Theorems in the form of “if and only if” such that the proof of one direction is extremely EASY to prove...
I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is:
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The contrast should be as apparent as possible.
Thanks in advance.
soft-question big-list
asked 2 hours ago
YuiTo ChengYuiTo Cheng
154112
add a comment |
I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is:
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The contrast should be as apparent as possible.
Thanks in advance.
soft-question big-list
asked 2 hours ago
YuiTo ChengYuiTo Cheng
154112
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
– Blue
2 hours ago
1
That's a good metaphor!
– YuiTo Cheng
2 hours ago
2
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
– bof
2 hours ago
@bof I've mentioned I don't know if this question is of any value. I just think this phenomenon is intriguing, and want to know more of that nature.
– YuiTo Cheng
1 hour ago
add a comment |
I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is:
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The contrast should be as apparent as possible.
Thanks in advance.
soft-question big-list
asked 2 hours ago
YuiTo ChengYuiTo Cheng
154112
I believe this is a common phenomenon in mathematics, but surprisingly, no such list has been created on this site. I don't know if it's of value, just out of curiosity, I want to see more examples. So my request is:
Theorems in the form of "if and only if" that the proof of one direction is extremely EASY, or intuitive, or make use of some standard techniques (e.g. diagram chasing), while the other one is extremely HARD, or counterintuitive, or require a certain amount of creativity. The contrast should be as apparent as possible.
Thanks in advance.
soft-question big-list
soft-question big-list
asked 2 hours ago
YuiTo ChengYuiTo Cheng
154112
asked 2 hours ago
YuiTo ChengYuiTo Cheng
154112
asked 2 hours ago
YuiTo ChengYuiTo Cheng
154112
asked 2 hours ago
YuiTo ChengYuiTo Cheng
154112
asked 2 hours ago
YuiTo ChengYuiTo Cheng
154112
154112
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
– Blue
2 hours ago
1
That's a good metaphor!
– YuiTo Cheng
2 hours ago
2
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
– bof
2 hours ago
@bof I've mentioned I don't know if this question is of any value. I just think this phenomenon is intriguing, and want to know more of that nature.
– YuiTo Cheng
1 hour ago
add a comment |
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
– Blue
2 hours ago
1
That's a good metaphor!
– YuiTo Cheng
2 hours ago
2
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
– bof
2 hours ago
@bof I've mentioned I don't know if this question is of any value. I just think this phenomenon is intriguing, and want to know more of that nature.
– YuiTo Cheng
1 hour ago
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
– Blue
2 hours ago
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
– Blue
2 hours ago
1
1
That's a good metaphor!
– YuiTo Cheng
2 hours ago
That's a good metaphor!
– YuiTo Cheng
2 hours ago
2
2
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
– bof
2 hours ago
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
– bof
2 hours ago
@bof I've mentioned I don't know if this question is of any value. I just think this phenomenon is intriguing, and want to know more of that nature.
– YuiTo Cheng
1 hour ago
@bof I've mentioned I don't know if this question is of any value. I just think this phenomenon is intriguing, and want to know more of that nature.
– YuiTo Cheng
1 hour ago
add a comment |
5 Answers
5
active
oldest
votes
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered 2 hours ago
Alvin JinAlvin Jin
2,127918
add a comment |
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered 33 mins ago
lonza leggieralonza leggiera
1534
New contributor
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered 57 mins ago
Carlos JiménezCarlos Jiménez
2,3551519
add a comment |
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered 5 mins ago
bofbof
50.5k457119
add a comment |
All planar graphs are $n$-colorable iff $nge4$.
answered 3 mins ago
bofbof
50.5k457119
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%2f3069488%2ftheorems-in-the-form-of-if-and-only-if-such-that-the-proof-of-one-direction-is%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered 2 hours ago
Alvin JinAlvin Jin
2,127918
add a comment |
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered 2 hours ago
Alvin JinAlvin Jin
2,127918
add a comment |
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered 2 hours ago
Alvin JinAlvin Jin
2,127918
"A compact 3-manifold is simply-connected if and only if it is homeomorphic to the 3-sphere."
