Mixing
Proving a theorem by proving a related one
$begingroup$
I have a question that might be stupid, but it bugs me for some time. It's quite simple:
Let's say we have a conjecture, such that if the conjecture is true then another theorem will be true. Does proving the latter theorem without relying on the initial conjecture at all imply that the conjecture is also true?
As an example, it is quite known that if Riemann Hypothesis is true then many theorems will hold true. (I read that somewhere, but I don't have a source right now.) Does proving one of those theorems without using in the proof any connection with the conjecture make the conjecture to be also true?
Put in other words, let's say if R-H is true then a equals b. Would proving a equals b (without touching R-H) transition the proof to R-H?
I would suspect that others thought of this thing too, so a link to that would suffice.
soft-question
edited Jan 28 at 3:25
J. W. Tanner
3,9071320
$endgroup$
add a comment |
$begingroup$
I have a question that might be stupid, but it bugs me for some time. It's quite simple:
Let's say we have a conjecture, such that if the conjecture is true then another theorem will be true. Does proving the latter theorem without relying on the initial conjecture at all imply that the conjecture is also true?
As an example, it is quite known that if Riemann Hypothesis is true then many theorems will hold true. (I read that somewhere, but I don't have a source right now.) Does proving one of those theorems without using in the proof any connection with the conjecture make the conjecture to be also true?
Put in other words, let's say if R-H is true then a equals b. Would proving a equals b (without touching R-H) transition the proof to R-H?
I would suspect that others thought of this thing too, so a link to that would suffice.
soft-question
edited Jan 28 at 3:25
J. W. Tanner
3,9071320
$endgroup$
$begingroup$
Apologies for bad exprimation of the idea. If anyone could do that better please freely edit in the question.
$endgroup$
– user625055
Jan 28 at 2:34
$begingroup$
Probably not. As a similar argument, if $-1=1$ then you can square both sides and conclude $1=1$. That is true, but the starting point is not.
$endgroup$
– Empy2
Jan 28 at 2:39
add a comment |
$begingroup$
I have a question that might be stupid, but it bugs me for some time. It's quite simple:
Let's say we have a conjecture, such that if the conjecture is true then another theorem will be true. Does proving the latter theorem without relying on the initial conjecture at all imply that the conjecture is also true?
As an example, it is quite known that if Riemann Hypothesis is true then many theorems will hold true. (I read that somewhere, but I don't have a source right now.) Does proving one of those theorems without using in the proof any connection with the conjecture make the conjecture to be also true?
Put in other words, let's say if R-H is true then a equals b. Would proving a equals b (without touching R-H) transition the proof to R-H?
I would suspect that others thought of this thing too, so a link to that would suffice.
soft-question
edited Jan 28 at 3:25
J. W. Tanner
3,9071320
$endgroup$
I have a question that might be stupid, but it bugs me for some time. It's quite simple:
Let's say we have a conjecture, such that if the conjecture is true then another theorem will be true. Does proving the latter theorem without relying on the initial conjecture at all imply that the conjecture is also true?
As an example, it is quite known that if Riemann Hypothesis is true then many theorems will hold true. (I read that somewhere, but I don't have a source right now.) Does proving one of those theorems without using in the proof any connection with the conjecture make the conjecture to be also true?
Put in other words, let's say if R-H is true then a equals b. Would proving a equals b (without touching R-H) transition the proof to R-H?
I would suspect that others thought of this thing too, so a link to that would suffice.
soft-question
soft-question
edited Jan 28 at 3:25
J. W. Tanner
3,9071320
edited Jan 28 at 3:25
J. W. Tanner
3,9071320
edited Jan 28 at 3:25
J. W. Tanner
3,9071320
edited Jan 28 at 3:25
J. W. Tanner
3,9071320
edited Jan 28 at 3:25
J. W. Tanner
3,9071320
3,9071320
asked Jan 28 at 2:23
user625055
$begingroup$
Apologies for bad exprimation of the idea. If anyone could do that better please freely edit in the question.
$endgroup$
– user625055
Jan 28 at 2:34
$begingroup$
Probably not. As a similar argument, if $-1=1$ then you can square both sides and conclude $1=1$. That is true, but the starting point is not.
$endgroup$
– Empy2
Jan 28 at 2:39
add a comment |
$begingroup$
Apologies for bad exprimation of the idea. If anyone could do that better please freely edit in the question.
