Mixing
Simple & Intuitive Statements that are Difficult to Prove
$begingroup$
Looking through the webcomic, I came across one of my favorite comics:
(from Saturday Morning Breakfast Cereal)
It seems that people have an ongoing interest in results in mathematics that are true, but highly unintuitive, like the Banach-Tarski Paradox. However, what results are there that are seemingly obvious and intuitive, but difficult to prove (or perhaps have non-obvious intricacies)?
I feel that such examples are important for helping people understand the necessity of rigor or that seemingly obvious results are not at all obvious from a mathematical perspective.
(Personally, I don't feel that 'simple' number theory conjectures/results like the ABC conjecture fall under this category, because while simple to state, they are often very removed from reality.)
soft-question big-list
edited Mar 29 '14 at 1:31
Rahul
33.1k568169
asked Mar 29 '14 at 1:24
HaydenHayden
13.7k12347
$endgroup$
add a comment |
$begingroup$
Looking through the webcomic, I came across one of my favorite comics:
(from Saturday Morning Breakfast Cereal)
It seems that people have an ongoing interest in results in mathematics that are true, but highly unintuitive, like the Banach-Tarski Paradox. However, what results are there that are seemingly obvious and intuitive, but difficult to prove (or perhaps have non-obvious intricacies)?
I feel that such examples are important for helping people understand the necessity of rigor or that seemingly obvious results are not at all obvious from a mathematical perspective.
(Personally, I don't feel that 'simple' number theory conjectures/results like the ABC conjecture fall under this category, because while simple to state, they are often very removed from reality.)
soft-question big-list
edited Mar 29 '14 at 1:31
Rahul
33.1k568169
asked Mar 29 '14 at 1:24
HaydenHayden
13.7k12347
$endgroup$
$begingroup$
I don't know that I'd call the ABC conjecture simple to state (at least compared to, say, FLT.)
$endgroup$
– Mike Miller
Mar 29 '14 at 1:31
1
$begingroup$
Relating to the statement in the comic more than your question: Obvious facts are generally easier to prove. However, when one is first learning mathematics, one generally does not develop the necessary machinery to prove anything from axioms, so proofs are done by reducing complex statements to the more obvious ones. In the case of already obvious statements there is no point to this, so the necessary machinery must be developed and used for there to be anything to do. However, this is still easier than proving more complex statements from equally basic principles.
$endgroup$
– Alex Becker
Mar 29 '14 at 1:33
$begingroup$
Agreed, but I wanted to reference an open question, though there are obviously other simpler to state open questions out there.
$endgroup$
– Hayden
Mar 29 '14 at 1:33
add a comment |
$begingroup$
Looking through the webcomic, I came across one of my favorite comics:
(from Saturday Morning Breakfast Cereal)
It seems that people have an ongoing interest in results in mathematics that are true, but highly unintuitive, like the Banach-Tarski Paradox. However, what results are there that are seemingly obvious and intuitive, but difficult to prove (or perhaps have non-obvious intricacies)?
I feel that such examples are important for helping people understand the necessity of rigor or that seemingly obvious results are not at all obvious from a mathematical perspective.
(Personally, I don't feel that 'simple' number theory conjectures/results like the ABC conjecture fall under this category, because while simple to state, they are often very removed from reality.)
soft-question big-list
edited Mar 29 '14 at 1:31
Rahul
33.1k568169
asked Mar 29 '14 at 1:24
HaydenHayden
13.7k12347
$endgroup$
Looking through the webcomic, I came across one of my favorite comics:
(from Saturday Morning Breakfast Cereal)
It seems that people have an ongoing interest in results in mathematics that are true, but highly unintuitive, like the Banach-Tarski Paradox. However, what results are there that are seemingly obvious and intuitive, but difficult to prove (or perhaps have non-obvious intricacies)?
I feel that such examples are important for helping people understand the necessity of rigor or that seemingly obvious results are not at all obvious from a mathematical perspective.
