Determining and justifying the validity of an argumentIs “$n$ is an integer and $fracnn+1$ is an integer” true or false? Prove that if $m$ and $n$ are positive integers, and $x$ is a real number, then: $ceiling(fracceiling(x)+nm) = ceiling(fracx+nm)$A simple floor function conditional proofStatement and Proposed NegationsWhat is the contra-positive of an equality.Truth value of a false negationDiscrete Mathematics, Predicates and NegationDetermining the order of quantified statement after negationNegating the statementProof by contradiction involving positive integer prime numbers
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Determining and justifying the validity of an argument
Is “$n$ is an integer and $fracnn+1$ is an integer” true or false? Prove that if $m$ and $n$ are positive integers, and $x$ is a real number, then: $ceiling(fracceiling(x)+nm) = ceiling(fracx+nm)$A simple floor function conditional proofStatement and Proposed NegationsWhat is the contra-positive of an equality.Truth value of a false negationDiscrete Mathematics, Predicates and NegationDetermining the order of quantified statement after negationNegating the statementProof by contradiction involving positive integer prime numbers
$begingroup$
Context: Question made up by uni lecturer
Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $fracxy>z$.
So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.
I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $fracxyle z$.
The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $fracxy$.
When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $fracxy=fracz+11=z+1>z$.
Can someone please help me to see the error in my answer.
Thanks
discrete-mathematics
asked 5 hours ago
Ruby PaRuby Pa
376
$endgroup$
add a comment |
$begingroup$
Context: Question made up by uni lecturer
Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $fracxy>z$.
So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.
I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $fracxyle z$.
The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $fracxy$.
When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $fracxy=fracz+11=z+1>z$.
Can someone please help me to see the error in my answer.
Thanks
discrete-mathematics
asked 5 hours ago
Ruby PaRuby Pa
376
$endgroup$
add a comment |
$begingroup$
Context: Question made up by uni lecturer
Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $fracxy>z$.
So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.
I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $fracxyle z$.
The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $fracxy$.
When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $fracxy=fracz+11=z+1>z$.
Can someone please help me to see the error in my answer.
Thanks
discrete-mathematics
asked 5 hours ago
Ruby PaRuby Pa
376
$endgroup$
Context: Question made up by uni lecturer
Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $fracxy>z$.
So the question was to find the negation of the statement, and then determine whether the original statement or its negation was true.
I found its negation to be: For all positive real numbers $x$ and $y$, there exists a positive integer $z$ such that $fracxyle z$.
The lecturer's solution to the question says that the negation is true since for any positive reals $x$ and $y$, you can choose $z$ to equal the ceiling of $fracxy$.
When I attempted the question myself, I said that the original statement is true because you can take $x=z+1$ (which would be a positive integer that still belongs to the set of all positive real numbers) and $y=1$ (which is a positive real number), as $fracxy=fracz+11=z+1>z$.
Can someone please help me to see the error in my answer.
Thanks
discrete-mathematics
discrete-mathematics
asked 5 hours ago
Ruby PaRuby Pa
376
asked 5 hours ago
Ruby PaRuby Pa
376
asked 5 hours ago
Ruby PaRuby Pa
376
asked 5 hours ago
Ruby PaRuby Pa
376
asked 5 hours ago
Ruby PaRuby Pa
376
376
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $fracxy>z$, so first you must pick an $x$ and a $y$, and then you must test whether $fracxy>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.
answered 4 hours ago
Floris ClaassensFloris Claassens
1,13927
$endgroup$
add a comment |
$begingroup$
"There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"
The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.
You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.
The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.
Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.
Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.
And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.
Consider as $z$ the number $r+1$ and it's done.
answered 4 hours ago
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449116
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $fracxy>z$, so first you must pick an $x$ and a $y$, and then you must test whether $fracxy>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.
answered 4 hours ago
Floris ClaassensFloris Claassens
1,13927
$endgroup$
add a comment |
$begingroup$
The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $fracxy>z$, so first you must pick an $x$ and a $y$, and then you must test whether $fracxy>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.
answered 4 hours ago
Floris ClaassensFloris Claassens
1,13927
$endgroup$
add a comment |
$begingroup$
The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $fracxy>z$, so first you must pick an $x$ and a $y$, and then you must test whether $fracxy>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.
answered 4 hours ago
Floris ClaassensFloris Claassens
1,13927
$endgroup$
The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $fracxy>z$, so first you must pick an $x$ and a $y$, and then you must test whether $fracxy>z$ for every positive integer $z$. Therefore you can't define $x=z+1$ as when you pick $x$ you don't know $z$ yet.
answered 4 hours ago
Floris ClaassensFloris Claassens
1,13927
answered 4 hours ago
Floris ClaassensFloris Claassens
1,13927
answered 4 hours ago
Floris ClaassensFloris Claassens
1,13927
answered 4 hours ago
Floris ClaassensFloris Claassens
1,13927
1,13927
add a comment |
add a comment |
$begingroup$
"There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"
The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.
You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.
The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.
Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.
Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.
And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.
Consider as $z$ the number $r+1$ and it's done.
answered 4 hours ago
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449116
$endgroup$
add a comment |
$begingroup$
"There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"
The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.
You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.
The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.
Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.
Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.
And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.
Consider as $z$ the number $r+1$ and it's done.
answered 4 hours ago
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449116
$endgroup$
add a comment |
$begingroup$
"There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"
The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.
You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.
The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.
Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.
Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.
And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.
Consider as $z$ the number $r+1$ and it's done.
answered 4 hours ago
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449116
$endgroup$
"There exists two positive real numbers $x$ and $y$ such that for all positive integer $z$ : $dfrac x y > z$"
The intuition says that it is False. $dfrac x y$ is a positive real; thus, the statement amounts to asserting that there is a real that is greater than every integer, which is not.
You reasoning is wrong because you have swapped the choice of the numbers : you start from $z$ and choose $x$ and $y$ accordingly.
The negation of the original statement is : $forall x forall y exists z (dfrac y y le z)$.
Thus, choose $x$ and $y$ positive whatever and what you get is a new positive real $dfrac x y$.
Now, you have to choose an integer $z$ (obviously positive) that is greater-or-equal to $dfrac x y$.
And this must be always possible, because $dfrac x y$ is a number $r.r_1 r_2 r_3 ldots$.
Consider as $z$ the number $r+1$ and it's done.
answered 4 hours ago
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449116
edited 4 hours ago
answered 4 hours ago
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449116
answered 4 hours ago
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449116
answered 4 hours ago
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449116
67.5k449116
add a comment |
add a comment |
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