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Proof by contradiction. Statement negation
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This should be an easy question. Yet, the provided solution confuses me.
The question comes from "Understanding analysis" by S. Abbot, 2nd edition (Exercise 1.2.11).
Negate the statement. Make an intuitive guess as to whether the claim or its negation is the true statement.
(b) There exists a real number $x > 0$ such that $x < 1/n;;forall n in mathbbN$.
The provided solution says:
The solution seems correct, apart from: shouldn't the negation be with $exists n in mathbbN$, i.e.:
$$forall x >0 ;; exists n in mathbbN: x geq 1/n$$
?
real-analysis
asked 1 hour ago
Sandu Ursu
146112
 |Â
show 2 more comments
up vote
2
down vote
favorite
This should be an easy question. Yet, the provided solution confuses me.
The question comes from "Understanding analysis" by S. Abbot, 2nd edition (Exercise 1.2.11).
Negate the statement. Make an intuitive guess as to whether the claim or its negation is the true statement.
(b) There exists a real number $x > 0$ such that $x < 1/n;;forall n in mathbbN$.
The provided solution says:
The solution seems correct, apart from: shouldn't the negation be with $exists n in mathbbN$, i.e.:
$$forall x >0 ;; exists n in mathbbN: x geq 1/n$$
?
real-analysis
asked 1 hour ago
Sandu Ursu
146112
1
The statement given is false, so it cannot be proved, not by contradiction nor by any other method. Your negation of the statement is correct, and is therefore a true statement.
– Simon
1 hour ago
I suspect that you have made a mistake in transcribing the problem from the book, or else that it is a typo in the book.
– Simon
59 mins ago
The statement has been correctly transcribed from the book.
– Sandu Ursu
57 mins ago
Thank you for the link in your comment below. Indeed you have transcribed the statement correctly, but the question is not "Provide a proof by contradiction for" the statement, it is simply to negate the statement and put your answer in a "positive" form, and then to make an educated guess as to which of the statement and its negation is true.
– Simon
40 mins ago
1
The solution that you quote is a flawed attempt at a proof of a true statement, that is not the same as the true statement that you correctly give as the negation of the statement given in the question. The flaw in the purported proof is that "for all n" should be "there exists n".
– Simon
26 mins ago
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This should be an easy question. Yet, the provided solution confuses me.
The question comes from "Understanding analysis" by S. Abbot, 2nd edition (Exercise 1.2.11).
Negate the statement. Make an intuitive guess as to whether the claim or its negation is the true statement.
(b) There exists a real number $x > 0$ such that $x < 1/n;;forall n in mathbbN$.
The provided solution says:
The solution seems correct, apart from: shouldn't the negation be with $exists n in mathbbN$, i.e.:
$$forall x >0 ;; exists n in mathbbN: x geq 1/n$$
?
real-analysis
asked 1 hour ago
Sandu Ursu
146112
This should be an easy question. Yet, the provided solution confuses me.
The question comes from "Understanding analysis" by S. Abbot, 2nd edition (Exercise 1.2.11).
Negate the statement. Make an intuitive guess as to whether the claim or its negation is the true statement.
(b) There exists a real number $x > 0$ such that $x < 1/n;;forall n in mathbbN$.
The provided solution says:
The solution seems correct, apart from: shouldn't the negation be with $exists n in mathbbN$, i.e.:
$$forall x >0 ;; exists n in mathbbN: x geq 1/n$$
?
real-analysis
real-analysis
asked 1 hour ago
Sandu Ursu
146112
asked 1 hour ago
Sandu Ursu
146112
edited 4 mins ago
asked 1 hour ago
Sandu Ursu
146112
asked 1 hour ago
Sandu Ursu
146112
asked 1 hour ago
Sandu Ursu
146112
146112
1
The statement given is false, so it cannot be proved, not by contradiction nor by any other method. Your negation of the statement is correct, and is therefore a true statement.
– Simon
1 hour ago
I suspect that you have made a mistake in transcribing the problem from the book, or else that it is a typo in the book.
– Simon
59 mins ago
The statement has been correctly transcribed from the book.
