Let k be an integer. Disprove: “The equation x^2 − x − k = 0 has no integer solution if and only if k...
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My problem is I keep ending up proving the statement true, instead of disproving it. I was getting it mixed up in my mind so I broke it down into very explicit steps but now I'm wondering if I'm overthinking it?
I started out proving that if there is no integer solution, then k is even (one negation of the statement, P and not Q where P = no integer solution and Q = k is odd). By contrapositive I try to prove that if k is odd then there is an integer solution (I quickly recognized this is the same as the other negation, Q and not P, so this is the only thing I need to prove).
So k = 2n + 1 for any integer n. Then x^2-x -(2n+1) = 0. So x^2-x=2n+1. But if x is an integer then x^2-x must always be even, and cannot equal an odd integer 2n+1. So this has led to a contradiction, which means I just proved that if k is odd then there is no integer solution.
Is my logic off somewhere, or should I be approaching it differently? Sorry if this has an obvious answer, I only started doing proofs this quarter. Thanks
proof-verification
asked 3 hours ago
jenellejenelle
132
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add a comment |
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My problem is I keep ending up proving the statement true, instead of disproving it. I was getting it mixed up in my mind so I broke it down into very explicit steps but now I'm wondering if I'm overthinking it?
I started out proving that if there is no integer solution, then k is even (one negation of the statement, P and not Q where P = no integer solution and Q = k is odd). By contrapositive I try to prove that if k is odd then there is an integer solution (I quickly recognized this is the same as the other negation, Q and not P, so this is the only thing I need to prove).
So k = 2n + 1 for any integer n. Then x^2-x -(2n+1) = 0. So x^2-x=2n+1. But if x is an integer then x^2-x must always be even, and cannot equal an odd integer 2n+1. So this has led to a contradiction, which means I just proved that if k is odd then there is no integer solution.
Is my logic off somewhere, or should I be approaching it differently? Sorry if this has an obvious answer, I only started doing proofs this quarter. Thanks
proof-verification
asked 3 hours ago
jenellejenelle
132
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jenelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
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– José Carlos Santos
3 hours ago
$begingroup$
You proved in the end that ($k$ is odd) $implies$ (no integer solutions). Now you still have to prove the sufficient condition and that is if there is (no integer solutions) $implies$ ($k$ is odd), which is equivalent to proving ($k$ is not odd) $implies$ (there exist an integer solution). I think you didn't prove this, am I right?
$endgroup$
– Fareed AF
3 hours ago
add a comment |
$begingroup$
My problem is I keep ending up proving the statement true, instead of disproving it. I was getting it mixed up in my mind so I broke it down into very explicit steps but now I'm wondering if I'm overthinking it?
I started out proving that if there is no integer solution, then k is even (one negation of the statement, P and not Q where P = no integer solution and Q = k is odd). By contrapositive I try to prove that if k is odd then there is an integer solution (I quickly recognized this is the same as the other negation, Q and not P, so this is the only thing I need to prove).
So k = 2n + 1 for any integer n. Then x^2-x -(2n+1) = 0. So x^2-x=2n+1. But if x is an integer then x^2-x must always be even, and cannot equal an odd integer 2n+1. So this has led to a contradiction, which means I just proved that if k is odd then there is no integer solution.
Is my logic off somewhere, or should I be approaching it differently? Sorry if this has an obvious answer, I only started doing proofs this quarter. Thanks
proof-verification
asked 3 hours ago
jenellejenelle
132
New contributor
jenelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
My problem is I keep ending up proving the statement true, instead of disproving it. I was getting it mixed up in my mind so I broke it down into very explicit steps but now I'm wondering if I'm overthinking it?
I started out proving that if there is no integer solution, then k is even (one negation of the statement, P and not Q where P = no integer solution and Q = k is odd). By contrapositive I try to prove that if k is odd then there is an integer solution (I quickly recognized this is the same as the other negation, Q and not P, so this is the only thing I need to prove).
So k = 2n + 1 for any integer n. Then x^2-x -(2n+1) = 0. So x^2-x=2n+1. But if x is an integer then x^2-x must always be even, and cannot equal an odd integer 2n+1. So this has led to a contradiction, which means I just proved that if k is odd then there is no integer solution.
