Home
Random
Log in
Settings
Donate
About Wikiversity
Disclaimers
Search
Real numbers/Ordering axioms/Archimedean/Implications/Fact
Language
Watch
Edit
<
Real numbers
For
x
,
y
∈
R
{\displaystyle {}x,y\in \mathbb {R} }
with
x
>
0
{\displaystyle {}x>0}
there exists
n
∈
N
{\displaystyle {}n\in \mathbb {N} }
such that
n
x
≥
y
{\displaystyle {}nx\geq y}
.
For
x
>
0
{\displaystyle {}x>0}
there exists a natural number
n
{\displaystyle {}n}
such that
1
n
<
x
{\displaystyle {}{\frac {1}{n}}<x}
.
For two real numbers
x
<
y
{\displaystyle {}x<y}
there exists a rational number
n
/
k
{\displaystyle {}n/k}
(with
n
∈
Z
{\displaystyle {}n\in \mathbb {Z} }
,
k
∈
N
+
{\displaystyle {}k\in \mathbb {N} _{+}}
) such that
x
<
n
k
<
y
.
{\displaystyle {}x<{\frac {n}{k}}<y\,.}
Proof 1
,
2
,
Write another proof