$endgroup$
– user625055
Jan 28 at 2:34
$begingroup$
Probably not. As a similar argument, if $-1=1$ then you can square both sides and conclude $1=1$. That is true, but the starting point is not.
$endgroup$
– Empy2
Jan 28 at 2:39
$begingroup$
Apologies for bad exprimation of the idea. If anyone could do that better please freely edit in the question.
$endgroup$
– user625055
Jan 28 at 2:34
$begingroup$
Apologies for bad exprimation of the idea. If anyone could do that better please freely edit in the question.
$endgroup$
– user625055
Jan 28 at 2:34
$begingroup$
Probably not. As a similar argument, if $-1=1$ then you can square both sides and conclude $1=1$. That is true, but the starting point is not.
$endgroup$
– Empy2
Jan 28 at 2:39
$begingroup$
Probably not. As a similar argument, if $-1=1$ then you can square both sides and conclude $1=1$. That is true, but the starting point is not.
$endgroup$
– Empy2
Jan 28 at 2:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, it doesn't.
Say $A$ implies $B$. If $B$ is true, that doesn't imply that $A$ is true.
Consider the statement "If it rains, then I will use an umbrella".
I might be someone who uses an umbrella regardless of the weather condition. Observing that I use an umbrella doesn't mean it is raining; I might be using it as a sunshield.
edited Jan 28 at 3:49
J. W. Tanner
3,9071320
answered Jan 28 at 2:37
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
Siong Thye Goh's answer demonstrates the flaw in the logic. I just want to add one comment.
When trying to think about the logic of proofs, don't reach for some big theorem or conjecture, or the thing you're trying to prove, as an example. Try the logic out on the simplest possible example you can think of. For example:
Suppose $(x=4)$ proves ($x$ is even). If I prove $x$ is even, does that prove $x=4$?
Since you already know that $x=4$ would make $x$ even, you don't need to be distracted by it and are free to focus just on the logic—which then becomes crystal clear.
Of course I could have chosen one of the many other theorems that would hold if $x=4$, such as "$x$ is a square", "$x$ is indivisible by $5$" . . . See what I mean? It's just easier to think about $x=4$ than the Riemann Hypothesis.
answered Jan 28 at 4:14
timtfjtimtfj
2,478420
$endgroup$
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%2f3090369%2fproving-a-theorem-by-proving-a-related-one%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
$begingroup$
No, it doesn't.
Say $A$ implies $B$. If $B$ is true, that doesn't imply that $A$ is true.
Consider the statement "If it rains, then I will use an umbrella".
I might be someone who uses an umbrella regardless of the weather condition. Observing that I use an umbrella doesn't mean it is raining; I might be using it as a sunshield.
edited Jan 28 at 3:49
J. W. Tanner
3,9071320
answered Jan 28 at 2:37
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
No, it doesn't.
Say $A$ implies $B$. If $B$ is true, that doesn't imply that $A$ is true.
Consider the statement "If it rains, then I will use an umbrella".
I might be someone who uses an umbrella regardless of the weather condition. Observing that I use an umbrella doesn't mean it is raining; I might be using it as a sunshield.
edited Jan 28 at 3:49
J. W. Tanner
3,9071320
answered Jan 28 at 2:37
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
No, it doesn't.
Say $A$ implies $B$. If $B$ is true, that doesn't imply that $A$ is true.
Consider the statement "If it rains, then I will use an umbrella".
I might be someone who uses an umbrella regardless of the weather condition. Observing that I use an umbrella doesn't mean it is raining; I might be using it as a sunshield.
edited Jan 28 at 3:49
J. W. Tanner
3,9071320
answered Jan 28 at 2:37
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
No, it doesn't.
Say $A$ implies $B$. If $B$ is true, that doesn't imply that $A$ is true.
Consider the statement "If it rains, then I will use an umbrella".
I might be someone who uses an umbrella regardless of the weather condition. Observing that I use an umbrella doesn't mean it is raining; I might be using it as a sunshield.
edited Jan 28 at 3:49
J. W. Tanner
3,9071320
answered Jan 28 at 2:37
Siong Thye GohSiong Thye Goh
103k1468119
edited Jan 28 at 3:49
J. W. Tanner
3,9071320
edited Jan 28 at 3:49
J. W. Tanner
3,9071320
edited Jan 28 at 3:49
J. W. Tanner
3,9071320
3,9071320
answered Jan 28 at 2:37
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 28 at 2:37
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 28 at 2:37
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
$begingroup$
Siong Thye Goh's answer demonstrates the flaw in the logic. I just want to add one comment.