One direction [only if] is the Poincaré conjecture. The other direction shouldn't be too bad hopefully.
answered 2 hours ago
Alvin JinAlvin Jin
2,127918
answered 2 hours ago
Alvin JinAlvin Jin
2,127918
answered 2 hours ago
Alvin JinAlvin Jin
2,127918
answered 2 hours ago
Alvin JinAlvin Jin
2,127918
2,127918
add a comment |
add a comment |
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered 33 mins ago
lonza leggieralonza leggiera
1534
New contributor
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered 33 mins ago
lonza leggieralonza leggiera
1534
New contributor
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered 33 mins ago
lonza leggieralonza leggiera
1534
New contributor
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The bisectors of two angles of a triangle are of equal length if and only if the two bisected angles are equal. If the two angles are equal, the triangle is isosceles and the proof is very easy. However proving that a triangle must be isosceles if the bisectors of two of its angles are of equal length seems to be quite difficult.
answered 33 mins ago
lonza leggieralonza leggiera
1534
New contributor
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 33 mins ago
lonza leggieralonza leggiera
1534
New contributor
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 33 mins ago
lonza leggieralonza leggiera
1534
answered 33 mins ago
lonza leggieralonza leggiera
1534
1534
New contributor
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
lonza leggiera is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered 57 mins ago
Carlos JiménezCarlos Jiménez
2,3551519
add a comment |
The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered 57 mins ago
Carlos JiménezCarlos Jiménez
2,3551519
add a comment |
The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered 57 mins ago
Carlos JiménezCarlos Jiménez
2,3551519
The product of topological spaces is compact if and only if all of them are compact. One direction is trivial because the projections are continuous and surjective functions. The another direction, well, is the axiom of choice.
answered 57 mins ago
Carlos JiménezCarlos Jiménez
2,3551519
answered 57 mins ago
Carlos JiménezCarlos Jiménez
2,3551519
answered 57 mins ago
Carlos JiménezCarlos Jiménez
2,3551519
answered 57 mins ago
Carlos JiménezCarlos Jiménez
2,3551519
2,3551519
add a comment |
add a comment |
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered 5 mins ago
bofbof
50.5k457119
add a comment |
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered 5 mins ago
bofbof
50.5k457119
add a comment |
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered 5 mins ago
bofbof
50.5k457119
Let $n$ be a positive integer. The equation $x^n+y^n=z^n$ is solvable in positive integers $x,y,z$ iff $nle2$.
answered 5 mins ago
bofbof
50.5k457119
answered 5 mins ago
bofbof
50.5k457119
answered 5 mins ago
bofbof
50.5k457119
answered 5 mins ago
bofbof
50.5k457119
50.5k457119
add a comment |
add a comment |
All planar graphs are $n$-colorable iff $nge4$.
answered 3 mins ago
bofbof
50.5k457119
add a comment |
All planar graphs are $n$-colorable iff $nge4$.
answered 3 mins ago
bofbof
50.5k457119
add a comment |
All planar graphs are $n$-colorable iff $nge4$.
answered 3 mins ago
bofbof
50.5k457119
All planar graphs are $n$-colorable iff $nge4$.
answered 3 mins ago
bofbof
50.5k457119
answered 3 mins ago
bofbof
50.5k457119
answered 3 mins ago
bofbof
50.5k457119
answered 3 mins ago
bofbof
50.5k457119
50.5k457119
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%2f3069488%2ftheorems-in-the-form-of-if-and-only-if-such-that-the-proof-of-one-direction-is%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
One of my college professors likened this phenomenon to running one's palm along one's face. Down is easy; up is not.
– Blue
2 hours ago
1
That's a good metaphor!
– YuiTo Cheng
2 hours ago
2
The reason nobody has created such a list may be the fact that most (nontrivial) "iff" theorems belong on your list. It would be more interesting to see a list of "iff" theorems where both directions are nontrivial.
– bof
2 hours ago
@bof I've mentioned I don't know if this question is of any value. I just think this phenomenon is intriguing, and want to know more of that nature.
– YuiTo Cheng
1 hour ago