Homework 3
Author
Brennan Gilmore
Last Updated
9 anni fa
License
Creative Commons CC BY 4.0
Abstract
MATH 108 - Introduction to Formal Mathematics
MATH 108 - Introduction to Formal Mathematics
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\documentclass[12pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
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\begin{document}
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\title{Homework 3}%replace X with the appropriate number
\author{Flynn Gilmore\\ %replace with your name
MATH 108 - Introduction to Formal Mathematics} %if necessary, replace with your course title
\maketitle
\begin{proposition}{1} %You can use theorem, proposition, exercise, or reflection here. Modify x.yz to be whatever number you are proving
If A is even and B is odd, then 3A+2B is even.
\end{proposition}
\begin{proof}
There exists a $K$ and $J$ that are integers such that $A=2K$ and $B=2J+1$.
Now, by plugging in our new $A$ and $B$, we have $3A+2B=3 \left( 2K \right) +2 \left( 2J+1 \right)$.
When multiplied out, we have $3A+2B=6K+4J+2=2 \left( 3K+2J+1 \right)$.
If we let $3K+2J+1=K_1$ where $K_1$ is an integer, $3A+2B=2 \left( K_1 \right)$.
So by definition, $2K_1$ is even, so $3A+2B$ is even.
\end{proof}
\begin{proposition}{2}
If $6\mid A$, then $36\mid A^2$.
\end{proposition}
\begin{proof}%Whatever you put in the square brackets will be the label for the block of text to follow in the proof environment.
Suppose $6\mid A$, so there exists an integer $K$ such that $6K=A$.
By squaring both sides, $A^2= \left( 6K \right) ^2=36K^2$.
Because $K^2$ is an integer, by definition $36\mid A^2$.
\end{proof}
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