When trying to think about the logic of proofs, don't reach for some big theorem or conjecture, or the thing you're trying to prove, as an example. Try the logic out on the simplest possible example you can think of. For example:
Suppose $(x=4)$ proves ($x$ is even). If I prove $x$ is even, does that prove $x=4$?
Since you already know that $x=4$ would make $x$ even, you don't need to be distracted by it and are free to focus just on the logic—which then becomes crystal clear.
Of course I could have chosen one of the many other theorems that would hold if $x=4$, such as "$x$ is a square", "$x$ is indivisible by $5$" . . . See what I mean? It's just easier to think about $x=4$ than the Riemann Hypothesis.
answered Jan 28 at 4:14
timtfjtimtfj
2,478420
$endgroup$
add a comment |
$begingroup$
Siong Thye Goh's answer demonstrates the flaw in the logic. I just want to add one comment.
When trying to think about the logic of proofs, don't reach for some big theorem or conjecture, or the thing you're trying to prove, as an example. Try the logic out on the simplest possible example you can think of. For example:
Suppose $(x=4)$ proves ($x$ is even). If I prove $x$ is even, does that prove $x=4$?
Since you already know that $x=4$ would make $x$ even, you don't need to be distracted by it and are free to focus just on the logic—which then becomes crystal clear.
Of course I could have chosen one of the many other theorems that would hold if $x=4$, such as "$x$ is a square", "$x$ is indivisible by $5$" . . . See what I mean? It's just easier to think about $x=4$ than the Riemann Hypothesis.
answered Jan 28 at 4:14
timtfjtimtfj
2,478420
$endgroup$
add a comment |
$begingroup$
Siong Thye Goh's answer demonstrates the flaw in the logic. I just want to add one comment.
When trying to think about the logic of proofs, don't reach for some big theorem or conjecture, or the thing you're trying to prove, as an example. Try the logic out on the simplest possible example you can think of. For example:
Suppose $(x=4)$ proves ($x$ is even). If I prove $x$ is even, does that prove $x=4$?
Since you already know that $x=4$ would make $x$ even, you don't need to be distracted by it and are free to focus just on the logic—which then becomes crystal clear.
Of course I could have chosen one of the many other theorems that would hold if $x=4$, such as "$x$ is a square", "$x$ is indivisible by $5$" . . . See what I mean? It's just easier to think about $x=4$ than the Riemann Hypothesis.
answered Jan 28 at 4:14
timtfjtimtfj
2,478420
$endgroup$
Siong Thye Goh's answer demonstrates the flaw in the logic. I just want to add one comment.
When trying to think about the logic of proofs, don't reach for some big theorem or conjecture, or the thing you're trying to prove, as an example. Try the logic out on the simplest possible example you can think of. For example:
Suppose $(x=4)$ proves ($x$ is even). If I prove $x$ is even, does that prove $x=4$?
Since you already know that $x=4$ would make $x$ even, you don't need to be distracted by it and are free to focus just on the logic—which then becomes crystal clear.
Of course I could have chosen one of the many other theorems that would hold if $x=4$, such as "$x$ is a square", "$x$ is indivisible by $5$" . . . See what I mean? It's just easier to think about $x=4$ than the Riemann Hypothesis.
answered Jan 28 at 4:14
timtfjtimtfj
2,478420
edited Jan 28 at 4:45
answered Jan 28 at 4:14
timtfjtimtfj
2,478420
answered Jan 28 at 4:14
timtfjtimtfj
2,478420
answered Jan 28 at 4:14
timtfjtimtfj
2,478420
2,478420
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%2f3090369%2fproving-a-theorem-by-proving-a-related-one%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
$begingroup$
Apologies for bad exprimation of the idea. If anyone could do that better please freely edit in the question.
$endgroup$
– user625055
Jan 28 at 2:34
$begingroup$
Probably not. As a similar argument, if $-1=1$ then you can square both sides and conclude $1=1$. That is true, but the starting point is not.
$endgroup$
– Empy2
Jan 28 at 2:39