Is my logic off somewhere, or should I be approaching it differently? Sorry if this has an obvious answer, I only started doing proofs this quarter. Thanks
proof-verification
proof-verification
asked 3 hours ago
jenellejenelle
132
New contributor
jenelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
jenellejenelle
132
New contributor
jenelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
jenellejenelle
132
New contributor
jenelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
jenellejenelle
132
asked 3 hours ago
jenellejenelle
132
132
New contributor
jenelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
jenelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
jenelle is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
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– José Carlos Santos
3 hours ago
$begingroup$
You proved in the end that ($k$ is odd) $implies$ (no integer solutions). Now you still have to prove the sufficient condition and that is if there is (no integer solutions) $implies$ ($k$ is odd), which is equivalent to proving ($k$ is not odd) $implies$ (there exist an integer solution). I think you didn't prove this, am I right?
$endgroup$
– Fareed AF
3 hours ago
add a comment |
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
You proved in the end that ($k$ is odd) $implies$ (no integer solutions). Now you still have to prove the sufficient condition and that is if there is (no integer solutions) $implies$ ($k$ is odd), which is equivalent to proving ($k$ is not odd) $implies$ (there exist an integer solution). I think you didn't prove this, am I right?
$endgroup$
– Fareed AF
3 hours ago
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
You proved in the end that ($k$ is odd) $implies$ (no integer solutions). Now you still have to prove the sufficient condition and that is if there is (no integer solutions) $implies$ ($k$ is odd), which is equivalent to proving ($k$ is not odd) $implies$ (there exist an integer solution). I think you didn't prove this, am I right?
$endgroup$
– Fareed AF
3 hours ago
$begingroup$
You proved in the end that ($k$ is odd) $implies$ (no integer solutions). Now you still have to prove the sufficient condition and that is if there is (no integer solutions) $implies$ ($k$ is odd), which is equivalent to proving ($k$ is not odd) $implies$ (there exist an integer solution). I think you didn't prove this, am I right?
$endgroup$
– Fareed AF
3 hours ago
add a comment |
2 Answers
2
active
oldest
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$begingroup$
You seem to be struggling with the logic involved here. The statement is
$$
ktext{ is odd}iff x^2-x-k=0text{ has no integer solutions}
$$
Yes, it seems that making $k$ odd makes the equation have no integer solutions (in other words, the $implies$ direction is true). Your proof of this looks fine.
However, the statement also claims that if there are no integer solutions, then $k$ is odd (the $Longleftarrow$ direction). The contrapositive of this claim is that making $k$ even will always result in an integer solution. This is disproven by providing a single counterexample.
answered 3 hours ago
ArthurArthur
112k7108192
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$begingroup$
Hint:
$x^2-x-k=0 Rightarrow x=frac{1 pm sqrt{4k+1}}{2}$
So $x$ can only be an integer if $4k+1$ is the square of an odd number.
This is true when $x=2$, in which case $sqrt{4k+1}=3$, and when $x=6$, in which case $sqrt{4k+1}=5$.
But what about when $k=4$ ?
answered 19 mins ago
gandalf61gandalf61
8,096625
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
You seem to be struggling with the logic involved here. The statement is
$$
ktext{ is odd}iff x^2-x-k=0text{ has no integer solutions}
$$
Yes, it seems that making $k$ odd makes the equation have no integer solutions (in other words, the $implies$ direction is true). Your proof of this looks fine.
However, the statement also claims that if there are no integer solutions, then $k$ is odd (the $Longleftarrow$ direction). The contrapositive of this claim is that making $k$ even will always result in an integer solution. This is disproven by providing a single counterexample.
answered 3 hours ago
ArthurArthur
112k7108192
$endgroup$
add a comment |
$begingroup$
You seem to be struggling with the logic involved here. The statement is
$$
ktext{ is odd}iff x^2-x-k=0text{ has no integer solutions}
$$
Yes, it seems that making $k$ odd makes the equation have no integer solutions (in other words, the $implies$ direction is true). Your proof of this looks fine.
