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Prove that if $a<A$ and $b<B$ then $a+b<A+B$ [closed]
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From the axioms from which $â„Â$ is usually constructed, how could one prove that for all $a, b, A, B in â„Â$ if $$a<A$$ and $$b<B$$ then $$a+b<A+B$$
Any help/thoughts would be really appreciated.
inequality real-numbers ordered-fields
asked Aug 1 at 19:57
Leo
687416
closed as off-topic by Jyrki Lahtonen, Simply Beautiful Art, John Ma, José Carlos Santos, Isaac Browne Aug 2 at 21:05
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." РJyrki Lahtonen, Simply Beautiful Art, John Ma, Jos̩ Carlos Santos, Isaac Browne
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up vote
-1
down vote
favorite
From the axioms from which $â„Â$ is usually constructed, how could one prove that for all $a, b, A, B in â„Â$ if $$a<A$$ and $$b<B$$ then $$a+b<A+B$$
Any help/thoughts would be really appreciated.
inequality real-numbers ordered-fields
asked Aug 1 at 19:57
Leo
687416
closed as off-topic by Jyrki Lahtonen, Simply Beautiful Art, John Ma, José Carlos Santos, Isaac Browne Aug 2 at 21:05
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." РJyrki Lahtonen, Simply Beautiful Art, John Ma, Jos̩ Carlos Santos, Isaac Browne
5
If you add the same thing to both sides of an equality the inequality holds. $A>a implies A+B > a + B$ and $B>b implies a + B > a+ b$ and inequality is transitive. $A+B > a + B > a+b$
– Doug M
Aug 1 at 20:02
@DougM - I was writing the same thing as you, but you have been faster than me. I think you should post it as an answer.
– Taroccoesbrocco
Aug 1 at 20:07
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
From the axioms from which $â„Â$ is usually constructed, how could one prove that for all $a, b, A, B in â„Â$ if $$a<A$$ and $$b<B$$ then $$a+b<A+B$$
Any help/thoughts would be really appreciated.
inequality real-numbers ordered-fields
asked Aug 1 at 19:57
Leo
687416
From the axioms from which $â„Â$ is usually constructed, how could one prove that for all $a, b, A, B in â„Â$ if $$a<A$$ and $$b<B$$ then $$a+b<A+B$$
Any help/thoughts would be really appreciated.
inequality real-numbers ordered-fields
asked Aug 1 at 19:57
Leo
687416
asked Aug 1 at 19:57
Leo
687416
asked Aug 1 at 19:57
Leo
687416
asked Aug 1 at 19:57
Leo
687416
687416
closed as off-topic by Jyrki Lahtonen, Simply Beautiful Art, John Ma, José Carlos Santos, Isaac Browne Aug 2 at 21:05
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." РJyrki Lahtonen, Simply Beautiful Art, John Ma, Jos̩ Carlos Santos, Isaac Browne
closed as off-topic by Jyrki Lahtonen, Simply Beautiful Art, John Ma, José Carlos Santos, Isaac Browne Aug 2 at 21:05
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." РJyrki Lahtonen, Simply Beautiful Art, John Ma, Jos̩ Carlos Santos, Isaac Browne
5
If you add the same thing to both sides of an equality the inequality holds. $A>a implies A+B > a + B$ and $B>b implies a + B > a+ b$ and inequality is transitive. $A+B > a + B > a+b$
– Doug M
Aug 1 at 20:02
@DougM - I was writing the same thing as you, but you have been faster than me. I think you should post it as an answer.
– Taroccoesbrocco
Aug 1 at 20:07
add a comment |Â
5
If you add the same thing to both sides of an equality the inequality holds. $A>a implies A+B > a + B$ and $B>b implies a + B > a+ b$ and inequality is transitive. $A+B > a + B > a+b$
– Doug M
Aug 1 at 20:02
@DougM - I was writing the same thing as you, but you have been faster than me. I think you should post it as an answer.
– Taroccoesbrocco
Aug 1 at 20:07
5
5
If you add the same thing to both sides of an equality the inequality holds. $A>a implies A+B > a + B$ and $B>b implies a + B > a+ b$ and inequality is transitive. $A+B > a + B > a+b$
– Doug M
Aug 1 at 20:02
If you add the same thing to both sides of an equality the inequality holds. $A>a implies A+B > a + B$ and $B>b implies a + B > a+ b$ and inequality is transitive. $A+B > a + B > a+b$
– Doug M
Aug 1 at 20:02
@DougM - I was writing the same thing as you, but you have been faster than me. I think you should post it as an answer.
– Taroccoesbrocco
Aug 1 at 20:07
@DougM - I was writing the same thing as you, but you have been faster than me. I think you should post it as an answer.