(Personally, I don't feel that 'simple' number theory conjectures/results like the ABC conjecture fall under this category, because while simple to state, they are often very removed from reality.)
soft-question big-list
soft-question big-list
edited Mar 29 '14 at 1:31
Rahul
33.1k568169
asked Mar 29 '14 at 1:24
HaydenHayden
13.7k12347
edited Mar 29 '14 at 1:31
Rahul
33.1k568169
asked Mar 29 '14 at 1:24
HaydenHayden
13.7k12347
edited Mar 29 '14 at 1:31
Rahul
33.1k568169
edited Mar 29 '14 at 1:31
Rahul
33.1k568169
edited Mar 29 '14 at 1:31
Rahul
33.1k568169
33.1k568169
asked Mar 29 '14 at 1:24
HaydenHayden
13.7k12347
asked Mar 29 '14 at 1:24
HaydenHayden
13.7k12347
asked Mar 29 '14 at 1:24
HaydenHayden
13.7k12347
13.7k12347
$begingroup$
I don't know that I'd call the ABC conjecture simple to state (at least compared to, say, FLT.)
$endgroup$
– Mike Miller
Mar 29 '14 at 1:31
1
$begingroup$
Relating to the statement in the comic more than your question: Obvious facts are generally easier to prove. However, when one is first learning mathematics, one generally does not develop the necessary machinery to prove anything from axioms, so proofs are done by reducing complex statements to the more obvious ones. In the case of already obvious statements there is no point to this, so the necessary machinery must be developed and used for there to be anything to do. However, this is still easier than proving more complex statements from equally basic principles.
$endgroup$
– Alex Becker
Mar 29 '14 at 1:33
$begingroup$
Agreed, but I wanted to reference an open question, though there are obviously other simpler to state open questions out there.
$endgroup$
– Hayden
Mar 29 '14 at 1:33
add a comment |
$begingroup$
I don't know that I'd call the ABC conjecture simple to state (at least compared to, say, FLT.)
$endgroup$
– Mike Miller
Mar 29 '14 at 1:31
1
$begingroup$
Relating to the statement in the comic more than your question: Obvious facts are generally easier to prove. However, when one is first learning mathematics, one generally does not develop the necessary machinery to prove anything from axioms, so proofs are done by reducing complex statements to the more obvious ones. In the case of already obvious statements there is no point to this, so the necessary machinery must be developed and used for there to be anything to do. However, this is still easier than proving more complex statements from equally basic principles.
$endgroup$
– Alex Becker
Mar 29 '14 at 1:33
$begingroup$
Agreed, but I wanted to reference an open question, though there are obviously other simpler to state open questions out there.
$endgroup$
– Hayden
Mar 29 '14 at 1:33
$begingroup$
I don't know that I'd call the ABC conjecture simple to state (at least compared to, say, FLT.)
$endgroup$
– Mike Miller
Mar 29 '14 at 1:31
$begingroup$
I don't know that I'd call the ABC conjecture simple to state (at least compared to, say, FLT.)
$endgroup$
– Mike Miller
Mar 29 '14 at 1:31
1
1
$begingroup$
Relating to the statement in the comic more than your question: Obvious facts are generally easier to prove. However, when one is first learning mathematics, one generally does not develop the necessary machinery to prove anything from axioms, so proofs are done by reducing complex statements to the more obvious ones. In the case of already obvious statements there is no point to this, so the necessary machinery must be developed and used for there to be anything to do. However, this is still easier than proving more complex statements from equally basic principles.
$endgroup$
– Alex Becker
Mar 29 '14 at 1:33
$begingroup$
Relating to the statement in the comic more than your question: Obvious facts are generally easier to prove. However, when one is first learning mathematics, one generally does not develop the necessary machinery to prove anything from axioms, so proofs are done by reducing complex statements to the more obvious ones. In the case of already obvious statements there is no point to this, so the necessary machinery must be developed and used for there to be anything to do. However, this is still easier than proving more complex statements from equally basic principles.
$endgroup$
– Alex Becker
Mar 29 '14 at 1:33
$begingroup$
Agreed, but I wanted to reference an open question, though there are obviously other simpler to state open questions out there.
