# 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.