Careless Mathematical Induction Fallacy












3












$begingroup$


This fallacy is given in Bartle's Introduction to Real Analysis (page 15) and I am trying to figure out where the problem is in the "fake proof".



Here we are using $mathbb{N} = {1,2,3 dots } $




If $n in mathbb{N}$ and if $max (p,q) = n $ for $p,q in mathbb{N}$, then $p=q$




The base case $n=1$ does check out. If $max (p,q) = 1$, then we do have that $p=q$ since $p,q in mathbb{N}$.



Then we assume the statement is true for $k ge 1$ and we want to prove the statement is true for $k+1$.



Suppose $max (p,q) = k+1$. Then we have $max (p-1, q-1) = k$. Thus, we have $p-1 = q-1 implies p=q $ $tag*{$square$}$



Obviously this is an absurd statement, as this is not true in general. It is clear that the base case is correct, so the flaw must be somewhere in the induction step.



My only guess is that it stems from saying $max(p-1,q-1) = k$ and that somehow we have a subtle violation going on. I would like to be able to clearly see the violation. Any help would be greatly appreciated.






induction fake-proofs







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Why can you take $p-1,q-1$? They might not be natural numbers. For instance, $max (1,2)=2$.
    $endgroup$
    – lulu
    3 hours ago






  • 1




    $begingroup$
    This depends on whether you include $0$ in $mathbf N$ or not.
    $endgroup$
    – Bernard
    3 hours ago
















3












$begingroup$


This fallacy is given in Bartle's Introduction to Real Analysis (page 15) and I am trying to figure out where the problem is in the "fake proof".



Here we are using $mathbb{N} = {1,2,3 dots } $




If $n in mathbb{N}$ and if $max (p,q) = n $ for $p,q in mathbb{N}$, then $p=q$




The base case $n=1$ does check out. If $max (p,q) = 1$, then we do have that $p=q$ since $p,q in mathbb{N}$.



Then we assume the statement is true for $k ge 1$ and we want to prove the statement is true for $k+1$.



Suppose $max (p,q) = k+1$. Then we have $max (p-1, q-1) = k$. Thus, we have $p-1 = q-1 implies p=q $ $tag*{$square$}$



Obviously this is an absurd statement, as this is not true in general. It is clear that the base case is correct, so the flaw must be somewhere in the induction step.



My only guess is that it stems from saying $max(p-1,q-1) = k$ and that somehow we have a subtle violation going on. I would like to be able to clearly see the violation. Any help would be greatly appreciated.






induction fake-proofs







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Why can you take $p-1,q-1$? They might not be natural numbers. For instance, $max (1,2)=2$.
    $endgroup$
    – lulu
    3 hours ago






  • 1




    $begingroup$
    This depends on whether you include $0$ in $mathbf N$ or not.
    $endgroup$
    – Bernard
    3 hours ago














3












3








3


1



$begingroup$


This fallacy is given in Bartle's Introduction to Real Analysis (page 15) and I am trying to figure out where the problem is in the "fake proof".



Here we are using $mathbb{N} = {1,2,3 dots } $




If $n in mathbb{N}$ and if $max (p,q) = n $ for $p,q in mathbb{N}$, then $p=q$




The base case $n=1$ does check out. If $max (p,q) = 1$, then we do have that $p=q$ since $p,q in mathbb{N}$.



Then we assume the statement is true for $k ge 1$ and we want to prove the statement is true for $k+1$.



Suppose $max (p,q) = k+1$. Then we have $max (p-1, q-1) = k$. Thus, we have $p-1 = q-1 implies p=q $ $tag*{$square$}$



Obviously this is an absurd statement, as this is not true in general. It is clear that the base case is correct, so the flaw must be somewhere in the induction step.



My only guess is that it stems from saying $max(p-1,q-1) = k$ and that somehow we have a subtle violation going on. I would like to be able to clearly see the violation. Any help would be greatly appreciated.






induction fake-proofs







share|cite|improve this question











$endgroup$




This fallacy is given in Bartle's Introduction to Real Analysis (page 15) and I am trying to figure out where the problem is in the "fake proof".



Here we are using $mathbb{N} = {1,2,3 dots } $




If $n in mathbb{N}$ and if $max (p,q) = n $ for $p,q in mathbb{N}$, then $p=q$




The base case $n=1$ does check out. If $max (p,q) = 1$, then we do have that $p=q$ since $p,q in mathbb{N}$.



Then we assume the statement is true for $k ge 1$ and we want to prove the statement is true for $k+1$.



Suppose $max (p,q) = k+1$. Then we have $max (p-1, q-1) = k$. Thus, we have $p-1 = q-1 implies p=q $ $tag*{$square$}$



Obviously this is an absurd statement, as this is not true in general. It is clear that the base case is correct, so the flaw must be somewhere in the induction step.



