An error in a proof due to variable creep?
I'm working through some exam practice questions and I came across this one:
Identify the error in the following “proof.”
Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.
Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.
Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.
discrete-mathematics
asked Nov 26 '18 at 19:08
metisMusings
161
add a comment |
I'm working through some exam practice questions and I came across this one:
Identify the error in the following “proof.”
Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.
Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.
Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.
discrete-mathematics
asked Nov 26 '18 at 19:08
metisMusings
161
add a comment |
I'm working through some exam practice questions and I came across this one:
Identify the error in the following “proof.”
Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.
Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.
Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.
discrete-mathematics
asked Nov 26 '18 at 19:08
metisMusings
161
I'm working through some exam practice questions and I came across this one:
Identify the error in the following “proof.”
Let u, m, n be three integers. If u|mn and gcd(u, m) = 1, then m = ±1.
If gcd(u, m) = 1, then 1 = us + mt for some integers s, t. If u|mn, then
us = mn for some integer s. Hence, 1 = mn + mt = m(n + t), which
implies that m|1, and therefore m = ±1.
Now I think the problem here is that we cannot go from the linear transformation statement 1 = us + mt and then state that u|mn => mn = us for some integer s, because we already have a value of s in the previous statement. so we would need to define u|mn => mn = uj for some integer j.
And now the substitution doesn't work and we cannot proceed.
Does this make sense to you guys? It makes sense to me that we cannot use s in two different places here. But I'm having a hard time trying to explain why in a clear manner.
discrete-mathematics
discrete-mathematics
asked Nov 26 '18 at 19:08
metisMusings
161
asked Nov 26 '18 at 19:08
metisMusings
161
asked Nov 26 '18 at 19:08
metisMusings
161
asked Nov 26 '18 at 19:08
metisMusings
161
asked Nov 26 '18 at 19:08
metisMusings
161
161
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
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%2f3014772%2fan-error-in-a-proof-due-to-variable-creep%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
add a comment |
You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
add a comment |
You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
You're right: the "proof" has defined $s$ in incompatible ways in two different places. You could explain it by analogy as follows:
We prove that $2=4$. Indeed, the number $2$ is even, so it can be expressed as $2 = 2n$ for some $n$. The number $4$ is even, so it can be expressed as $4 = 2n$ for some $n$. Hence $2 = 2n = 4$, so $2 = 4$.
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
answered Nov 26 '18 at 19:16
Patrick Stevens
28.5k52874
28.5k52874
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
add a comment |
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
Thanks, this helps a lot :)
– metisMusings
Nov 26 '18 at 19:20
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3014772%2fan-error-in-a-proof-due-to-variable-creep%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
