rational number

Rational Numbers

$$\mathbb{Q}:=\left\{\frac{p}{q}:p\in \mathbb{Z},q\in \mathbb{N}\right\}$$

$p$ is the numerator, $q$ the denominator.

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Basic operations:

$$\frac{a}{b}\pm \frac{c}{d}=\frac{ad\pm bc}{bd}$$

This can be simplified if there are common factors in numerator and denominator by finding the greatest common divisor.

$$\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$$ $$\frac{a}{b}÷\frac{c}{d}=\frac{a\cdot d}{b\cdot c}$$

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Lemma: Let $n\in \mathbb{N}$, then $n|2⟺n^2|2$.

If a number is even, then its square is even, and vice versa.
Proof:
$⟹$: Assume $n|2$, i.e. $\exists k\in \mathbb{N}:n=2k⟹n^2=4k^2=2(2k^2)$, thus $n^2|2$.
$⟸$: By contradiction: Assume $n\nmid 2$, and show that $n^2$ is odd.
$\exists k\in \mathbb{N}:n=2k+1⟹n^2=4k^2+4k+1=2(2k^2+2k)+1$, thus $n^2\nmid 2$. □

There is no $x\in \mathbb{Q}$ such that $x^2=2$, i.e. $\sqrt{2}\notin \mathbb{Q}$

Proof by contradiction:
Assume $x=\frac{p}{q}$ and $x^2=2$, and at least one of $p,q$ is odd (i.e. $\frac{p}{q}$ is in lowest terms); WLOG.
Then:
$x^2=\frac{p^2}{q^2}=2⟺p^2=2q^2$
By the lemma above, $p^2$ is an even number ($⟹p=2k$), and $q$ has to be odd.

$$\begin{array}{rl}(2k{)}^2& =2q^2\\ 2(2k^2)& =2q^2\\ 2k^2& =q^2\end{array}$$

By the lemma again, $q^2$ is even, thus $q$ is even.
This contradicts the assumption that at least one of $p,q$ is odd. □

Rational numbers are dense in the reals.

a)

Transclude of archimedian-property#^814f46

b)

$$\forall ϵ>0,\exists n\in \mathbb{N}:\frac{1}{n}<ϵ$$

c)

$$\forall x,y\in \mathbb{R},x<y,\exists \frac{m}{n}\in \mathbb{Q}:x\le \frac{m}{n}\le y$$

→ If I pick two real numbers, there is always a rational number in between them.

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