$endgroup$
– Hayden
Mar 29 '14 at 1:33
$begingroup$
Agreed, but I wanted to reference an open question, though there are obviously other simpler to state open questions out there.
$endgroup$
– Hayden
Mar 29 '14 at 1:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The Jordan curve theorem asserts that every a non-self-intersecting continuous loop divides the plane into an "interior" region bounded by the curve and an "exterior" region.
Another nice example is the P vs NP problem, which basically says verifying is easier than finding solutions. But it is still unsolved , one of the Clay Millennium problems.
answered Mar 29 '14 at 1:37
MohanMohan
5,9541252101
$endgroup$
add a comment |
$begingroup$
Prove the reflexive property, or that $x = x$.
http://www.tondering.dk/claus/sur16.pdf
Crazy stuff.
answered Mar 29 '14 at 1:28
louie mcconnelllouie mcconnell
1,4731723
$endgroup$
$begingroup$
This is misleading, because if you're asking for a formal proof you need to give a formal statement; and a formal statement of "$x=x$ where $x$ is a surreal number" is definitely not simple,
$endgroup$
– Alex Becker
Mar 29 '14 at 1:35
$begingroup$
Yes. A more obvious fact, according to the expert graph, has a more difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 1:35
$begingroup$
I'm not sure what your comment means. Can you clarify?
$endgroup$
– Alex Becker
Mar 29 '14 at 2:14
$begingroup$
The xkcd comic is implying that a very simple axiom can be very difficult to prove. This is one of the simplest axioms imaginable, with an extremely difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 2:40
add a comment |
$begingroup$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x + y) + z = x + (y + z)$
- $forall x in mathbb{R}forall y in mathbb{R}x + y = y + x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x times y) times z = x times (y times z)$
- $forall x in mathbb{R}forall y in mathbb{R}x times y = y times x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} x times (y + z) = (x times y) + (x times z)$
answered Jan 10 at 20:39
TimothyTimothy
306213
$endgroup$
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%2f730830%2fsimple-intuitive-statements-that-are-difficult-to-prove%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The Jordan curve theorem asserts that every a non-self-intersecting continuous loop divides the plane into an "interior" region bounded by the curve and an "exterior" region.
Another nice example is the P vs NP problem, which basically says verifying is easier than finding solutions. But it is still unsolved , one of the Clay Millennium problems.
answered Mar 29 '14 at 1:37
MohanMohan
5,9541252101
$endgroup$
add a comment |
$begingroup$
The Jordan curve theorem asserts that every a non-self-intersecting continuous loop divides the plane into an "interior" region bounded by the curve and an "exterior" region.
Another nice example is the P vs NP problem, which basically says verifying is easier than finding solutions. But it is still unsolved , one of the Clay Millennium problems.
answered Mar 29 '14 at 1:37
MohanMohan
5,9541252101
$endgroup$
add a comment |
$begingroup$
The Jordan curve theorem asserts that every a non-self-intersecting continuous loop divides the plane into an "interior" region bounded by the curve and an "exterior" region.
Another nice example is the P vs NP problem, which basically says verifying is easier than finding solutions. But it is still unsolved , one of the Clay Millennium problems.
answered Mar 29 '14 at 1:37
MohanMohan
5,9541252101
$endgroup$
The Jordan curve theorem asserts that every a non-self-intersecting continuous loop divides the plane into an "interior" region bounded by the curve and an "exterior" region.
Another nice example is the P vs NP problem, which basically says verifying is easier than finding solutions. But it is still unsolved , one of the Clay Millennium problems.
answered Mar 29 '14 at 1:37
MohanMohan
5,9541252101
answered Mar 29 '14 at 1:37
MohanMohan
5,9541252101
answered Mar 29 '14 at 1:37
MohanMohan
5,9541252101
answered Mar 29 '14 at 1:37
MohanMohan
5,9541252101
5,9541252101
add a comment |
add a comment |
$begingroup$
Prove the reflexive property, or that $x = x$.
http://www.tondering.dk/claus/sur16.pdf
Crazy stuff.
answered Mar 29 '14 at 1:28
louie mcconnelllouie mcconnell
1,4731723
$endgroup$
$begingroup$
This is misleading, because if you're asking for a formal proof you need to give a formal statement; and a formal statement of "$x=x$ where $x$ is a surreal number" is definitely not simple,
$endgroup$
– Alex Becker
Mar 29 '14 at 1:35
$begingroup$
Yes. A more obvious fact, according to the expert graph, has a more difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 1:35
$begingroup$
I'm not sure what your comment means. Can you clarify?
