Equivalent of Archimedean Property












2












$begingroup$


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







share|cite|improve this question









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  • 1




    $begingroup$
    Its trivial since all $n$ satisfy.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 16:38
















2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Its trivial since all $n$ satisfy.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 16:38














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 '18 at 16:33









Amanda.MAmanda.M

1,62411435




1,62411435








  • 1




    $begingroup$
    Its trivial since all $n$ satisfy.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 16:38














  • 1




    $begingroup$
    Its trivial since all $n$ satisfy.
    $endgroup$
    – Yadati Kiran
    Nov 25 '18 at 16:38








1




1




$begingroup$
Its trivial since all $n$ satisfy.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 16:38




$begingroup$
Its trivial since all $n$ satisfy.
$endgroup$
– Yadati Kiran
Nov 25 '18 at 16:38










3 Answers
3






active

oldest

votes


















4












$begingroup$

If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I think I did not read the statement carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19



















4












$begingroup$

If $x>0$ and $yle 0$, then



$$color{red}{1}times x>0>-1>-2>-3...>y>...$$



so $n=color{red}{1}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! I think I did not read the problem carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19



















1












$begingroup$

Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your alternative answer.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:49











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I think I did not read the statement carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19
















4












$begingroup$

If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you! I think I did not read the statement carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19














4












4








4





$begingroup$

If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.






share|cite|improve this answer









$endgroup$



If $x>0$ but $yle0$, then any natural number $nge1$ satisfies $nx>y$, so it is a trivial case.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 '18 at 16:37









BerciBerci

61.2k23674




61.2k23674












  • $begingroup$
    Thank you! I think I did not read the statement carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19


















  • $begingroup$
    Thank you! I think I did not read the statement carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19
















$begingroup$
Thank you! I think I did not read the statement carefully enough.
$endgroup$
– Amanda.M
Nov 25 '18 at 16:43




$begingroup$
Thank you! I think I did not read the statement carefully enough.
$endgroup$
– Amanda.M
Nov 25 '18 at 16:43












$begingroup$
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
$endgroup$
– Amanda.M
Nov 25 '18 at 19:19




$begingroup$
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
$endgroup$
– Amanda.M
Nov 25 '18 at 19:19











4












$begingroup$

If $x>0$ and $yle 0$, then



$$color{red}{1}times x>0>-1>-2>-3...>y>...$$



so $n=color{red}{1}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! I think I did not read the problem carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19
















4












$begingroup$

If $x>0$ and $yle 0$, then



$$color{red}{1}times x>0>-1>-2>-3...>y>...$$



so $n=color{red}{1}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! I think I did not read the problem carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19














4












4








4





$begingroup$

If $x>0$ and $yle 0$, then



$$color{red}{1}times x>0>-1>-2>-3...>y>...$$



so $n=color{red}{1}$.






share|cite|improve this answer











$endgroup$



If $x>0$ and $yle 0$, then



$$color{red}{1}times x>0>-1>-2>-3...>y>...$$



so $n=color{red}{1}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 '18 at 16:44

























answered Nov 25 '18 at 16:41









hamam_Abdallahhamam_Abdallah

38k21634




38k21634












  • $begingroup$
    Thank you! I think I did not read the problem carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19


















  • $begingroup$
    Thank you! I think I did not read the problem carefully enough.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:43










  • $begingroup$
    I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 19:19
















$begingroup$
Thank you! I think I did not read the problem carefully enough.
$endgroup$
– Amanda.M
Nov 25 '18 at 16:43




$begingroup$
Thank you! I think I did not read the problem carefully enough.
$endgroup$
– Amanda.M
Nov 25 '18 at 16:43












$begingroup$
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
$endgroup$
– Amanda.M
Nov 25 '18 at 19:19




$begingroup$
I am so embarrassed - turns out the answer is so simple. Thanks again to all.
$endgroup$
– Amanda.M
Nov 25 '18 at 19:19











1












$begingroup$

Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your alternative answer.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:49
















1












$begingroup$

Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your alternative answer.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:49














1












1








1





$begingroup$

Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)






share|cite|improve this answer









$endgroup$



Choose simply $$n=lceilfrac{y}{x}rceil+1$$
You will then have $$nx=lceilfrac{y}{x}rceil x+x>y$$
(since $x>0$)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 '18 at 16:48









Dr. MathvaDr. Mathva

2,086324




2,086324












  • $begingroup$
    Thank you for your alternative answer.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:49


















  • $begingroup$
    Thank you for your alternative answer.
    $endgroup$
    – Amanda.M
    Nov 25 '18 at 16:49
















$begingroup$
Thank you for your alternative answer.
$endgroup$
– Amanda.M
Nov 25 '18 at 16:49




$begingroup$
Thank you for your alternative answer.
$endgroup$
– Amanda.M
Nov 25 '18 at 16:49


















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