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\begin{document}

\title{Structured Proof}


\author{Dayo Ogundipe}
\affil{Affiliation not available}



\date{\today}

\maketitle




\section{Structured proof that $\sqrt{2}$ is not a rational number.}

Rational numbers $\in$ $\mathbb{R}$ can be written as $\frac{a}{b}$, where a and b are positive integers.  
\\The quotient of two rational numbers is a rational number.  
\paragraph {Outline of Proof by contradiction}

$P = \sqrt{2}$ is not a rational number.
\\Suppose $\sim P$ which means not $P$, we have  $\sim P$ = $\sqrt{2}$ is a rational number.
\\Is so $\sqrt{2} = \frac{a}{b}$ and for any integer lets say $I$, $I - \sqrt{2} = \frac{c}{d}$ where a,b,c and d are positive integers and b,d cannot equal 0.   
\\If $I = 5$ using integer $5$,
\\ \[5-\sqrt{2}= \frac{c}{d},\] 
\\ adding the square root of $2$ shows \[5 = \frac{c}{d} + \sqrt{2} \] 
\\ subtracting the fraction we have \[5-\frac{c}{d} = \sqrt{2} \]  
\\ \[ \frac{5}{1} - \frac{c}{d} = \sqrt{2} \]
\\ \[\sqrt{2} = \frac{5d-c}{d} \]
\\ There are no integers for c and d to make this equation true.
\\ Therefore $\sim P$ is false and $P = \sqrt{2}$ is not a rational number is true.   




  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

\section{Two functions $f,g$  have contact order k at x_0}
\paragraph{(a) Construct a structured proof to show that $f(x):=\cos(x)$ and $g(x):=\sqrt{1-x^2}$ have contact order 4 at $x_0=0$.}
\\ \\ \\ We have $f(x_0)=\cos(0)= 1$ and $g(x_0):=\sqrt{1-0^2}=1$ So it obeys the first rule.
 

\[\frac{d^i f(x)}{dx^i} |_{x=x_0} = \frac{d^i g(x)}{dx^i} |_{x=x_0},\]
 
 Lets say $i = 1, x_0 = 0, f(x) = cos(x), g(x) =  \sqrt{1-x^2}$
  
  \begin{equation}
\frac{d' f(x)}{dx'} \vert_{x=x_0} = \frac{d' g(x)}{dx'} \vert_{x=x_0},  i=1.
 \end{equation}
 
 \begin{equation}
\frac{d' cos(x)}{dx'} \vert_{x=x_0} = \frac{d' \sqrt{1-x^2}}{dx'} \vert_{x=x_0},  i=1.
 \end{equation}
 
  
 \begin{equation}
- \sin(0)  = -\frac{0} {\sqrt{1-0^2}}, \\  0 = 0
 \end{equation}

It obeys the second rule.

 \\
 For $1 \geq i \geq k, x_0 = 0 $ we have
\begin{equation}
\frac{d^{k+1}f(x)}{dx^{k+1}}|x=x_0 \ne \frac{d^{k+1}g(x)}{dx^{k+1}}|x=x_0 
\end{equation}
 Lets say $k = 4, f(x) = \cos(x), g(x) =  \sqrt{1-x^2}$
 \begin{equation}
\frac{d^{5}\cos(x)}{dx^{5}}|x=x_0 \ne \frac{d^{5}\sqrt{1-x^2}}{dx^{5}}|x=x_0 
\end{equation}

 \begin{equation}
-sin(0) \ne - \frac{15(0)(4(0)^2+3)}{(1-(0)^{2})^{\frac{9}{2}}}|x=x_0 
\end{equation}
 
 \paragraph{(b)  Construct a structured proof to show that $f(x):=\sin(x)$ and $g(x):=x$ have contact order $3$ at $x_0=0$.}
  
 
  \\ \\ \\ We have $f(x_0)=\sin(0)= 0$ and $g(x_0):=0=0$ So it obeys the first rule.
 
 \begin{equation}
\frac{d^i f(x)}{dx^i} |_{x=x_0} = \frac{d^i g(x)}{dx^i} |_{x=x_0}, .
 \end{equation}
 Lets say $i = 1, x_0 = 0, f(x) = sin(x), g(x) =x$
  
  \begin{equation}
\frac{d' f(x)}{dx'} \vert_{x=x_0} = \frac{d' g(x)}{dx'} \vert_{x=x_0},  i=1.
 \end{equation}
 
 \begin{equation}
\frac{d' sin(x)}{dx'} \vert_{x=x_0} = \frac{d' x}{dx'} \vert_{x=x_0},  i=1.
 \end{equation}
 
  
 \begin{equation}
\cos(0)  = 1 \\  1 = 1
 \end{equation}

It obeys the second rule.

 \\
 For $1 \geq i \geq k, x_0 = 0 $ we have
\begin{equation}
\frac{d^{k+1}f(x)}{dx^{k+1}}|x=x_0 \ne \frac{d^{k+1}g(x)}{dx^{k+1}}|x=x_0 
\end{equation}
 Lets say $k = 3, f(x) = \sin(x), g(x) =x$
 \begin{equation}
\frac{d^{4}\sin(x)}{dx^{4}}|x=x_0 \ne \frac{d^{4}x}{dx^{4}}|x=x_0 
\end{equation}

 \begin{equation}
sin(0) \ne \frac{d^{4}0}{d(0)^{4}}|x=x_0 
\end{equation}
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  



  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  

\section{(a)Approximating $\sqrt{2}$}

We have 

\[ \begin{bmatrix}
1 & 2 \\
1 & 1 
\end{bmatrix}
\begin{bmatrix} 
p \\ q 
\end{bmatrix} = 
\begin{bmatrix} 
p + 2q \\ p+q 
\end{bmatrix} \] 

\\ By squaring the rational number $\frac{p}{q}$ with p and q being positive integers, thus $\frac{p^2}{q^2} < 2$
\\ After multiplication we have a new rational number lets call it $\frac{r}{s}$ and $\frac{r^2}{s^2} > 2$
\[
\begin{bmatrix} 
p^2 \\ q^2 
\end{bmatrix}<2 ,  
\begin{bmatrix} 
p^2 + 4pq +4q^2 \\ p^2+2pq+2q^2 
\end{bmatrix} > 2 
\]

We can use 1 to be our rational number. Cube root of 1, 1^2, is 1 and is less than 2.

\[\begin{bmatrix}
1 & 2 \\
1 & 1 
\end{bmatrix}
\begin{bmatrix} 
1 \\ 1 
\end{bmatrix} = 

\begin{bmatrix} 
3 \\ 2 
\end{bmatrix}\]  \[=\frac{3}{2} \]
\[
(\frac{3}{2}) ^2 = \frac{9}{4} \]
\\
\\ Here $\frac{9}{4} > 2$ therefore our proof is true.

\\ \\ \paragraph{(b) Construct a structured proof of the following (true) statement, using only rational numbers in your proof:}

Lets say a = 0.5 and b = 0.6
 \\ \\\\ $ a = 0.5 > 0 \\ (0.5)^2 < 2 \\ 0.25 < 2 $ 

\\ \\That works, Now for b = 0.6 does it apply?

\\ \\1. $a < b, 0.5 < 0.6$
\\ \\2. $b^2 < 2, 0.6^2 < 2, 0.36 < 2 $


\\ \\It satisfies our proof therefore being true.

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