However, the statement also claims that if there are no integer solutions, then $k$ is odd (the $Longleftarrow$ direction). The contrapositive of this claim is that making $k$ even will always result in an integer solution. This is disproven by providing a single counterexample.
answered 3 hours ago
ArthurArthur
112k7108192
$endgroup$
add a comment |
$begingroup$
You seem to be struggling with the logic involved here. The statement is
$$
ktext{ is odd}iff x^2-x-k=0text{ has no integer solutions}
$$
Yes, it seems that making $k$ odd makes the equation have no integer solutions (in other words, the $implies$ direction is true). Your proof of this looks fine.
However, the statement also claims that if there are no integer solutions, then $k$ is odd (the $Longleftarrow$ direction). The contrapositive of this claim is that making $k$ even will always result in an integer solution. This is disproven by providing a single counterexample.
answered 3 hours ago
ArthurArthur
112k7108192
$endgroup$
You seem to be struggling with the logic involved here. The statement is
$$
ktext{ is odd}iff x^2-x-k=0text{ has no integer solutions}
$$
Yes, it seems that making $k$ odd makes the equation have no integer solutions (in other words, the $implies$ direction is true). Your proof of this looks fine.
However, the statement also claims that if there are no integer solutions, then $k$ is odd (the $Longleftarrow$ direction). The contrapositive of this claim is that making $k$ even will always result in an integer solution. This is disproven by providing a single counterexample.
answered 3 hours ago
ArthurArthur
112k7108192
edited 3 hours ago
answered 3 hours ago
ArthurArthur
112k7108192
answered 3 hours ago
ArthurArthur
112k7108192
answered 3 hours ago
ArthurArthur
112k7108192
112k7108192
add a comment |
add a comment |
$begingroup$
Hint:
$x^2-x-k=0 Rightarrow x=frac{1 pm sqrt{4k+1}}{2}$
So $x$ can only be an integer if $4k+1$ is the square of an odd number.
This is true when $x=2$, in which case $sqrt{4k+1}=3$, and when $x=6$, in which case $sqrt{4k+1}=5$.
But what about when $k=4$ ?
answered 19 mins ago
gandalf61gandalf61
8,096625
$endgroup$
add a comment |
$begingroup$
Hint:
$x^2-x-k=0 Rightarrow x=frac{1 pm sqrt{4k+1}}{2}$
So $x$ can only be an integer if $4k+1$ is the square of an odd number.
This is true when $x=2$, in which case $sqrt{4k+1}=3$, and when $x=6$, in which case $sqrt{4k+1}=5$.
But what about when $k=4$ ?
answered 19 mins ago
gandalf61gandalf61
8,096625
$endgroup$
add a comment |
$begingroup$
Hint:
$x^2-x-k=0 Rightarrow x=frac{1 pm sqrt{4k+1}}{2}$
So $x$ can only be an integer if $4k+1$ is the square of an odd number.
This is true when $x=2$, in which case $sqrt{4k+1}=3$, and when $x=6$, in which case $sqrt{4k+1}=5$.
But what about when $k=4$ ?
answered 19 mins ago
gandalf61gandalf61
8,096625
$endgroup$
Hint:
$x^2-x-k=0 Rightarrow x=frac{1 pm sqrt{4k+1}}{2}$
So $x$ can only be an integer if $4k+1$ is the square of an odd number.
This is true when $x=2$, in which case $sqrt{4k+1}=3$, and when $x=6$, in which case $sqrt{4k+1}=5$.
But what about when $k=4$ ?
answered 19 mins ago
gandalf61gandalf61
8,096625
answered 19 mins ago
gandalf61gandalf61
8,096625
answered 19 mins ago
gandalf61gandalf61
8,096625
answered 19 mins ago
gandalf61gandalf61
8,096625
8,096625
add a comment |
add a comment |
jenelle is a new contributor. Be nice, and check out our Code of Conduct.
jenelle is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
You proved in the end that ($k$ is odd) $implies$ (no integer solutions). Now you still have to prove the sufficient condition and that is if there is (no integer solutions) $implies$ ($k$ is odd), which is equivalent to proving ($k$ is not odd) $implies$ (there exist an integer solution). I think you didn't prove this, am I right?
$endgroup$
– Fareed AF
3 hours ago