Can a valid proof by contradiction contradict the opposite proposition?
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Sorry if the wording in the title is a bit off, an example will better illustrate my query. Say I wanted to prove the following: $$ (x_n) text{ and } (y_n) text{ are both real sequences such that } forall n in mathbb{N}, x_n leq y_n. text{If } x_n rightarrow x text{ and } y_n rightarrow y, text{ then } x leq y.$$
To prove this by contradiction we first assume the opposite proposition $ x > y $ and show that this leads to the statement $ x_n > y_n $, which contradicts one of our initial conditions (namely $ x_n leq y_n) $.
Could an equally valid proof use the opposite proposition $ x > y $ and show that this leads to $ x = y $, which contradicts only the opposite proposition and not any initial conditions?
real-analysis proof-writing
asked Dec 27 '18 at 14:10
Michael UdembaMichael Udemba
31
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add a comment |
$begingroup$
Sorry if the wording in the title is a bit off, an example will better illustrate my query. Say I wanted to prove the following: $$ (x_n) text{ and } (y_n) text{ are both real sequences such that } forall n in mathbb{N}, x_n leq y_n. text{If } x_n rightarrow x text{ and } y_n rightarrow y, text{ then } x leq y.$$
To prove this by contradiction we first assume the opposite proposition $ x > y $ and show that this leads to the statement $ x_n > y_n $, which contradicts one of our initial conditions (namely $ x_n leq y_n) $.
Could an equally valid proof use the opposite proposition $ x > y $ and show that this leads to $ x = y $, which contradicts only the opposite proposition and not any initial conditions?
real-analysis proof-writing
asked Dec 27 '18 at 14:10
Michael UdembaMichael Udemba
31
$endgroup$
2
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Yes, it works. Assume as hypothesis : $x > y$. Then derive $x=y$; now, we have a contradiction, because $x > y$ and $x=y$ cannot both be true. Finally, conclude with the negation of the assumption, i.e. with $x le y$.
$endgroup$
– Mauro ALLEGRANZA
Dec 27 '18 at 14:14
add a comment |
$begingroup$
Sorry if the wording in the title is a bit off, an example will better illustrate my query. Say I wanted to prove the following: $$ (x_n) text{ and } (y_n) text{ are both real sequences such that } forall n in mathbb{N}, x_n leq y_n. text{If } x_n rightarrow x text{ and } y_n rightarrow y, text{ then } x leq y.$$
To prove this by contradiction we first assume the opposite proposition $ x > y $ and show that this leads to the statement $ x_n > y_n $, which contradicts one of our initial conditions (namely $ x_n leq y_n) $.
Could an equally valid proof use the opposite proposition $ x > y $ and show that this leads to $ x = y $, which contradicts only the opposite proposition and not any initial conditions?
real-analysis proof-writing
asked Dec 27 '18 at 14:10
Michael UdembaMichael Udemba
31
$endgroup$
Sorry if the wording in the title is a bit off, an example will better illustrate my query. Say I wanted to prove the following: $$ (x_n) text{ and } (y_n) text{ are both real sequences such that } forall n in mathbb{N}, x_n leq y_n. text{If } x_n rightarrow x text{ and } y_n rightarrow y, text{ then } x leq y.$$
To prove this by contradiction we first assume the opposite proposition $ x > y $ and show that this leads to the statement $ x_n > y_n $, which contradicts one of our initial conditions (namely $ x_n leq y_n) $.
Could an equally valid proof use the opposite proposition $ x > y $ and show that this leads to $ x = y $, which contradicts only the opposite proposition and not any initial conditions?
real-analysis proof-writing
real-analysis proof-writing
asked Dec 27 '18 at 14:10
Michael UdembaMichael Udemba
31
asked Dec 27 '18 at 14:10
Michael UdembaMichael Udemba
31
asked Dec 27 '18 at 14:10
Michael UdembaMichael Udemba
31
asked Dec 27 '18 at 14:10
Michael UdembaMichael Udemba
31
asked Dec 27 '18 at 14:10
Michael UdembaMichael Udemba
31
31
2
$begingroup$
Yes, it works. Assume as hypothesis : $x > y$. Then derive $x=y$; now, we have a contradiction, because $x > y$ and $x=y$ cannot both be true. Finally, conclude with the negation of the assumption, i.e. with $x le y$.
$endgroup$
– Mauro ALLEGRANZA
Dec 27 '18 at 14:14
add a comment |
2
$begingroup$
Yes, it works. Assume as hypothesis : $x > y$. Then derive $x=y$; now, we have a contradiction, because $x > y$ and $x=y$ cannot both be true. Finally, conclude with the negation of the assumption, i.e. with $x le y$.
$endgroup$
– Mauro ALLEGRANZA
Dec 27 '18 at 14:14
2
2
$begingroup$
Yes, it works. Assume as hypothesis : $x > y$. Then derive $x=y$; now, we have a contradiction, because $x > y$ and $x=y$ cannot both be true. Finally, conclude with the negation of the assumption, i.e. with $x le y$.
