Which version of the Pigeonhole principle is correct? One is far stronger than the otherWhat is your favorite application of the Pigeonhole Principle?Is Pigeonhole Principle the negation of Dedekind-infinite?About the Pigeonhole principleProof of the pigeonhole principle by contradictionAre there rigorous formulation and proof of the pigeonhole principle?$1, 2, . . . , 49$ is partitioned into $3$ subsets. Prove that at least $1$ of them contains $3$ different numbers $a,b,c$ such that $a+b=c$How to use generalized pigeonhole principle to be sure that at least one of the integers picked is even?Prove by using the Pigeonhole Principle that there are at least $5$ of the $41$ chess pieces on the $10×10$ board that are not on the same row.Examples of the Pigeonhole PrincipleA “distinguishing” family of subsets
Why are there no programmes / playbills for movies?
Is it safe to unplug a blinking USB drive after 'safely' ejecting it?
Who are the people reviewing far more papers than they're submitting for review?
How would you translate Evangelii Nuntiandi?
Why are two-stroke engines nearly unheard of in aviation?
Why does JavaScript convert an array of one string to a string, when used as an object key?
Answer Not A Fool, or Answer A Fool?
Test to know when to use GLM over Linear Regression?
What's the benefit of prohibiting the use of techniques/language constructs that have not been taught?
Statistical tests for benchmark comparison
Why does dd not make working bootable USB sticks for Microsoft?
L and epsilon factors of Gelbart-Jacquet lifts
Wrong Schengen Visa exit stamp on my passport, who can I complain to?
Other than good shoes and a stick, what are some ways to preserve your knees on long hikes?
Persuading players to be less attached to a pre-session 0 character concept
What is a "major country" as named in Bernie Sanders' Healthcare debate answers?
How to generate short fixed length cryptographic hashs?
Very lazy puppy
What is this WWII four-engine plane on skis?
Wouldn't Kreacher have been able to escape even without following an order?
Is there a generally agreed upon solution to Bradley's Infinite Regress without appeal to Paraconsistent Logic?
In what state are satellites left in when they are left in a graveyard orbit?
How to convert Mn2O3 to Mn3O4?
Asked to Not Use Transactions and to Use A Workaround to Simulate One
Which version of the Pigeonhole principle is correct? One is far stronger than the other
What is your favorite application of the Pigeonhole Principle?Is Pigeonhole Principle the negation of Dedekind-infinite?About the Pigeonhole principleProof of the pigeonhole principle by contradictionAre there rigorous formulation and proof of the pigeonhole principle?$1, 2, . . . , 49$ is partitioned into $3$ subsets. Prove that at least $1$ of them contains $3$ different numbers $a,b,c$ such that $a+b=c$How to use generalized pigeonhole principle to be sure that at least one of the integers picked is even?Prove by using the Pigeonhole Principle that there are at least $5$ of the $41$ chess pieces on the $10×10$ board that are not on the same row.Examples of the Pigeonhole PrincipleA “distinguishing” family of subsets
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
One version of the Pigeonhole principle says that if the cardinality of a set $A$ is greater than that of a set $B$, then there can be no one-to-one function that maps from $A$ to $B$.
Another version says: If $n$ elements are partitioned into $m$ subsets, then at least one subset must contain at least $lceil n/m rceil$ elements. Note that the $lceil rceil$ enclosing the $n/m$ tells us to round the value up.
The latter version is just a more powerful version, no? It seems strange to me that one version of a theorem can objectively give more information than another, so it's likely that either the latter version is not correct or I am incorrect in thinking that it somehow has more utility.
Or is this normal? Is it common for certain versions of theorems to simply have less utility than other versions?
combinatorics discrete-mathematics pigeonhole-principle
asked 8 hours ago
James RonaldJames Ronald
5051 silver badge8 bronze badges
$endgroup$
add a comment
|
$begingroup$
One version of the Pigeonhole principle says that if the cardinality of a set $A$ is greater than that of a set $B$, then there can be no one-to-one function that maps from $A$ to $B$.
Another version says: If $n$ elements are partitioned into $m$ subsets, then at least one subset must contain at least $lceil n/m rceil$ elements. Note that the $lceil rceil$ enclosing the $n/m$ tells us to round the value up.