My only guess is that it stems from saying $max(p-1,q-1) = k$ and that somehow we have a subtle violation going on. I would like to be able to clearly see the violation. Any help would be greatly appreciated.






induction fake-proofs




induction fake-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







WaveX

















asked 3 hours ago









WaveXWaveX

2,5622722




2,5622722








  • 4




    $begingroup$
    Why can you take $p-1,q-1$? They might not be natural numbers. For instance, $max (1,2)=2$.
    $endgroup$
    – lulu
    3 hours ago






  • 1




    $begingroup$
    This depends on whether you include $0$ in $mathbf N$ or not.
    $endgroup$
    – Bernard
    3 hours ago














  • 4




    $begingroup$
    Why can you take $p-1,q-1$? They might not be natural numbers. For instance, $max (1,2)=2$.
    $endgroup$
    – lulu
    3 hours ago






  • 1




    $begingroup$
    This depends on whether you include $0$ in $mathbf N$ or not.
    $endgroup$
    – Bernard
    3 hours ago








4




4




$begingroup$
Why can you take $p-1,q-1$? They might not be natural numbers. For instance, $max (1,2)=2$.
$endgroup$
– lulu
3 hours ago




$begingroup$
Why can you take $p-1,q-1$? They might not be natural numbers. For instance, $max (1,2)=2$.
$endgroup$
– lulu
3 hours ago




1




1




$begingroup$
This depends on whether you include $0$ in $mathbf N$ or not.
$endgroup$
– Bernard
3 hours ago




$begingroup$
This depends on whether you include $0$ in $mathbf N$ or not.
$endgroup$
– Bernard
3 hours ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

The problem is that $p-1,q-1$ might not be in $mathbb N$. For instance, consider $max (1,2)=2$ . The induction would direct us to look at $max(0,1)=1$ but that was not covered in the base case.



Note: if we considered $0$ as a natural number then the base case is false as presented (since $max (0,1)=1$ is a counterexample). Of course, we could consider the base case $n=0$ and that would still be correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Going by the logic this approach would also work if we did include $0$ in our definition of $mathbb{N}$. Then we would be looking at $max (-1,0)$ and once again we fall out of $mathbb{N}$, am I correct?
    $endgroup$
    – WaveX
    3 hours ago






  • 2




    $begingroup$
    Exactly right. We'd have $max (a,b)=0implies a=b=0$ but you couldn't do the induction, as the example $max (0,1)=1$ demonstrates.
    $endgroup$
    – lulu
    3 hours ago











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088884%2fcareless-mathematical-induction-fallacy%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









7












$begingroup$

The problem is that $p-1,q-1$ might not be in $mathbb N$. For instance, consider $max (1,2)=2$ . The induction would direct us to look at $max(0,1)=1$ but that was not covered in the base case.



Note: if we considered $0$ as a natural number then the base case is false as presented (since $max (0,1)=1$ is a counterexample). Of course, we could consider the base case $n=0$ and that would still be correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Going by the logic this approach would also work if we did include $0$ in our definition of $mathbb{N}$. Then we would be looking at $max (-1,0)$ and once again we fall out of $mathbb{N}$, am I correct?
    $endgroup$
    – WaveX
    3 hours ago






  • 2




    $begingroup$
    Exactly right. We'd have $max (a,b)=0implies a=b=0$ but you couldn't do the induction, as the example $max (0,1)=1$ demonstrates.
    $endgroup$
    – lulu
    3 hours ago
















7












$begingroup$

The problem is that $p-1,q-1$ might not be in $mathbb N$. For instance, consider $max (1,2)=2$ . The induction would direct us to look at $max(0,1)=1$ but that was not covered in the base case.



Note: if we considered $0$ as a natural number then the base case is false as presented (since $max (0,1)=1$ is a counterexample). Of course, we could consider the base case $n=0$ and that would still be correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Going by the logic this approach would also work if we did include $0$ in our definition of $mathbb{N}$. Then we would be looking at $max (-1,0)$ and once again we fall out of $mathbb{N}$, am I correct?
    $endgroup$
    – WaveX
    3 hours ago






  • 2




    $begingroup$
    Exactly right. We'd have $max (a,b)=0implies a=b=0$ but you couldn't do the induction, as the example $max (0,1)=1$ demonstrates.
    $endgroup$
    – lulu
    3 hours ago














7












7








7





$begingroup$

The problem is that $p-1,q-1$ might not be in $mathbb N$. For instance, consider $max (1,2)=2$ . The induction would direct us to look at $max(0,1)=1$ but that was not covered in the base case.



Note: if we considered $0$ as a natural number then the base case is false as presented (since $max (0,1)=1$ is a counterexample). Of course, we could consider the base case $n=0$ and that would still be correct.






share|cite|improve this answer









$endgroup$



The problem is that $p-1,q-1$ might not be in $mathbb N$. For instance, consider $max (1,2)=2$ . The induction would direct us to look at $max(0,1)=1$ but that was not covered in the base case.