$endgroup$
– Mauro ALLEGRANZA
Dec 27 '18 at 14:14
$begingroup$
Yes, it works. Assume as hypothesis : $x > y$. Then derive $x=y$; now, we have a contradiction, because $x > y$ and $x=y$ cannot both be true. Finally, conclude with the negation of the assumption, i.e. with $x le y$.
$endgroup$
– Mauro ALLEGRANZA
Dec 27 '18 at 14:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think the short answer is yes: You may have a (very) common misconception about proof by contradiction.
To prove something by contradiction you merely derive any logical contradiction. In basic analysis proofs, it is often clean for your contradiction to be of the form $1 < 0$ or something concrete, as opposed to contradicting the hypotheses.
answered Dec 27 '18 at 14:14
T_MT_M
1,12827
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add a comment |
$begingroup$
As you can typically prove anything when working in an inconsistent system, any contradiction, whether it's (most obviously) to some condition in the formulation of the statement, to the assumption made when beginning the proof, to basic arithmetic (i.e $1>2$), ... will be the contradiction you look for (and - in most cases - can be used to "prove" any other contradiction).
answered Dec 27 '18 at 14:22
HenrikHenrik
6,03592030
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the short answer is yes: You may have a (very) common misconception about proof by contradiction.
To prove something by contradiction you merely derive any logical contradiction. In basic analysis proofs, it is often clean for your contradiction to be of the form $1 < 0$ or something concrete, as opposed to contradicting the hypotheses.
answered Dec 27 '18 at 14:14
T_MT_M
1,12827
$endgroup$
add a comment |
$begingroup$
I think the short answer is yes: You may have a (very) common misconception about proof by contradiction.
To prove something by contradiction you merely derive any logical contradiction. In basic analysis proofs, it is often clean for your contradiction to be of the form $1 < 0$ or something concrete, as opposed to contradicting the hypotheses.
answered Dec 27 '18 at 14:14
T_MT_M
1,12827
$endgroup$
add a comment |
$begingroup$
I think the short answer is yes: You may have a (very) common misconception about proof by contradiction.
To prove something by contradiction you merely derive any logical contradiction. In basic analysis proofs, it is often clean for your contradiction to be of the form $1 < 0$ or something concrete, as opposed to contradicting the hypotheses.
answered Dec 27 '18 at 14:14
T_MT_M
1,12827
$endgroup$
I think the short answer is yes: You may have a (very) common misconception about proof by contradiction.
To prove something by contradiction you merely derive any logical contradiction. In basic analysis proofs, it is often clean for your contradiction to be of the form $1 < 0$ or something concrete, as opposed to contradicting the hypotheses.
answered Dec 27 '18 at 14:14
T_MT_M
1,12827
answered Dec 27 '18 at 14:14
T_MT_M
1,12827
answered Dec 27 '18 at 14:14
T_MT_M
1,12827
answered Dec 27 '18 at 14:14
T_MT_M
1,12827
1,12827
add a comment |
add a comment |
$begingroup$
As you can typically prove anything when working in an inconsistent system, any contradiction, whether it's (most obviously) to some condition in the formulation of the statement, to the assumption made when beginning the proof, to basic arithmetic (i.e $1>2$), ... will be the contradiction you look for (and - in most cases - can be used to "prove" any other contradiction).
answered Dec 27 '18 at 14:22
HenrikHenrik
6,03592030
$endgroup$
add a comment |
$begingroup$
As you can typically prove anything when working in an inconsistent system, any contradiction, whether it's (most obviously) to some condition in the formulation of the statement, to the assumption made when beginning the proof, to basic arithmetic (i.e $1>2$), ... will be the contradiction you look for (and - in most cases - can be used to "prove" any other contradiction).
answered Dec 27 '18 at 14:22
HenrikHenrik
6,03592030
$endgroup$
add a comment |
$begingroup$
As you can typically prove anything when working in an inconsistent system, any contradiction, whether it's (most obviously) to some condition in the formulation of the statement, to the assumption made when beginning the proof, to basic arithmetic (i.e $1>2$), ... will be the contradiction you look for (and - in most cases - can be used to "prove" any other contradiction).
answered Dec 27 '18 at 14:22
HenrikHenrik
6,03592030
$endgroup$
As you can typically prove anything when working in an inconsistent system, any contradiction, whether it's (most obviously) to some condition in the formulation of the statement, to the assumption made when beginning the proof, to basic arithmetic (i.e $1>2$), ... will be the contradiction you look for (and - in most cases - can be used to "prove" any other contradiction).
answered Dec 27 '18 at 14:22
HenrikHenrik
6,03592030
answered Dec 27 '18 at 14:22
HenrikHenrik
6,03592030
answered Dec 27 '18 at 14:22
HenrikHenrik
6,03592030
answered Dec 27 '18 at 14:22
HenrikHenrik
6,03592030
6,03592030
add a comment |
add a comment |
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$begingroup$
Yes, it works. Assume as hypothesis : $x > y$. Then derive $x=y$; now, we have a contradiction, because $x > y$ and $x=y$ cannot both be true. Finally, conclude with the negation of the assumption, i.e. with $x le y$.
$endgroup$
– Mauro ALLEGRANZA
Dec 27 '18 at 14:14