– Sandu Ursu
57 mins ago
Thank you for the link in your comment below. Indeed you have transcribed the statement correctly, but the question is not "Provide a proof by contradiction for" the statement, it is simply to negate the statement and put your answer in a "positive" form, and then to make an educated guess as to which of the statement and its negation is true.
– Simon
40 mins ago
1
The solution that you quote is a flawed attempt at a proof of a true statement, that is not the same as the true statement that you correctly give as the negation of the statement given in the question. The flaw in the purported proof is that "for all n" should be "there exists n".
– Simon
26 mins ago
 |Â
show 2 more comments
1
The statement given is false, so it cannot be proved, not by contradiction nor by any other method. Your negation of the statement is correct, and is therefore a true statement.
– Simon
1 hour ago
I suspect that you have made a mistake in transcribing the problem from the book, or else that it is a typo in the book.
– Simon
59 mins ago
The statement has been correctly transcribed from the book.
– Sandu Ursu
57 mins ago
Thank you for the link in your comment below. Indeed you have transcribed the statement correctly, but the question is not "Provide a proof by contradiction for" the statement, it is simply to negate the statement and put your answer in a "positive" form, and then to make an educated guess as to which of the statement and its negation is true.
– Simon
40 mins ago
1
The solution that you quote is a flawed attempt at a proof of a true statement, that is not the same as the true statement that you correctly give as the negation of the statement given in the question. The flaw in the purported proof is that "for all n" should be "there exists n".
– Simon
26 mins ago
1
1
The statement given is false, so it cannot be proved, not by contradiction nor by any other method. Your negation of the statement is correct, and is therefore a true statement.
– Simon
1 hour ago
The statement given is false, so it cannot be proved, not by contradiction nor by any other method. Your negation of the statement is correct, and is therefore a true statement.
– Simon
1 hour ago
I suspect that you have made a mistake in transcribing the problem from the book, or else that it is a typo in the book.
– Simon
59 mins ago
I suspect that you have made a mistake in transcribing the problem from the book, or else that it is a typo in the book.
– Simon
59 mins ago
The statement has been correctly transcribed from the book.
– Sandu Ursu
57 mins ago
The statement has been correctly transcribed from the book.
– Sandu Ursu
57 mins ago
Thank you for the link in your comment below. Indeed you have transcribed the statement correctly, but the question is not "Provide a proof by contradiction for" the statement, it is simply to negate the statement and put your answer in a "positive" form, and then to make an educated guess as to which of the statement and its negation is true.
– Simon
40 mins ago
Thank you for the link in your comment below. Indeed you have transcribed the statement correctly, but the question is not "Provide a proof by contradiction for" the statement, it is simply to negate the statement and put your answer in a "positive" form, and then to make an educated guess as to which of the statement and its negation is true.
– Simon
40 mins ago
1
1
The solution that you quote is a flawed attempt at a proof of a true statement, that is not the same as the true statement that you correctly give as the negation of the statement given in the question. The flaw in the purported proof is that "for all n" should be "there exists n".
– Simon
26 mins ago
The solution that you quote is a flawed attempt at a proof of a true statement, that is not the same as the true statement that you correctly give as the negation of the statement given in the question. The flaw in the purported proof is that "for all n" should be "there exists n".
– Simon
26 mins ago
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
Regarding Ex.1.2.11 (b) :
Form the logical negation of [...] there exists a real number $x > 0$ such that $x < dfrac 1 n$ for all $n in mathbb N$,
the formula to be negated is :
$exists x > 0 forall n in mathbb N (x < dfrac 1 n)$.
Thus, tou are right. The correct negation will be :
$forall x > 0 exists n in mathbb N (x ge dfrac 1 n)$.
answered 52 mins ago
Mauro ALLEGRANZA
62.3k447106
Is this in the 2nd edition of the book? (I've just edited my description to mention that)
– Sandu Ursu
50 mins ago
1
Exercise 1.2.11 here
– Sandu Ursu
45 mins ago
add a comment |Â
up vote
1
down vote
This is why putting quantifiers at the end of a formula is a bad practice. It creates ambiguity. The statements
- $(forall n in mathbbN)(exists x > 0)(x < frac1n)$
$(exists x > 0)(forall n in mathbbN)(x < frac1n)$
are not equivalent. The second one is obviously false, however it's more likely to interpret your formulation as the second statement. Undoubtedly, the first statement is what's actually meant. For proving by contradiction, we need its negation which goes as follows:
$$(exists n in mathbbN)(forall x > 0)left(x ge frac1nright)$$
answered 46 mins ago
Drinkwater
32029
The second statement (which is the one from the question) implies the first one (which is the one from the provided solution).