$endgroup$
– Alex Becker
Mar 29 '14 at 2:14
$begingroup$
The xkcd comic is implying that a very simple axiom can be very difficult to prove. This is one of the simplest axioms imaginable, with an extremely difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 2:40
add a comment |
$begingroup$
Prove the reflexive property, or that $x = x$.
http://www.tondering.dk/claus/sur16.pdf
Crazy stuff.
answered Mar 29 '14 at 1:28
louie mcconnelllouie mcconnell
1,4731723
$endgroup$
$begingroup$
This is misleading, because if you're asking for a formal proof you need to give a formal statement; and a formal statement of "$x=x$ where $x$ is a surreal number" is definitely not simple,
$endgroup$
– Alex Becker
Mar 29 '14 at 1:35
$begingroup$
Yes. A more obvious fact, according to the expert graph, has a more difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 1:35
$begingroup$
I'm not sure what your comment means. Can you clarify?
$endgroup$
– Alex Becker
Mar 29 '14 at 2:14
$begingroup$
The xkcd comic is implying that a very simple axiom can be very difficult to prove. This is one of the simplest axioms imaginable, with an extremely difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 2:40
add a comment |
$begingroup$
Prove the reflexive property, or that $x = x$.
http://www.tondering.dk/claus/sur16.pdf
Crazy stuff.
answered Mar 29 '14 at 1:28
louie mcconnelllouie mcconnell
1,4731723
$endgroup$
Prove the reflexive property, or that $x = x$.
http://www.tondering.dk/claus/sur16.pdf
Crazy stuff.
answered Mar 29 '14 at 1:28
louie mcconnelllouie mcconnell
1,4731723
edited Mar 29 '14 at 1:34
answered Mar 29 '14 at 1:28
louie mcconnelllouie mcconnell
1,4731723
answered Mar 29 '14 at 1:28
louie mcconnelllouie mcconnell
1,4731723
answered Mar 29 '14 at 1:28
louie mcconnelllouie mcconnell
1,4731723
1,4731723
$begingroup$
This is misleading, because if you're asking for a formal proof you need to give a formal statement; and a formal statement of "$x=x$ where $x$ is a surreal number" is definitely not simple,
$endgroup$
– Alex Becker
Mar 29 '14 at 1:35
$begingroup$
Yes. A more obvious fact, according to the expert graph, has a more difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 1:35
$begingroup$
I'm not sure what your comment means. Can you clarify?
$endgroup$
– Alex Becker
Mar 29 '14 at 2:14
$begingroup$
The xkcd comic is implying that a very simple axiom can be very difficult to prove. This is one of the simplest axioms imaginable, with an extremely difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 2:40
add a comment |
$begingroup$
This is misleading, because if you're asking for a formal proof you need to give a formal statement; and a formal statement of "$x=x$ where $x$ is a surreal number" is definitely not simple,
$endgroup$
– Alex Becker
Mar 29 '14 at 1:35
$begingroup$
Yes. A more obvious fact, according to the expert graph, has a more difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 1:35
$begingroup$
I'm not sure what your comment means. Can you clarify?
