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 $r<x,s<y$?
real-analysis
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
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.
add a comment |
up vote
-4
down vote
favorite
up vote
-4
down vote
favorite
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
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
asked yesterday
kemin
114
114
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.
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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
1
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.
add a comment |
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.
add a comment |
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.
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.
answered yesterday
TheSilverDoe
1,934111
1,934111
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