Archimedean Property Corollary proof
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I am trying to prove one of the Archmedian properties corollaries.
The corollary states:
If $x$ is a real number such that $x>0$ then there exists a natural number $nin mathbb{N}$ such that $n−1le x <n$.
From the Archmedian property, I know that for $xin mathbb{R}$, then there is a positive integer $n$ such that $n>x$ and from the Well Ordering Principle I know that for any nonempty subset of $mathbb{N}$ there is a least element, however, how do I show that $n-1le x$?
real-analysis
asked Dec 20 '18 at 6:15
RobbenRobben
1415
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|
show 2 more comments
$begingroup$
I am trying to prove one of the Archmedian properties corollaries.
The corollary states:
If $x$ is a real number such that $x>0$ then there exists a natural number $nin mathbb{N}$ such that $n−1le x <n$.
From the Archmedian property, I know that for $xin mathbb{R}$, then there is a positive integer $n$ such that $n>x$ and from the Well Ordering Principle I know that for any nonempty subset of $mathbb{N}$ there is a least element, however, how do I show that $n-1le x$?
real-analysis
asked Dec 20 '18 at 6:15
RobbenRobben
1415
$endgroup$
$begingroup$
You got the archimedian property wrong, for every x in R you can find n, s.t $x<n$.
$endgroup$
– Riquelme
Dec 20 '18 at 6:43
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@Riquelme but that is what I said.
$endgroup$
– Robben
Dec 20 '18 at 6:45
$begingroup$
No, you said there exists x in R s. t x <n
$endgroup$
– Riquelme
Dec 20 '18 at 6:48
$begingroup$
This is from my book, "If $x in mathbb{R}$, then there is a positive integer n such that $n > x$." Not sure what the difference is?
$endgroup$
– Robben
Dec 20 '18 at 6:50
$begingroup$
This is true, what you said in your next paragraph is what I was referring to
$endgroup$
– Riquelme
Dec 20 '18 at 6:52
|
show 2 more comments
$begingroup$
I am trying to prove one of the Archmedian properties corollaries.
The corollary states:
If $x$ is a real number such that $x>0$ then there exists a natural number $nin mathbb{N}$ such that $n−1le x <n$.
From the Archmedian property, I know that for $xin mathbb{R}$, then there is a positive integer $n$ such that $n>x$ and from the Well Ordering Principle I know that for any nonempty subset of $mathbb{N}$ there is a least element, however, how do I show that $n-1le x$?
real-analysis
asked Dec 20 '18 at 6:15
RobbenRobben
1415
$endgroup$
I am trying to prove one of the Archmedian properties corollaries.
The corollary states:
If $x$ is a real number such that $x>0$ then there exists a natural number $nin mathbb{N}$ such that $n−1le x <n$.
From the Archmedian property, I know that for $xin mathbb{R}$, then there is a positive integer $n$ such that $n>x$ and from the Well Ordering Principle I know that for any nonempty subset of $mathbb{N}$ there is a least element, however, how do I show that $n-1le x$?
real-analysis
real-analysis
asked Dec 20 '18 at 6:15
RobbenRobben
1415
asked Dec 20 '18 at 6:15
RobbenRobben
1415
edited Dec 20 '18 at 18:47
Robben
asked Dec 20 '18 at 6:15
RobbenRobben
1415
asked Dec 20 '18 at 6:15
RobbenRobben
1415
asked Dec 20 '18 at 6:15
RobbenRobben
1415
1415
$begingroup$
You got the archimedian property wrong, for every x in R you can find n, s.t $x<n$.
$endgroup$
– Riquelme
Dec 20 '18 at 6:43
$begingroup$
@Riquelme but that is what I said.
$endgroup$
– Robben
Dec 20 '18 at 6:45
$begingroup$
No, you said there exists x in R s. t x <n
$endgroup$
– Riquelme
Dec 20 '18 at 6:48
$begingroup$
This is from my book, "If $x in mathbb{R}$, then there is a positive integer n such that $n > x$." Not sure what the difference is?
$endgroup$
– Robben
Dec 20 '18 at 6:50
$begingroup$
This is true, what you said in your next paragraph is what I was referring to
$endgroup$
– Riquelme
Dec 20 '18 at 6:52
|
show 2 more comments
$begingroup$
You got the archimedian property wrong, for every x in R you can find n, s.t $x<n$.
$endgroup$
– Riquelme
Dec 20 '18 at 6:43
$begingroup$
@Riquelme but that is what I said.
$endgroup$
– Robben
Dec 20 '18 at 6:45
$begingroup$
No, you said there exists x in R s. t x <n
$endgroup$
– Riquelme
Dec 20 '18 at 6:48
$begingroup$
This is from my book, "If $x in mathbb{R}$, then there is a positive integer n such that $n > x$." Not sure what the difference is?
$endgroup$
– Robben
Dec 20 '18 at 6:50
$begingroup$
This is true, what you said in your next paragraph is what I was referring to
$endgroup$
– Riquelme
Dec 20 '18 at 6:52
$begingroup$
You got the archimedian property wrong, for every x in R you can find n, s.t $x<n$.
