A Question about existence of rationals [on hold]











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If $x,y in mathbb{R}$, $t in mathbb{Q}$ and $t<x+y$. How to proof :Exists two rationals $r,s in mathbb{Q}$ such that $t=r+s$ and $r<x,s<y$?






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put on hold as off-topic by Xander Henderson, rschwieb, amWhy, user21820, José Carlos Santos yesterday


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    If $x,y in mathbb{R}$, $t in mathbb{Q}$ and $t<x+y$. How to proof :Exists two rationals $r,s in mathbb{Q}$ such that $t=r+s$ and $r<x,s<y$?






    real-analysis







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    put on hold as off-topic by Xander Henderson, rschwieb, amWhy, user21820, José Carlos Santos yesterday


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, rschwieb, amWhy, user21820, José Carlos Santos

    If this question can be reworded to fit the rules in the help center, please edit the question.















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      If $x,y in mathbb{R}$, $t in mathbb{Q}$ and $t<x+y$. How to proof :Exists two rationals $r,s in mathbb{Q}$ such that $t=r+s$ and $r<x,s<y$?






      real-analysis







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      If $x,y in mathbb{R}$, $t in mathbb{Q}$ and $t<x+y$. How to proof :Exists two rationals $r,s in mathbb{Q}$ such that $t=r+s$ and $r<x,s<y$?






      real-analysis




      real-analysis






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      asked yesterday









      kemin

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      put on hold as off-topic by Xander Henderson, rschwieb, amWhy, user21820, José Carlos Santos yesterday


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, rschwieb, amWhy, user21820, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.




      put on hold as off-topic by Xander Henderson, rschwieb, amWhy, user21820, José Carlos Santos yesterday


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, rschwieb, amWhy, user21820, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          Let's take $x, y in mathbb{R}$ and $t in mathbb{Q}$ such that $t<x+y$.



          Therefore you have $t-y < x$, so there exists a rationnal $r in mathbb{Q}$ such that $t-y < r <x$.



          You can now consider $s = t-r$. Of course you have $s in mathbb{Q}$ because $t,r in mathbb{Q}$. Moreover, $t=r+s$ obviously. And finally, you have $s = t-r < t-(t-y) = y$.



          You get the result.






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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            -3
            down vote













            Let's take $x, y in mathbb{R}$ and $t in mathbb{Q}$ such that $t<x+y$.



            Therefore you have $t-y < x$, so there exists a rationnal $r in mathbb{Q}$ such that $t-y < r <x$.



            You can now consider $s = t-r$. Of course you have $s in mathbb{Q}$ because $t,r in mathbb{Q}$. Moreover, $t=r+s$ obviously. And finally, you have $s = t-r < t-(t-y) = y$.



            You get the result.






            share|cite|improve this answer

























              up vote
              -3
              down vote













              Let's take $x, y in mathbb{R}$ and $t in mathbb{Q}$ such that $t<x+y$.



              Therefore you have $t-y < x$, so there exists a rationnal $r in mathbb{Q}$ such that $t-y < r <x$.



              You can now consider $s = t-r$. Of course you have $s in mathbb{Q}$ because $t,r in mathbb{Q}$. Moreover, $t=r+s$ obviously. And finally, you have $s = t-r < t-(t-y) = y$.



              You get the result.






              share|cite|improve this answer























                up vote
                -3
                down vote










                up vote
                -3
                down vote









                Let's take $x, y in mathbb{R}$ and $t in mathbb{Q}$ such that $t<x+y$.



                Therefore you have $t-y < x$, so there exists a rationnal $r in mathbb{Q}$ such that $t-y < r <x$.



                You can now consider $s = t-r$. Of course you have $s in mathbb{Q}$ because $t,r in mathbb{Q}$. Moreover, $t=r+s$ obviously. And finally, you have $s = t-r < t-(t-y) = y$.



                You get the result.






                share|cite|improve this answer












                Let's take $x, y in mathbb{R}$ and $t in mathbb{Q}$ such that $t<x+y$.



                Therefore you have $t-y < x$, so there exists a rationnal $r in mathbb{Q}$ such that $t-y < r <x$.



                You can now consider $s = t-r$. Of course you have $s in mathbb{Q}$ because $t,r in mathbb{Q}$. Moreover, $t=r+s$ obviously. And finally, you have $s = t-r < t-(t-y) = y$.



                You get the result.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                TheSilverDoe

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