Determining and justifying the validity of an argument
$begingroup$
Context: Question made up by uni lecturer
Original statement: There exists two positive real numbers $x$ and $y$ such that for all positive integers $z$, $frac{x}{y}>z$.
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 $frac{x}{y}le z$.
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 $frac{x}{y}$.
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 $frac{x}{y}=frac{z+1}{1}=z+1>z$.
Can someone please help me to see the error in my answer.
Thanks
discrete-mathematics
asked Mar 26 at 13:14
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 $z$, $frac{x}{y}>z$.
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 $frac{x}{y}le z$.
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 $frac{x}{y}$.
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 $frac{x}{y}=frac{z+1}{1}=z+1>z$.
Can someone please help me to see the error in my answer.
Thanks
discrete-mathematics
asked Mar 26 at 13:14
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 $z$, $frac{x}{y}>z$.
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 $frac{x}{y}le z$.
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 $frac{x}{y}$.
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 $frac{x}{y}=frac{z+1}{1}=z+1>z$.
Can someone please help me to see the error in my answer.
Thanks
discrete-mathematics
asked Mar 26 at 13:14
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 $z$, $frac{x}{y}>z$.
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 $frac{x}{y}le z$.
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 $frac{x}{y}$.
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 $frac{x}{y}=frac{z+1}{1}=z+1>z$.
Can someone please help me to see the error in my answer.
Thanks
discrete-mathematics
discrete-mathematics
asked Mar 26 at 13:14
Ruby PaRuby Pa
376
asked Mar 26 at 13:14
Ruby PaRuby Pa
376
asked Mar 26 at 13:14
Ruby PaRuby Pa
376
asked Mar 26 at 13:14
Ruby PaRuby Pa
376
asked Mar 26 at 13:14
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 $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ 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 Mar 26 at 13:33
Floris ClaassensFloris Claassens
1,33229
$endgroup$
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 Mar 26 at 13:29
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163151%2fdetermining-and-justifying-the-validity-of-an-argument%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ 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 Mar 26 at 13:33
Floris ClaassensFloris Claassens
1,33229
$endgroup$
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 $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ 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 Mar 26 at 13:33
Floris ClaassensFloris Claassens
1,33229
$endgroup$
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 $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ 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 Mar 26 at 13:33
Floris ClaassensFloris Claassens
1,33229
$endgroup$
The problem with your reasoning is the order. There exist $x,y$ positive integers such that for all positive integers $z$ we have $frac{x}{y}>z$, so first you must pick an $x$ and a $y$, and then you must test whether $frac{x}{y}>z$ 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 Mar 26 at 13:33
Floris ClaassensFloris Claassens
1,33229
answered Mar 26 at 13:33
Floris ClaassensFloris Claassens
1,33229
answered Mar 26 at 13:33
Floris ClaassensFloris Claassens
1,33229
answered Mar 26 at 13:33
Floris ClaassensFloris Claassens
1,33229
1,33229
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 Mar 26 at 13:29
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
$endgroup$
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 Mar 26 at 13:29
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
$endgroup$
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 Mar 26 at 13:29
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
$endgroup$
"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 Mar 26 at 13:29
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
edited Mar 26 at 13:44
answered Mar 26 at 13:29
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
answered Mar 26 at 13:29
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
answered Mar 26 at 13:29
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
67.7k449117
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3163151%2fdetermining-and-justifying-the-validity-of-an-argument%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
