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Constructive intermediate value theorem
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I have given real numbers $x_1,x_2,y_1,y_2$ such that $x_1 > x_2$ and $y_1 < y_2$. The the claim is that there exists some $lambda in (0,1)$ such that $lambda (x_1 - x_2) + (1-lambda)(y_1-y_2) = 0$. In order to proof this, one needs ( at least in my opinion) the intermediate value theorem. But the intermediate value theorem does not hold in constructive mathematics (that is without the law of excluded middle; or constructive mathematics acts in intuitionistic logic). For a proof of this c.f. this paper.
Is there any constructive way to show the above equation?
analysis logic constructive-mathematics
asked 5 hours ago
Diamir
145111
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up vote
2
down vote
favorite
I have given real numbers $x_1,x_2,y_1,y_2$ such that $x_1 > x_2$ and $y_1 < y_2$. The the claim is that there exists some $lambda in (0,1)$ such that $lambda (x_1 - x_2) + (1-lambda)(y_1-y_2) = 0$. In order to proof this, one needs ( at least in my opinion) the intermediate value theorem. But the intermediate value theorem does not hold in constructive mathematics (that is without the law of excluded middle; or constructive mathematics acts in intuitionistic logic). For a proof of this c.f. this paper.
Is there any constructive way to show the above equation?
analysis logic constructive-mathematics
asked 5 hours ago
Diamir
145111
Anyway, the intermediate value theorem can be proved in a "constructive" way. (I put the "..." because I don't know a word about constructive mathematics). By bisecting iteratively, you construct a sequence that turns out to be Cauchy, and so it has a limit.
– Giuseppe Negro
3 hours ago
@GiuseppeNegro the "construction" is not "constructive" enough for this purpose; it uses "either the left or the right works, so pick one, and ..." but constructions of this kind (especially with an infinite number of such choices) are forbidden in that context
– Richard Rast
20 secs ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have given real numbers $x_1,x_2,y_1,y_2$ such that $x_1 > x_2$ and $y_1 < y_2$. The the claim is that there exists some $lambda in (0,1)$ such that $lambda (x_1 - x_2) + (1-lambda)(y_1-y_2) = 0$. In order to proof this, one needs ( at least in my opinion) the intermediate value theorem. But the intermediate value theorem does not hold in constructive mathematics (that is without the law of excluded middle; or constructive mathematics acts in intuitionistic logic). For a proof of this c.f. this paper.
Is there any constructive way to show the above equation?
analysis logic constructive-mathematics
asked 5 hours ago
Diamir
145111
I have given real numbers $x_1,x_2,y_1,y_2$ such that $x_1 > x_2$ and $y_1 < y_2$. The the claim is that there exists some $lambda in (0,1)$ such that $lambda (x_1 - x_2) + (1-lambda)(y_1-y_2) = 0$. In order to proof this, one needs ( at least in my opinion) the intermediate value theorem. But the intermediate value theorem does not hold in constructive mathematics (that is without the law of excluded middle; or constructive mathematics acts in intuitionistic logic). For a proof of this c.f. this paper.
Is there any constructive way to show the above equation?
analysis logic constructive-mathematics
analysis logic constructive-mathematics
asked 5 hours ago
Diamir
145111
asked 5 hours ago
Diamir
145111
asked 5 hours ago
Diamir
145111
asked 5 hours ago
Diamir
145111
asked 5 hours ago
Diamir
145111
145111
Anyway, the intermediate value theorem can be proved in a "constructive" way. (I put the "..." because I don't know a word about constructive mathematics). By bisecting iteratively, you construct a sequence that turns out to be Cauchy, and so it has a limit.
– Giuseppe Negro
3 hours ago
@GiuseppeNegro the "construction" is not "constructive" enough for this purpose; it uses "either the left or the right works, so pick one, and ..." but constructions of this kind (especially with an infinite number of such choices) are forbidden in that context
– Richard Rast
20 secs ago
add a comment |Â
Anyway, the intermediate value theorem can be proved in a "constructive" way. (I put the "..." because I don't know a word about constructive mathematics). By bisecting iteratively, you construct a sequence that turns out to be Cauchy, and so it has a limit.