$endgroup$
– Alex Becker
Mar 29 '14 at 2:14
$begingroup$
The xkcd comic is implying that a very simple axiom can be very difficult to prove. This is one of the simplest axioms imaginable, with an extremely difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 2:40
$begingroup$
This is misleading, because if you're asking for a formal proof you need to give a formal statement; and a formal statement of "$x=x$ where $x$ is a surreal number" is definitely not simple,
$endgroup$
– Alex Becker
Mar 29 '14 at 1:35
$begingroup$
This is misleading, because if you're asking for a formal proof you need to give a formal statement; and a formal statement of "$x=x$ where $x$ is a surreal number" is definitely not simple,
$endgroup$
– Alex Becker
Mar 29 '14 at 1:35
$begingroup$
Yes. A more obvious fact, according to the expert graph, has a more difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 1:35
$begingroup$
Yes. A more obvious fact, according to the expert graph, has a more difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 1:35
$begingroup$
I'm not sure what your comment means. Can you clarify?
$endgroup$
– Alex Becker
Mar 29 '14 at 2:14
$begingroup$
I'm not sure what your comment means. Can you clarify?
$endgroup$
– Alex Becker
Mar 29 '14 at 2:14
$begingroup$
The xkcd comic is implying that a very simple axiom can be very difficult to prove. This is one of the simplest axioms imaginable, with an extremely difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 2:40
$begingroup$
The xkcd comic is implying that a very simple axiom can be very difficult to prove. This is one of the simplest axioms imaginable, with an extremely difficult proof.
$endgroup$
– louie mcconnell
Mar 29 '14 at 2:40
add a comment |
$begingroup$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x + y) + z = x + (y + z)$
- $forall x in mathbb{R}forall y in mathbb{R}x + y = y + x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x times y) times z = x times (y times z)$
- $forall x in mathbb{R}forall y in mathbb{R}x times y = y times x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} x times (y + z) = (x times y) + (x times z)$
answered Jan 10 at 20:39
TimothyTimothy
306213
$endgroup$
add a comment |
$begingroup$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x + y) + z = x + (y + z)$
- $forall x in mathbb{R}forall y in mathbb{R}x + y = y + x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x times y) times z = x times (y times z)$
- $forall x in mathbb{R}forall y in mathbb{R}x times y = y times x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} x times (y + z) = (x times y) + (x times z)$
answered Jan 10 at 20:39
TimothyTimothy
306213
$endgroup$
add a comment |
$begingroup$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x + y) + z = x + (y + z)$
- $forall x in mathbb{R}forall y in mathbb{R}x + y = y + x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x times y) times z = x times (y times z)$
- $forall x in mathbb{R}forall y in mathbb{R}x times y = y times x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} x times (y + z) = (x times y) + (x times z)$
answered Jan 10 at 20:39
TimothyTimothy
306213
$endgroup$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x + y) + z = x + (y + z)$
- $forall x in mathbb{R}forall y in mathbb{R}x + y = y + x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} (x times y) times z = x times (y times z)$
- $forall x in mathbb{R}forall y in mathbb{R}x times y = y times x$
- $forall x in mathbb{R}forall y in mathbb{R}forall z in mathbb{R} x times (y + z) = (x times y) + (x times z)$
answered Jan 10 at 20:39
TimothyTimothy
306213
answered Jan 10 at 20:39
TimothyTimothy
306213
answered Jan 10 at 20:39
TimothyTimothy
306213
answered Jan 10 at 20:39
TimothyTimothy
306213
306213
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%2f730830%2fsimple-intuitive-statements-that-are-difficult-to-prove%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
$begingroup$
I don't know that I'd call the ABC conjecture simple to state (at least compared to, say, FLT.)
$endgroup$
– Mike Miller
Mar 29 '14 at 1:31
1
$begingroup$
Relating to the statement in the comic more than your question: Obvious facts are generally easier to prove. However, when one is first learning mathematics, one generally does not develop the necessary machinery to prove anything from axioms, so proofs are done by reducing complex statements to the more obvious ones. In the case of already obvious statements there is no point to this, so the necessary machinery must be developed and used for there to be anything to do. However, this is still easier than proving more complex statements from equally basic principles.
$endgroup$
– Alex Becker
Mar 29 '14 at 1:33
$begingroup$
Agreed, but I wanted to reference an open question, though there are obviously other simpler to state open questions out there.
$endgroup$
– Hayden
Mar 29 '14 at 1:33