– Taroccoesbrocco
Aug 1 at 20:07
add a comment |Â
1 Answer
1
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up vote
0
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We have
$a<Aimplies A-a>0$
$b<Bimplies B-b>0$
therefore
$$A-a+B-b>B-b>0implies A-a+B-b>0 implies a+b<A+B$$
answered Aug 1 at 20:05
gimusi
63.9k73480
1
But, question was to use the axioms stated in the link. I don't think this approach does that
– Rumpelstiltskin
Aug 1 at 20:08
1
@gimusi. I can see how $a-A<0>b-B$. However, how did you get to $a-A+b-B<0$?
– Leo
Aug 1 at 20:09
@Adam Now it should be fixed! Thanks
– gimusi
Aug 1 at 21:09
@Leo Now it should be more clear.
– gimusi
Aug 1 at 21:10
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We have
$a<Aimplies A-a>0$
$b<Bimplies B-b>0$
therefore
$$A-a+B-b>B-b>0implies A-a+B-b>0 implies a+b<A+B$$
answered Aug 1 at 20:05
gimusi
63.9k73480
1
But, question was to use the axioms stated in the link. I don't think this approach does that
– Rumpelstiltskin
Aug 1 at 20:08
1
@gimusi. I can see how $a-A<0>b-B$. However, how did you get to $a-A+b-B<0$?
– Leo
Aug 1 at 20:09
@Adam Now it should be fixed! Thanks
– gimusi
Aug 1 at 21:09
@Leo Now it should be more clear.
– gimusi
Aug 1 at 21:10
add a comment |Â
up vote
0
down vote
We have
$a<Aimplies A-a>0$
$b<Bimplies B-b>0$
therefore
$$A-a+B-b>B-b>0implies A-a+B-b>0 implies a+b<A+B$$
answered Aug 1 at 20:05
gimusi
63.9k73480
1
But, question was to use the axioms stated in the link. I don't think this approach does that
– Rumpelstiltskin
Aug 1 at 20:08
1
@gimusi. I can see how $a-A<0>b-B$. However, how did you get to $a-A+b-B<0$?
– Leo
Aug 1 at 20:09
@Adam Now it should be fixed! Thanks
– gimusi
Aug 1 at 21:09
@Leo Now it should be more clear.
– gimusi
Aug 1 at 21:10
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have
$a<Aimplies A-a>0$
$b<Bimplies B-b>0$
therefore
$$A-a+B-b>B-b>0implies A-a+B-b>0 implies a+b<A+B$$
answered Aug 1 at 20:05
gimusi
63.9k73480
We have
$a<Aimplies A-a>0$
$b<Bimplies B-b>0$
therefore
$$A-a+B-b>B-b>0implies A-a+B-b>0 implies a+b<A+B$$
answered Aug 1 at 20:05
gimusi
63.9k73480
edited Aug 1 at 20:12
answered Aug 1 at 20:05
gimusi
63.9k73480
answered Aug 1 at 20:05
gimusi
63.9k73480
answered Aug 1 at 20:05
gimusi
63.9k73480
63.9k73480
1
But, question was to use the axioms stated in the link. I don't think this approach does that
– Rumpelstiltskin
Aug 1 at 20:08
1
@gimusi. I can see how $a-A<0>b-B$. However, how did you get to $a-A+b-B<0$?
– Leo
Aug 1 at 20:09
@Adam Now it should be fixed! Thanks
– gimusi
Aug 1 at 21:09
@Leo Now it should be more clear.
– gimusi
Aug 1 at 21:10
add a comment |Â
1
But, question was to use the axioms stated in the link. I don't think this approach does that
– Rumpelstiltskin
Aug 1 at 20:08
1
@gimusi. I can see how $a-A<0>b-B$. However, how did you get to $a-A+b-B<0$?
– Leo
Aug 1 at 20:09
@Adam Now it should be fixed! Thanks
– gimusi
Aug 1 at 21:09
@Leo Now it should be more clear.
– gimusi
Aug 1 at 21:10
1
1
But, question was to use the axioms stated in the link. I don't think this approach does that
– Rumpelstiltskin
Aug 1 at 20:08
But, question was to use the axioms stated in the link. I don't think this approach does that
– Rumpelstiltskin
Aug 1 at 20:08
1
1
@gimusi. I can see how $a-A<0>b-B$. However, how did you get to $a-A+b-B<0$?
– Leo
Aug 1 at 20:09
@gimusi. I can see how $a-A<0>b-B$. However, how did you get to $a-A+b-B<0$?
– Leo
Aug 1 at 20:09
@Adam Now it should be fixed! Thanks
– gimusi
Aug 1 at 21:09
@Adam Now it should be fixed! Thanks
– gimusi
Aug 1 at 21:09
@Leo Now it should be more clear.
– gimusi
Aug 1 at 21:10
@Leo Now it should be more clear.
– gimusi
Aug 1 at 21:10
add a comment |Â
5
If you add the same thing to both sides of an equality the inequality holds. $A>a implies A+B > a + B$ and $B>b implies a + B > a+ b$ and inequality is transitive. $A+B > a + B > a+b$
– Doug M
Aug 1 at 20:02
@DougM - I was writing the same thing as you, but you have been faster than me. I think you should post it as an answer.
– Taroccoesbrocco
Aug 1 at 20:07