– Giuseppe Negro
3 hours ago
@GiuseppeNegro the "construction" is not "constructive" enough for this purpose; it uses "either the left or the right works, so pick one, and ..." but constructions of this kind (especially with an infinite number of such choices) are forbidden in that context
– Richard Rast
20 secs ago
Anyway, the intermediate value theorem can be proved in a "constructive" way. (I put the "..." because I don't know a word about constructive mathematics). By bisecting iteratively, you construct a sequence that turns out to be Cauchy, and so it has a limit.
– Giuseppe Negro
3 hours ago
Anyway, the intermediate value theorem can be proved in a "constructive" way. (I put the "..." because I don't know a word about constructive mathematics). By bisecting iteratively, you construct a sequence that turns out to be Cauchy, and so it has a limit.
– Giuseppe Negro
3 hours ago
@GiuseppeNegro the "construction" is not "constructive" enough for this purpose; it uses "either the left or the right works, so pick one, and ..." but constructions of this kind (especially with an infinite number of such choices) are forbidden in that context
– Richard Rast
20 secs ago
@GiuseppeNegro the "construction" is not "constructive" enough for this purpose; it uses "either the left or the right works, so pick one, and ..." but constructions of this kind (especially with an infinite number of such choices) are forbidden in that context
– Richard Rast
20 secs ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
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accepted
Just solve the equation for $lambda$. You get $lambda =frac y_2-y_1x_1-x_2+y_2-y_1$.
answered 5 hours ago
Kavi Rama Murthy
26.3k31437
add a comment |Â
up vote
3
down vote
You can explicitly solve the equation;
$$lambda= fracy_2-y_1x_1-x_2-y_1+y_2.$$
answered 5 hours ago
Giuseppe Negro
16.2k328118
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Just solve the equation for $lambda$. You get $lambda =frac y_2-y_1x_1-x_2+y_2-y_1$.
answered 5 hours ago
Kavi Rama Murthy
26.3k31437
add a comment |Â
up vote
5
down vote
accepted
Just solve the equation for $lambda$. You get $lambda =frac y_2-y_1x_1-x_2+y_2-y_1$.
answered 5 hours ago
Kavi Rama Murthy
26.3k31437
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Just solve the equation for $lambda$. You get $lambda =frac y_2-y_1x_1-x_2+y_2-y_1$.
answered 5 hours ago
Kavi Rama Murthy
26.3k31437
Just solve the equation for $lambda$. You get $lambda =frac y_2-y_1x_1-x_2+y_2-y_1$.
answered 5 hours ago
Kavi Rama Murthy
26.3k31437
answered 5 hours ago
Kavi Rama Murthy
26.3k31437
answered 5 hours ago
Kavi Rama Murthy
26.3k31437
answered 5 hours ago
Kavi Rama Murthy
26.3k31437
26.3k31437
add a comment |Â
add a comment |Â
up vote
3
down vote
You can explicitly solve the equation;
$$lambda= fracy_2-y_1x_1-x_2-y_1+y_2.$$
answered 5 hours ago
Giuseppe Negro
16.2k328118
add a comment |Â
up vote
3
down vote
You can explicitly solve the equation;
$$lambda= fracy_2-y_1x_1-x_2-y_1+y_2.$$
answered 5 hours ago
Giuseppe Negro
16.2k328118
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You can explicitly solve the equation;
$$lambda= fracy_2-y_1x_1-x_2-y_1+y_2.$$
answered 5 hours ago
Giuseppe Negro
16.2k328118
You can explicitly solve the equation;
$$lambda= fracy_2-y_1x_1-x_2-y_1+y_2.$$
answered 5 hours ago
Giuseppe Negro
16.2k328118
answered 5 hours ago
Giuseppe Negro
16.2k328118
answered 5 hours ago
Giuseppe Negro
16.2k328118
answered 5 hours ago
Giuseppe Negro
16.2k328118
16.2k328118
add a comment |Â
add a comment |Â
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Anyway, the intermediate value theorem can be proved in a "constructive" way. (I put the "..." because I don't know a word about constructive mathematics). By bisecting iteratively, you construct a sequence that turns out to be Cauchy, and so it has a limit.
– Giuseppe Negro
3 hours ago
@GiuseppeNegro the "construction" is not "constructive" enough for this purpose; it uses "either the left or the right works, so pick one, and ..." but constructions of this kind (especially with an infinite number of such choices) are forbidden in that context
– Richard Rast
20 secs ago