– Sandu Ursu
33 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Regarding Ex.1.2.11 (b) :
Form the logical negation of [...] there exists a real number $x > 0$ such that $x < dfrac 1 n$ for all $n in mathbb N$,
the formula to be negated is :
$exists x > 0 forall n in mathbb N (x < dfrac 1 n)$.
Thus, tou are right. The correct negation will be :
$forall x > 0 exists n in mathbb N (x ge dfrac 1 n)$.
answered 52 mins ago
Mauro ALLEGRANZA
62.3k447106
Is this in the 2nd edition of the book? (I've just edited my description to mention that)
– Sandu Ursu
50 mins ago
1
Exercise 1.2.11 here
– Sandu Ursu
45 mins ago
add a comment |Â
up vote
2
down vote
Regarding Ex.1.2.11 (b) :
Form the logical negation of [...] there exists a real number $x > 0$ such that $x < dfrac 1 n$ for all $n in mathbb N$,
the formula to be negated is :
$exists x > 0 forall n in mathbb N (x < dfrac 1 n)$.
Thus, tou are right. The correct negation will be :
$forall x > 0 exists n in mathbb N (x ge dfrac 1 n)$.
answered 52 mins ago
Mauro ALLEGRANZA
62.3k447106
Is this in the 2nd edition of the book? (I've just edited my description to mention that)
– Sandu Ursu
50 mins ago
1
Exercise 1.2.11 here
– Sandu Ursu
45 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Regarding Ex.1.2.11 (b) :
Form the logical negation of [...] there exists a real number $x > 0$ such that $x < dfrac 1 n$ for all $n in mathbb N$,
the formula to be negated is :
$exists x > 0 forall n in mathbb N (x < dfrac 1 n)$.
Thus, tou are right. The correct negation will be :
$forall x > 0 exists n in mathbb N (x ge dfrac 1 n)$.
answered 52 mins ago
Mauro ALLEGRANZA
62.3k447106
Regarding Ex.1.2.11 (b) :
Form the logical negation of [...] there exists a real number $x > 0$ such that $x < dfrac 1 n$ for all $n in mathbb N$,
the formula to be negated is :
$exists x > 0 forall n in mathbb N (x < dfrac 1 n)$.
Thus, tou are right. The correct negation will be :
$forall x > 0 exists n in mathbb N (x ge dfrac 1 n)$.
answered 52 mins ago
Mauro ALLEGRANZA
62.3k447106
edited 36 mins ago
answered 52 mins ago
Mauro ALLEGRANZA
62.3k447106
answered 52 mins ago
Mauro ALLEGRANZA
62.3k447106
answered 52 mins ago
Mauro ALLEGRANZA
62.3k447106
62.3k447106
Is this in the 2nd edition of the book? (I've just edited my description to mention that)
– Sandu Ursu
50 mins ago
1
Exercise 1.2.11 here
– Sandu Ursu
45 mins ago
add a comment |Â
Is this in the 2nd edition of the book? (I've just edited my description to mention that)
– Sandu Ursu
50 mins ago
1
Exercise 1.2.11 here
– Sandu Ursu
45 mins ago
Is this in the 2nd edition of the book? (I've just edited my description to mention that)
– Sandu Ursu
50 mins ago
Is this in the 2nd edition of the book? (I've just edited my description to mention that)
– Sandu Ursu
50 mins ago
1
1
Exercise 1.2.11 here
– Sandu Ursu
45 mins ago
Exercise 1.2.11 here
– Sandu Ursu
45 mins ago
add a comment |Â
up vote
1
down vote
This is why putting quantifiers at the end of a formula is a bad practice. It creates ambiguity. The statements
- $(forall n in mathbbN)(exists x > 0)(x < frac1n)$
$(exists x > 0)(forall n in mathbbN)(x < frac1n)$
are not equivalent. The second one is obviously false, however it's more likely to interpret your formulation as the second statement. Undoubtedly, the first statement is what's actually meant. For proving by contradiction, we need its negation which goes as follows:
$$(exists n in mathbbN)(forall x > 0)left(x ge frac1nright)$$
answered 46 mins ago
Drinkwater
32029
The second statement (which is the one from the question) implies the first one (which is the one from the provided solution).