Note: if we considered $0$ as a natural number then the base case is false as presented (since $max (0,1)=1$ is a counterexample). Of course, we could consider the base case $n=0$ and that would still be correct.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









lulululu

40.3k24778




40.3k24778












  • $begingroup$
    Going by the logic this approach would also work if we did include $0$ in our definition of $mathbb{N}$. Then we would be looking at $max (-1,0)$ and once again we fall out of $mathbb{N}$, am I correct?
    $endgroup$
    – WaveX
    3 hours ago






  • 2




    $begingroup$
    Exactly right. We'd have $max (a,b)=0implies a=b=0$ but you couldn't do the induction, as the example $max (0,1)=1$ demonstrates.
    $endgroup$
    – lulu
    3 hours ago


















  • $begingroup$
    Going by the logic this approach would also work if we did include $0$ in our definition of $mathbb{N}$. Then we would be looking at $max (-1,0)$ and once again we fall out of $mathbb{N}$, am I correct?
    $endgroup$
    – WaveX
    3 hours ago






  • 2




    $begingroup$
    Exactly right. We'd have $max (a,b)=0implies a=b=0$ but you couldn't do the induction, as the example $max (0,1)=1$ demonstrates.
    $endgroup$
    – lulu
    3 hours ago
















$begingroup$
Going by the logic this approach would also work if we did include $0$ in our definition of $mathbb{N}$. Then we would be looking at $max (-1,0)$ and once again we fall out of $mathbb{N}$, am I correct?
$endgroup$
– WaveX
3 hours ago




$begingroup$
Going by the logic this approach would also work if we did include $0$ in our definition of $mathbb{N}$. Then we would be looking at $max (-1,0)$ and once again we fall out of $mathbb{N}$, am I correct?
$endgroup$
– WaveX
3 hours ago




2




2




$begingroup$
Exactly right. We'd have $max (a,b)=0implies a=b=0$ but you couldn't do the induction, as the example $max (0,1)=1$ demonstrates.
$endgroup$
– lulu
3 hours ago




$begingroup$
Exactly right. We'd have $max (a,b)=0implies a=b=0$ but you couldn't do the induction, as the example $max (0,1)=1$ demonstrates.
$endgroup$
– lulu
3 hours ago


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3088884%2fcareless-mathematical-induction-fallacy%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Loup dans la culture

Image
Loups sur une enluminure du Bestiaire d'Aberdeen Le loup est l'animal le plus emblématique de l'histoire de l'Eurasie, il était à l'honneur durant l'Antiquité chez la totalité des anciens peuples européens [ 1 ] . Sommaire 1 Le loup dans la mythologie 2 Les loups dans le folklore 2.1 L'évolution des mentalités 2.2 Le loup dans la psychanalyse 3 Le loup dans les œuvres culturelles 3.1 Le loup dans la littérature 3.1.1 Fables 3.1.2 Poésie 3.1.3 Romans 3.1.4 Littérature jeunesse 3.1.4.1 Quelques titres d'albums contemporains 3.1.5 Bande dessinée 3.2 Le loup à l'écran 3.2.1 Le cinéma et la télévision 3.2.2 Documentaires 3.2.3 Dessins animés 3.2.4 Jeux vidéo 3.3 Le loup en peinture et en sculpture 3.4 Le loup dans les œuvres musicales 4 Le loup et les noms propres 5 Bibliographie 6 Notes et références 7 Voir aussi 7.1 Articles connexes
Read more

How to solve the problem of ntp “Unable to contact time server” from KDE?

Image
1 0 I want to set date and time automatically but this error appears when I try to .. I'm using openSUSE, I tried different servers but it didn't work either, help please! It doesn't seem to be a network problem: $ ping pool.ntp.org 64 bytes from 41.78.128.17: icmp_seq=1 ttl=41 time=260 ms To the question of whether ntpd is installed and running: $ ps -C ntpd PID TTY TIME CMD i.e. there's no ntpd process running. $ rpm -qa | grep ntp yast2-ntp-client-3.1.12-1.7.noarch ntp-4.2.6p5-25.2.1.i586 Further information from the comments: $ sudo ntpdate pool.ntp.org 25 Sep 19:41:51 ntpdate[3162]: no server suitable for synchronization found networking kde opensuse ntp ntpd sh
Read more

Connection limited (no internet access)

Image
-1 My laptop is running on Windows 10 ، My wifi adapter is tenda V7TW311M ,My router is D-Link , When I connect to the internet I get connection limited with yellow triangle , and no ipv4 address in ipconfig ,Ihave tried manually setting the ipv4 address When dhcp works there is no ipv4 , When I select ipv4 manually dhcp is turned off , I tried an update my wifi to the Latest version But the same problem remained . networking wifi internet net share | improve this question asked 13 mins ago Souchi Souchi 1 1
Read more