The latter version is just a more powerful version, no? It seems strange to me that one version of a theorem can objectively give more information than another, so it's likely that either the latter version is not correct or I am incorrect in thinking that it somehow has more utility.
Or is this normal? Is it common for certain versions of theorems to simply have less utility than other versions?
combinatorics discrete-mathematics pigeonhole-principle
asked 8 hours ago
James RonaldJames Ronald
5051 silver badge8 bronze badges
$endgroup$
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
5 mins ago
add a comment
|
$begingroup$
One version of the Pigeonhole principle says that if the cardinality of a set $A$ is greater than that of a set $B$, then there can be no one-to-one function that maps from $A$ to $B$.
Another version says: If $n$ elements are partitioned into $m$ subsets, then at least one subset must contain at least $lceil n/m rceil$ elements. Note that the $lceil rceil$ enclosing the $n/m$ tells us to round the value up.
The latter version is just a more powerful version, no? It seems strange to me that one version of a theorem can objectively give more information than another, so it's likely that either the latter version is not correct or I am incorrect in thinking that it somehow has more utility.
Or is this normal? Is it common for certain versions of theorems to simply have less utility than other versions?
combinatorics discrete-mathematics pigeonhole-principle
asked 8 hours ago
James RonaldJames Ronald
5051 silver badge8 bronze badges
$endgroup$
One version of the Pigeonhole principle says that if the cardinality of a set $A$ is greater than that of a set $B$, then there can be no one-to-one function that maps from $A$ to $B$.
Another version says: If $n$ elements are partitioned into $m$ subsets, then at least one subset must contain at least $lceil n/m rceil$ elements. Note that the $lceil rceil$ enclosing the $n/m$ tells us to round the value up.
The latter version is just a more powerful version, no? It seems strange to me that one version of a theorem can objectively give more information than another, so it's likely that either the latter version is not correct or I am incorrect in thinking that it somehow has more utility.
Or is this normal? Is it common for certain versions of theorems to simply have less utility than other versions?
combinatorics discrete-mathematics pigeonhole-principle
combinatorics discrete-mathematics pigeonhole-principle
asked 8 hours ago
James RonaldJames Ronald
5051 silver badge8 bronze badges
asked 8 hours ago
James RonaldJames Ronald
5051 silver badge8 bronze badges
asked 8 hours ago
James RonaldJames Ronald
5051 silver badge8 bronze badges
asked 8 hours ago
James RonaldJames Ronald
5051 silver badge8 bronze badges
asked 8 hours ago
James RonaldJames Ronald
5051 silver badge8 bronze badges
5051 silver badge8 bronze badges
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
5 mins ago
add a comment
|
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
5 mins ago
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
5 mins ago
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
5 mins ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true, this is pretty much useless information.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
answered 8 hours ago
Eric TowersEric Towers
37.6k2 gold badges24 silver badges73 bronze badges
$endgroup$
1
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
$begingroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
answered 8 hours ago
Matthew DalyMatthew Daly
5,0671 gold badge8 silver badges27 bronze badges
$endgroup$
1
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
$endgroup$
– Ethan Bolker
8 hours ago
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
8 hours ago
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
Your Answer
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/4.0/"u003ecc by-sa 4.0 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%2f3357414%2fwhich-version-of-the-pigeonhole-principle-is-correct-one-is-far-stronger-than-t%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 first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true, this is pretty much useless information.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
answered 8 hours ago
Eric TowersEric Towers
37.6k2 gold badges24 silver badges73 bronze badges
$endgroup$
1
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
$begingroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true, this is pretty much useless information.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
answered 8 hours ago
Eric TowersEric Towers
37.6k2 gold badges24 silver badges73 bronze badges
$endgroup$
1
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
$begingroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true, this is pretty much useless information.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
answered 8 hours ago
Eric TowersEric Towers
37.6k2 gold badges24 silver badges73 bronze badges
$endgroup$
The first version also applies to infinite sets. The latter only applies to finite sets. How can you say the latter is more powerful? It applies to fewer situations.