$endgroup$
– Riquelme
Dec 20 '18 at 6:43
$begingroup$
You got the archimedian property wrong, for every x in R you can find n, s.t $x<n$.
$endgroup$
– Riquelme
Dec 20 '18 at 6:43
$begingroup$
@Riquelme but that is what I said.
$endgroup$
– Robben
Dec 20 '18 at 6:45
$begingroup$
@Riquelme but that is what I said.
$endgroup$
– Robben
Dec 20 '18 at 6:45
$begingroup$
No, you said there exists x in R s. t x <n
$endgroup$
– Riquelme
Dec 20 '18 at 6:48
$begingroup$
No, you said there exists x in R s. t x <n
$endgroup$
– Riquelme
Dec 20 '18 at 6:48
$begingroup$
This is from my book, "If $x in mathbb{R}$, then there is a positive integer n such that $n > x$." Not sure what the difference is?
$endgroup$
– Robben
Dec 20 '18 at 6:50
$begingroup$
This is from my book, "If $x in mathbb{R}$, then there is a positive integer n such that $n > x$." Not sure what the difference is?
$endgroup$
– Robben
Dec 20 '18 at 6:50
$begingroup$
This is true, what you said in your next paragraph is what I was referring to
$endgroup$
– Riquelme
Dec 20 '18 at 6:52
$begingroup$
This is true, what you said in your next paragraph is what I was referring to
$endgroup$
– Riquelme
Dec 20 '18 at 6:52
|
show 2 more comments
1 Answer
1
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oldest
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look at all those natural numbers $n$ which is larger than $x$, and choose the smallest one (e.g. $n_0$) by well-ordering principle, then argue that $n_0-1$ is also a natural number(shall make use of the fact that $x>0$). Hence it must follow that $n_0-1le x$.
answered Dec 20 '18 at 6:37
NickNick
1,7151417
$endgroup$
$begingroup$
Can you elaborate please?
$endgroup$
– Robben
Dec 20 '18 at 6:58
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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votes
$begingroup$
look at all those natural numbers $n$ which is larger than $x$, and choose the smallest one (e.g. $n_0$) by well-ordering principle, then argue that $n_0-1$ is also a natural number(shall make use of the fact that $x>0$). Hence it must follow that $n_0-1le x$.
answered Dec 20 '18 at 6:37
NickNick
1,7151417
$endgroup$
$begingroup$
Can you elaborate please?
$endgroup$
– Robben
Dec 20 '18 at 6:58
add a comment |
$begingroup$
look at all those natural numbers $n$ which is larger than $x$, and choose the smallest one (e.g. $n_0$) by well-ordering principle, then argue that $n_0-1$ is also a natural number(shall make use of the fact that $x>0$). Hence it must follow that $n_0-1le x$.
answered Dec 20 '18 at 6:37
NickNick
1,7151417
$endgroup$
$begingroup$
Can you elaborate please?
$endgroup$
– Robben
Dec 20 '18 at 6:58
add a comment |
$begingroup$
look at all those natural numbers $n$ which is larger than $x$, and choose the smallest one (e.g. $n_0$) by well-ordering principle, then argue that $n_0-1$ is also a natural number(shall make use of the fact that $x>0$). Hence it must follow that $n_0-1le x$.
answered Dec 20 '18 at 6:37
NickNick
1,7151417
$endgroup$
look at all those natural numbers $n$ which is larger than $x$, and choose the smallest one (e.g. $n_0$) by well-ordering principle, then argue that $n_0-1$ is also a natural number(shall make use of the fact that $x>0$). Hence it must follow that $n_0-1le x$.
answered Dec 20 '18 at 6:37
NickNick
1,7151417
edited Dec 20 '18 at 6:44
answered Dec 20 '18 at 6:37
NickNick
1,7151417
answered Dec 20 '18 at 6:37
NickNick
1,7151417
answered Dec 20 '18 at 6:37
NickNick
1,7151417
1,7151417
$begingroup$
Can you elaborate please?
$endgroup$
– Robben
Dec 20 '18 at 6:58
add a comment |
$begingroup$
Can you elaborate please?
$endgroup$
– Robben
Dec 20 '18 at 6:58
$begingroup$
Can you elaborate please?
$endgroup$
– Robben
Dec 20 '18 at 6:58
$begingroup$
Can you elaborate please?
$endgroup$
– Robben
Dec 20 '18 at 6:58
add a comment |
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$begingroup$
You got the archimedian property wrong, for every x in R you can find n, s.t $x<n$.
$endgroup$
– Riquelme
Dec 20 '18 at 6:43
$begingroup$
@Riquelme but that is what I said.
$endgroup$
– Robben
Dec 20 '18 at 6:45
$begingroup$
No, you said there exists x in R s. t x <n
$endgroup$
– Riquelme
Dec 20 '18 at 6:48
$begingroup$
This is from my book, "If $x in mathbb{R}$, then there is a positive integer n such that $n > x$." Not sure what the difference is?
$endgroup$
– Robben
Dec 20 '18 at 6:50
$begingroup$
This is true, what you said in your next paragraph is what I was referring to
$endgroup$
– Riquelme
Dec 20 '18 at 6:52