real number

Real numbers

$$\mathbb{R}:=\{g+r:g\in \mathbb{Z},r=\underset{n\to \infty }{\lim }\frac{a_n}{b_n},a_n,b_n\in \mathbb{N}\}$$

Here, $g$ is the integer (whole number) part and $r$ is the fractional part, $n$ is the number of digits after the decimal point and it can be finite or infinite.
The rational numbers $\mathbb{Q}\subset \mathbb{R}$, while $\mathbb{R}\setminus \mathbb{Q}$ are the irrational numbers.
The real numbers are a field under addition and multiplication.

Completeness axiom of the real numbers $\mathbb{R}$

Every non-empty subset $A\subset \mathbb{R}$ that is bounded from above has a supremum in $\mathbb{R}$, i.e. $\exists T\in \mathbb{R}$ such that $T=\sup A$ (similarily for infimum).

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Not every subset of $\mathbb{R}$ has a supremum in $\mathbb{R}$

For example: The set $A=\{x\in \mathbb{Q}:x^2\le 2\}$ is bounded, i.e. $-\sqrt{2}\le x\le \sqrt{2}$.
So $\sup A=\sqrt{2}\notin \mathbb{Q}$, and similarily $\inf A=-\sqrt{2}\notin \mathbb{Q}$, so they do not exist in $\mathbb{Q}$.
→ Infimum and Supremum are always defined in a certain space.
We can approach $\sqrt{2}$ abritrarly closely with rational numbers, but never reach it.
But, we can define $\sup A=\sqrt{2}$ in $\mathbb{R}$, because $\sqrt{2}\in \mathbb{R}$.

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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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