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







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  • 1




    Its trivial since all $n$ satisfy.
    – Yadati Kiran
    2 days ago















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







share|cite|improve this question


















  • 1




    Its trivial since all $n$ satisfy.
    – Yadati Kiran
    2 days ago













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







share|cite|improve this question













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 2 days ago









Amanda.M

1,60411432




1,60411432








  • 1




    Its trivial since all $n$ satisfy.
    – Yadati Kiran
    2 days ago














  • 1




    Its trivial since all $n$ satisfy.
    – Yadati Kiran
    2 days ago








1




1




Its trivial since all $n$ satisfy.
– Yadati Kiran
2 days ago




Its trivial since all $n$ satisfy.
– Yadati Kiran
2 days ago










3 Answers
3






active

oldest

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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.






share|cite|improve this answer





















  • Thank you! I think I did not read the statement carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago


















up vote
4
down vote













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



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



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






share|cite|improve this answer























  • Thank you! I think I did not read the problem carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago


















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 $x>0$)






share|cite|improve this answer





















  • Thank you for your alternative answer.
    – Amanda.M
    2 days ago











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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.






share|cite|improve this answer





















  • Thank you! I think I did not read the statement carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago















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.






share|cite|improve this answer





















  • Thank you! I think I did not read the statement carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago













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.






share|cite|improve this answer












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 2 days ago









Berci

59k23671




59k23671












  • Thank you! I think I did not read the statement carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago


















  • Thank you! I think I did not read the statement carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago
















Thank you! I think I did not read the statement carefully enough.
– Amanda.M
2 days ago




Thank you! I think I did not read the statement carefully enough.
– Amanda.M
2 days ago












I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
2 days ago




I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
2 days ago










up vote
4
down vote













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



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



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






share|cite|improve this answer























  • Thank you! I think I did not read the problem carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago















up vote
4
down vote













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



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



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






share|cite|improve this answer























  • Thank you! I think I did not read the problem carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago













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






share|cite|improve this answer














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 2 days ago

























answered 2 days ago









hamam_Abdallah

36.8k21533




36.8k21533












  • Thank you! I think I did not read the problem carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago


















  • Thank you! I think I did not read the problem carefully enough.
    – Amanda.M
    2 days ago










  • I am so embarrassed - turns out the answer is so simple. Thanks again to all.
    – Amanda.M
    2 days ago
















Thank you! I think I did not read the problem carefully enough.
– Amanda.M
2 days ago




Thank you! I think I did not read the problem carefully enough.
– Amanda.M
2 days ago












I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
2 days ago




I am so embarrassed - turns out the answer is so simple. Thanks again to all.
– Amanda.M
2 days ago










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 $x>0$)






share|cite|improve this answer





















  • Thank you for your alternative answer.
    – Amanda.M
    2 days ago















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 $x>0$)






share|cite|improve this answer





















  • Thank you for your alternative answer.
    – Amanda.M
    2 days ago













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 $x>0$)






share|cite|improve this answer












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 2 days ago









Dr. Mathva

558110




558110












  • Thank you for your alternative answer.
    – Amanda.M
    2 days ago


















  • Thank you for your alternative answer.
    – Amanda.M
    2 days ago
















Thank you for your alternative answer.
– Amanda.M
2 days ago




Thank you for your alternative answer.
– Amanda.M
2 days ago


















 

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