Equivalent of Archimedean Property
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I am reading real analysis textbook by Stephen C. Lay on the Archimedean property of $mathbb R$. One of the three equivalents is stated as follow:
For each $x > 0$ and for each $y in mathbb R$, there exists an $n in mathbb N$ such that $nx > y$.
At least to my untrained novice eyes, it is counter intuitive. I can understand if $y$ is positive, but what happens when it is not? For example, when $x = 1$ and $y = -1$, since $0 notin mathbb N$?
I have searched this site for the answer under "Archimedean Property" but could not find one. I hope someone could give me intuition and perhaps some examples. Thank you for your time and helps.
real-analysis real-numbers
asked Nov 25 at 16:33
Amanda.M
1,60611432
add a comment |
up vote
2
down vote
favorite
I am reading real analysis textbook by Stephen C. Lay on the Archimedean property of $mathbb R$. One of the three equivalents is stated as follow:
For each $x > 0$ and for each $y in mathbb R$, there exists an $n in mathbb N$ such that $nx > y$.
At least to my untrained novice eyes, it is counter intuitive. I can understand if $y$ is positive, but what happens when it is not? For example, when $x = 1$ and $y = -1$, since $0 notin mathbb N$?
I have searched this site for the answer under "Archimedean Property" but could not find one. I hope someone could give me intuition and perhaps some examples. Thank you for your time and helps.
real-analysis real-numbers
asked Nov 25 at 16:33
Amanda.M
1,60611432
1
Its trivial since all $n$ satisfy.
– Yadati Kiran
Nov 25 at 16:38
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading real analysis textbook by Stephen C. Lay on the Archimedean property of $mathbb R$. One of the three equivalents is stated as follow:
For each $x > 0$ and for each $y in mathbb R$, there exists an $n in mathbb N$ such that $nx > y$.
At least to my untrained novice eyes, it is counter intuitive. I can understand if $y$ is positive, but what happens when it is not? For example, when $x = 1$ and $y = -1$, since $0 notin mathbb N$?
I have searched this site for the answer under "Archimedean Property" but could not find one. I hope someone could give me intuition and perhaps some examples. Thank you for your time and helps.
real-analysis real-numbers
asked Nov 25 at 16:33
Amanda.M
1,60611432
I am reading real analysis textbook by Stephen C. Lay on the Archimedean property of $mathbb R$. One of the three equivalents is stated as follow:
For each $x > 0$ and for each $y in mathbb R$, there exists an $n in mathbb N$ such that $nx > y$.
At least to my untrained novice eyes, it is counter intuitive. I can understand if $y$ is positive, but what happens when it is not? For example, when $x = 1$ and $y = -1$, since $0 notin mathbb N$?
I have searched this site for the answer under "Archimedean Property" but could not find one. I hope someone could give me intuition and perhaps some examples. Thank you for your time and helps.
real-analysis real-numbers
real-analysis real-numbers
asked Nov 25 at 16:33
Amanda.M
1,60611432
asked Nov 25 at 16:33
Amanda.M
1,60611432
asked Nov 25 at 16:33
Amanda.M
1,60611432
asked Nov 25 at 16:33
Amanda.M
1,60611432
asked Nov 25 at 16:33
Amanda.M
1,60611432
1,60611432
1
Its trivial since all $n$ satisfy.
– Yadati Kiran
Nov 25 at 16:38
add a comment |
1
Its trivial since all $n$ satisfy.
– Yadati Kiran
Nov 25 at 16:38
1
1
Its trivial since all $n$ satisfy.
– Yadati Kiran
Nov 25 at 16:38
Its trivial since all $n$ satisfy.
– Yadati Kiran
Nov 25 at 16:38
add a comment |
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.
answered Nov 25 at 16:37
Berci
59.2k23671
Thank you! I think I did not read the statement carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
up vote
4
down vote
If $x>0$ and $yle 0$, then
$$color{red}{1}times x>0>-1>-2>-3...>y>...$$
so $n=color{red}{1}$.
answered Nov 25 at 16:41
hamam_Abdallah
37k21534
Thank you! I think I did not read the problem carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
up vote
1
down vote
Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)
answered Nov 25 at 16:48
Dr. Mathva
645110
Thank you for your alternative answer.
