+ c = (3**(p-2))%p 
+
+ print(f"Question 1: a={a} b={b} c={c}")
+
+
+def question_two():
+ """
+ 2. For S = {0,1,2,3,4,5,6}
+ Can we consider 'S' and the operation '+' to be a group?
+ """
+ notes = """
+ Simply put a group is a set of elements set(a, b, c, ...) plus a binary operation
+ To be concidered a group the combination needs to have the following Properties:
+ 1. Closure:
+ For all a,b in set S, the result of a + b is also in S
+ 2. Associativity:
+ For all a,b and c in set S, (a+b)+c = a+(b+c)
+ 3. Identity element
+ There exists an element e in set S such that, every element a in set S, the equation e + a = a + e = a holds.
+ Such an element is unique and thus on speaks of the identity element.
+ 4. Inverse element
+ For each a in set S, there exists an element b in set S, commonly denoted -a (if operations is +) or a**-1 if operator is *, such that a + b = b + a = e, where e is the identity element. 
+ """
+
+def questions_three():
+ """
+ What is -13 mod 5?
+ """
+ res = -13 % 5
+ print(f"Question 3: {res}")
+
+
+def question_four():
+ """
+ Polynomials
+ For the polynomials x**3 - x**2 + 4x - 12
+ Find a the positive root?
+ what is the degree of this polymial?
+ """
+ # def f(x): return x**2 - x + 4
+ # res = fsolve(f,2)
+
+ # print(f"Question 4: {res}")
+
+
+def main():
+ question_one()
+ question_two()
+ questions_three()
+ question_four()
+
+
+main()

answers/zk_solution_1.tex

@@ -32,13 +32,13 @@ \section*{Question 2}
 \end{itemize} 
 
 \section*{Question 3}
-$-13 \mod 5 \equiv 2 \mod 5$. Answer is 2.
+$-13 \mod 5 \equiv (-13 + 15 )\mod 5 \equiv 2 \mod 5$. Answer is 2.
 
 \section*{Question 4}
 
-We can simplify our function $f(x) = x^{3}-x^{2}+4x-12$ to factors $f(x)= (x-2)\cdot (x^{2}+x+6)$
+The degree of our polynomial is $3$. We can simplify our function $f(x) = x^{3}-x^{2}+4x-12$ to factors $f(x)= (x-2)\cdot (x^{2}+x+6)$. Using factorization method for $(x-2)=0$ our answer is 2.
 
-Using Polynomial Remainder Theorem $\frac{P(x)}{x-a}\xrightarrow{} r=P(a)$ where $P(x) = x^{3}-x^{2}+4x-12$. If we divided $P(x)$ with a first degree polynomial $(x-a)$, and we don't get a remainder $r=0$, this verifies $(x-a)$ is a valid factor of our polynomial $P(x)$. We can solve this with our factor $(x-2)$ for $P(a)$. When $x=2$, $P(2)=2^{3}-2^{2}+4\cdot2-12=0$. Solving for $x$ using $(x-2)=0$ or $(x^{2}+x+6)=0$. Answer is $2$. The degree of this polynomial is $3$.
+Using Polynomial Remainder Theorem formula $\frac{P(x)}{x-a}\xrightarrow{} r=P(a)$ where $P(x) = x^{3}-x^{2}+4x-12$. If we divided $P(x)$ with a first degree polynomial $(x-a)$, and we don't get a remainder $r=0$, this verifies $(x-a)$ is a valid factor of our polynomial $P(x)$. We can solve this with our factor $(x-2)$ for $P(a)$. When $x=2$, $P(2)=2^{3}-2^{2}+4\cdot2-12=0$. Solving for $x$ using $(x-2)=0$ give us $2$. While solving for $x$ using $(x^{2}+x+6)=0$ gives us complex numbers.
 
 See: \href{https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-div/x2ec2f6f830c9fb89:remainder-theorem/v/polynomial-remainder-theorem-to-test-factor}{Remainder theorem: checking factors}