Careless Mathematical Induction Fallacy












2












$begingroup$


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











$endgroup$








  • 3




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










  • $begingroup$
    This depends on whether you include $0$ in $mathbf N$ or not.
    $endgroup$
    – Bernard
    1 hour ago
















2












$begingroup$


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











$endgroup$








  • 3




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










  • $begingroup$
    This depends on whether you include $0$ in $mathbf N$ or not.
    $endgroup$
    – Bernard
    1 hour ago














2












2








2





$begingroup$


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 1 hour ago







WaveX

















asked 1 hour ago









WaveXWaveX

2,5572722




2,5572722








  • 3




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










  • $begingroup$
    This depends on whether you include $0$ in $mathbf N$ or not.
    $endgroup$
    – Bernard
    1 hour ago














  • 3




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










  • $begingroup$
    This depends on whether you include $0$ in $mathbf N$ or not.
    $endgroup$
    – Bernard
    1 hour ago








3




3




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




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












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




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










1 Answer
1






active

oldest

votes


















5












$begingroup$

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
    1 hour ago






  • 1




    $begingroup$
    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
    1 hour ago











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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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5












$begingroup$

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
    1 hour ago






  • 1




    $begingroup$
    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
    1 hour ago
















5












$begingroup$

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
    1 hour ago






  • 1




    $begingroup$
    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
    1 hour ago














5












5








5





$begingroup$

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 1 hour ago









lulululu

40.2k24778




40.2k24778












  • $begingroup$
    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
    1 hour ago






  • 1




    $begingroup$
    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
    1 hour 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
    1 hour ago






  • 1




    $begingroup$
    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
    1 hour 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
1 hour 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
1 hour ago




1




1




$begingroup$
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
1 hour 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
1 hour ago


















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