– Amanda.M
Nov 25 at 16:49
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.
answered Nov 25 at 16:37
Berci
59.2k23671
Thank you! I think I did not read the statement carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
up vote
4
down vote
accepted
If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.
answered Nov 25 at 16:37
Berci
59.2k23671
Thank you! I think I did not read the statement carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.
answered Nov 25 at 16:37
Berci
59.2k23671
If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.
answered Nov 25 at 16:37
Berci
59.2k23671
answered Nov 25 at 16:37
Berci
59.2k23671
answered Nov 25 at 16:37
Berci
59.2k23671
answered Nov 25 at 16:37
Berci
59.2k23671
59.2k23671
Thank you! I think I did not read the statement carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
Thank you! I think I did not read the statement carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
Thank you! I think I did not read the statement carefully enough.
– Amanda.M
Nov 25 at 16:43
Thank you! I think I did not read the statement carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
up vote
4
down vote
If $x>0$ and $yle 0$, then
$$color{red}{1}times x>0>-1>-2>-3...>y>...$$
so $n=color{red}{1}$.
answered Nov 25 at 16:41
hamam_Abdallah
37k21534
Thank you! I think I did not read the problem carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
up vote
4
down vote
If $x>0$ and $yle 0$, then
$$color{red}{1}times x>0>-1>-2>-3...>y>...$$
so $n=color{red}{1}$.
answered Nov 25 at 16:41
hamam_Abdallah
37k21534
Thank you! I think I did not read the problem carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
up vote
4
down vote
up vote
4
down vote
If $x>0$ and $yle 0$, then
$$color{red}{1}times x>0>-1>-2>-3...>y>...$$
so $n=color{red}{1}$.
answered Nov 25 at 16:41
hamam_Abdallah
37k21534
If $x>0$ and $yle 0$, then
$$color{red}{1}times x>0>-1>-2>-3...>y>...$$
so $n=color{red}{1}$.
answered Nov 25 at 16:41
hamam_Abdallah
37k21534
edited Nov 25 at 16:44
answered Nov 25 at 16:41
hamam_Abdallah
37k21534
answered Nov 25 at 16:41
hamam_Abdallah
37k21534
answered Nov 25 at 16:41
hamam_Abdallah
37k21534
37k21534
Thank you! I think I did not read the problem carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
Thank you! I think I did not read the problem carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
Thank you! I think I did not read the problem carefully enough.
– Amanda.M
Nov 25 at 16:43
Thank you! I think I did not read the problem carefully enough.
– Amanda.M
Nov 25 at 16:43
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
Nov 25 at 19:19
add a comment |
up vote
1
down vote
Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)
answered Nov 25 at 16:48
Dr. Mathva
645110
Thank you for your alternative answer.
– Amanda.M
Nov 25 at 16:49
add a comment |
up vote
1
down vote
Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)
answered Nov 25 at 16:48
Dr. Mathva
645110
Thank you for your alternative answer.
– Amanda.M
Nov 25 at 16:49
add a comment |
up vote
1
down vote
up vote
1
down vote
Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)
answered Nov 25 at 16:48
Dr. Mathva
645110
Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)
answered Nov 25 at 16:48
Dr. Mathva
645110
answered Nov 25 at 16:48
Dr. Mathva
645110
answered Nov 25 at 16:48
Dr. Mathva
645110
answered Nov 25 at 16:48
Dr. Mathva
645110
645110
Thank you for your alternative answer.
– Amanda.M
Nov 25 at 16:49
add a comment |
Thank you for your alternative answer.
– Amanda.M
Nov 25 at 16:49
Thank you for your alternative answer.
– Amanda.M
Nov 25 at 16:49
Thank you for your alternative answer.
– Amanda.M
Nov 25 at 16:49
add a comment |
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Its trivial since all $n$ satisfy.
– Yadati Kiran
Nov 25 at 16:38