Structured Proof

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.

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.

Two functions \(f,g\) have contact order k at x_0

(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 \(i = 1, x_0 = 0, f(x) = cos(x), g(x) = \sqrt{1-x^2}\)

\[\frac{d' f(x)}{dx'} \vert_{x=x_0} = \frac{d' g(x)}{dx'} \vert_{x=x_0}, i=1.\]

\[\frac{d' cos(x)}{dx'} \vert_{x=x_0} = \frac{d' \sqrt{1-x^2}}{dx'} \vert_{x=x_0}, i=1.\]

\[- \sin(0) = -\frac{0} {\sqrt{1-0^2}}, \\ 0 = 0\]

It obeys the second rule.


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

\[-sin(0) \ne - \frac{15(0)(4(0)^2+3)}{(1-(0)^{2})^{\frac{9}{2}}}|x=x_0\]

(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.

\[\frac{d^i f(x)}{dx^i} |_{x=x_0} = \frac{d^i g(x)}{dx^i} |_{x=x_0}, .\]

Lets say \(i = 1, x_0 = 0, f(x) = sin(x), g(x) =x\)

\[\frac{d' f(x)}{dx'} \vert_{x=x_0} = \frac{d' g(x)}{dx'} \vert_{x=x_0}, i=1.\]

\[\frac{d' sin(x)}{dx'} \vert_{x=x_0} = \frac{d' x}{dx'} \vert_{x=x_0}, i=1.\]

\[\cos(0) = 1 \\ 1 = 1\]

It obeys the second rule.


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

\[sin(0) \ne \frac{d^{4}0}{d(0)^{4}}|x=x_0\]

(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.



(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.