It is fairly common that a little more information can be extracted from a finite version of a theorem. But that information is frequently useless when applied to the infinite version. For instance, since the cardinality of the real numbers is strictly greater than the cardinality of the integers, there is no injection from the reals to the integers. Using $|mathbbR|$ and $|mathbbZ|$ to represent the cardinalities of the reals and integers, respectively, your second version would say that there are $lceil |mathbbR|/|mathbbZ| rceil = |mathbbR|$ reals sent to at least one integer. While true, this is pretty much useless information.
Also, it is not unusual to find there is one easily proved version of a theorem and another difficult to prove version that is a little sharper. The Whitney embedding theorem has a number of sharpenings that are much more difficult to prove.
answered 8 hours ago
Eric TowersEric Towers
37.6k2 gold badges24 silver badges73 bronze badges
answered 8 hours ago
Eric TowersEric Towers
37.6k2 gold badges24 silver badges73 bronze badges
answered 8 hours ago
Eric TowersEric Towers
37.6k2 gold badges24 silver badges73 bronze badges
answered 8 hours ago
Eric TowersEric Towers
37.6k2 gold badges24 silver badges73 bronze badges
37.6k2 gold badges24 silver badges73 bronze badges
1
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
1
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
8 hours ago
1
1
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
8 hours ago
$begingroup$
Completely forgot about how the latter only applies to finite sets. This was really insightful, thank you!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
$begingroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
answered 8 hours ago
Matthew DalyMatthew Daly
5,0671 gold badge8 silver badges27 bronze badges
$endgroup$
1
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
$endgroup$
– Ethan Bolker
8 hours ago
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
8 hours ago
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
$begingroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
answered 8 hours ago
Matthew DalyMatthew Daly
5,0671 gold badge8 silver badges27 bronze badges
$endgroup$
1
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
$endgroup$
– Ethan Bolker
8 hours ago
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
8 hours ago
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
$begingroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
answered 8 hours ago
Matthew DalyMatthew Daly
5,0671 gold badge8 silver badges27 bronze badges
$endgroup$
This is perfectly normal. If you like, the first version is a corollary of the second one (assuming you are interested in finite sets). It has less utility in the sense that it is less powerful but more utility in the sense that it actually gets applied more often.
answered 8 hours ago
Matthew DalyMatthew Daly
5,0671 gold badge8 silver badges27 bronze badges
edited 8 hours ago
answered 8 hours ago
Matthew DalyMatthew Daly
5,0671 gold badge8 silver badges27 bronze badges
answered 8 hours ago
Matthew DalyMatthew Daly
5,0671 gold badge8 silver badges27 bronze badges
answered 8 hours ago
Matthew DalyMatthew Daly
5,0671 gold badge8 silver badges27 bronze badges
5,0671 gold badge8 silver badges27 bronze badges
1
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
$endgroup$
– Ethan Bolker
8 hours ago
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
8 hours ago
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
8 hours ago
add a comment
|
1
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
$endgroup$
– Ethan Bolker
8 hours ago
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
8 hours ago
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
8 hours ago
1
1
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
$endgroup$
– Ethan Bolker
8 hours ago
$begingroup$
This argument is correct for finite sets. See @EricTowers answer.
$endgroup$
– Ethan Bolker
8 hours ago
1
1
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
8 hours ago
$begingroup$
@EthanBolker As this problem was tagged as discrete math and combinatorics, it seemed appropriate to answer in a finite context. (And thank you for the edit!)
$endgroup$
– Matthew Daly
8 hours ago
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
8 hours ago
$begingroup$
That makes sense, I was too narrow in my thinking: thanks!
$endgroup$
– James Ronald
8 hours ago
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%2f3357414%2fwhich-version-of-the-pigeonhole-principle-is-correct-one-is-far-stronger-than-t%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
$begingroup$
It's quite normal, actually, for two slightly different results to be called the same name, and sometimes one of them happens to be slightly stronger than the other (although that's not really the case here). An example (though perhaps not the best) is the fundamental theorem of algebra: it is sometimes stated as "every nonconstant complex polynomial has at least one root" and other times as "every nonconstant complex polynomial of degree $n$ has exactly $n$ roots". The second obviously implies the first, although it's easy to see that the first also implies the second by induction.
$endgroup$
– YiFan
5 mins ago