- #1
zeronem
- 117
- 1
- Homework Statement
- If x is any real number, prove that
$$x=sup\{ r \in Q: r<x\}$$ $$=inf\{ s \in Q: x<s\}$$
- Relevant Equations
- $$r<x<s$$
$$s-r>0$$
$$n(s-r)>1$$
$$nr\in[m,m+1)$$
$$r<x<s$$
$$s-r>0$$
We enploy the Archimedean principle where
$$n(s-r)>1$$
We employ density of rationals where
$$\exists [m,m+1] \in Q$$
Such that
$$nr\in [m,m+1)$$
Therefore
$$m\leq nr \lt m+1$$$$ \frac m n \leq r \lt \frac m n + \frac 1 n $$
Since
$$ \frac m n \leq r $$
Then
$$ \frac m n \leq r \lt r + \frac 1 n \lt s $$
Since $$r \lt x \lt s$$
two possibilities:
First possibility:
$$ r < x \leq \frac m n + \frac 1 n \lt s $$
Where
$$ \forall x \in R $$
$$ \exists\{ \frac m n + \frac1 n\} \in Q$$ such that $$x \leq \frac m n + \frac 1 n $$
Or
Second possibility:
$$ r < \frac m n + \frac 1 n \leq x \lt s $$
Where
$$ \forall x \in R$$
There $$ \exists \{ \frac m n + \frac 1 n \} \in Q $$ such that $$ \frac m n + \frac 1 n \leq x $$
Therefore
$$x=sup\{ r \in Q : r<x\}= inf\{ s \in Q : x<s\}$$
QED
Is this a valid argument?
$$s-r>0$$
We enploy the Archimedean principle where
$$n(s-r)>1$$
We employ density of rationals where
$$\exists [m,m+1] \in Q$$
Such that
$$nr\in [m,m+1)$$
Therefore
$$m\leq nr \lt m+1$$$$ \frac m n \leq r \lt \frac m n + \frac 1 n $$
Since
$$ \frac m n \leq r $$
Then
$$ \frac m n \leq r \lt r + \frac 1 n \lt s $$
Since $$r \lt x \lt s$$
two possibilities:
First possibility:
$$ r < x \leq \frac m n + \frac 1 n \lt s $$
Where
$$ \forall x \in R $$
$$ \exists\{ \frac m n + \frac1 n\} \in Q$$ such that $$x \leq \frac m n + \frac 1 n $$
Or
Second possibility:
$$ r < \frac m n + \frac 1 n \leq x \lt s $$
Where
$$ \forall x \in R$$
There $$ \exists \{ \frac m n + \frac 1 n \} \in Q $$ such that $$ \frac m n + \frac 1 n \leq x $$
Therefore
$$x=sup\{ r \in Q : r<x\}= inf\{ s \in Q : x<s\}$$
QED
Is this a valid argument?
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