– Sandu Ursu
33 mins ago
add a comment |Â
up vote
1
down vote
This is why putting quantifiers at the end of a formula is a bad practice. It creates ambiguity. The statements
- $(forall n in mathbbN)(exists x > 0)(x < frac1n)$
$(exists x > 0)(forall n in mathbbN)(x < frac1n)$
are not equivalent. The second one is obviously false, however it's more likely to interpret your formulation as the second statement. Undoubtedly, the first statement is what's actually meant. For proving by contradiction, we need its negation which goes as follows:
$$(exists n in mathbbN)(forall x > 0)left(x ge frac1nright)$$
answered 46 mins ago
Drinkwater
32029
The second statement (which is the one from the question) implies the first one (which is the one from the provided solution).
– Sandu Ursu
33 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is why putting quantifiers at the end of a formula is a bad practice. It creates ambiguity. The statements
- $(forall n in mathbbN)(exists x > 0)(x < frac1n)$
$(exists x > 0)(forall n in mathbbN)(x < frac1n)$
are not equivalent. The second one is obviously false, however it's more likely to interpret your formulation as the second statement. Undoubtedly, the first statement is what's actually meant. For proving by contradiction, we need its negation which goes as follows:
$$(exists n in mathbbN)(forall x > 0)left(x ge frac1nright)$$
answered 46 mins ago
Drinkwater
32029
This is why putting quantifiers at the end of a formula is a bad practice. It creates ambiguity. The statements
- $(forall n in mathbbN)(exists x > 0)(x < frac1n)$
$(exists x > 0)(forall n in mathbbN)(x < frac1n)$
are not equivalent. The second one is obviously false, however it's more likely to interpret your formulation as the second statement. Undoubtedly, the first statement is what's actually meant. For proving by contradiction, we need its negation which goes as follows:
$$(exists n in mathbbN)(forall x > 0)left(x ge frac1nright)$$
answered 46 mins ago
Drinkwater
32029
answered 46 mins ago
Drinkwater
32029
answered 46 mins ago
Drinkwater
32029
answered 46 mins ago
Drinkwater
32029
32029
The second statement (which is the one from the question) implies the first one (which is the one from the provided solution).
– Sandu Ursu
33 mins ago
add a comment |Â
The second statement (which is the one from the question) implies the first one (which is the one from the provided solution).
– Sandu Ursu
33 mins ago
The second statement (which is the one from the question) implies the first one (which is the one from the provided solution).
– Sandu Ursu
33 mins ago
The second statement (which is the one from the question) implies the first one (which is the one from the provided solution).
– Sandu Ursu
33 mins ago
add a comment |Â
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1
The statement given is false, so it cannot be proved, not by contradiction nor by any other method. Your negation of the statement is correct, and is therefore a true statement.
– Simon
1 hour ago
I suspect that you have made a mistake in transcribing the problem from the book, or else that it is a typo in the book.
– Simon
59 mins ago
The statement has been correctly transcribed from the book.
– Sandu Ursu
57 mins ago
Thank you for the link in your comment below. Indeed you have transcribed the statement correctly, but the question is not "Provide a proof by contradiction for" the statement, it is simply to negate the statement and put your answer in a "positive" form, and then to make an educated guess as to which of the statement and its negation is true.
– Simon
40 mins ago
1
The solution that you quote is a flawed attempt at a proof of a true statement, that is not the same as the true statement that you correctly give as the negation of the statement given in the question. The flaw in the purported proof is that "for all n" should be "there exists n".
– Simon
26 mins ago