Prove constructively that $log_2 3$ is irrational.












2












$begingroup$


The usual proof that $log_2 3$ is irrational is by contradiction. For instance:




Assume the negation: that $log_2 3 = m/n$ for some integers $m$ and $n$. Then, by the property of logarithms, $2^{m/n} = 3$, which implies that $2^m = 3^n$. However, $2^m$ is even and $3^n$ is odd and an even number cannot be equal to an odd number. Therefore the assumption that $log_2 3$ is rational is wrong.




My understanding is that this form of proof by contradiction (assume the negation and arrive at a contradiction) is using the law of excluded middle (that proving $lnot lnot A$ is the same as proving $A$) and is therefore not a valid constructive proof.



So that leads to my two-part question:




  • Is the proof actually okay as a constructive proof (i.e., is my understanding wrong) and if so, why is it okay?

  • If it is not valid, what is a constructive proof that $log_2 3$ is irrational?






alternative-proof irrational-numbers constructive-mathematics intuitionistic-logic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What exactly do you mean by "constructively"?
    $endgroup$
    – Frpzzd
    Dec 21 '18 at 18:42










  • $begingroup$
    @Frpzzd - See en.wikipedia.org/wiki/Constructive_proof
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 18:43
















2












$begingroup$


The usual proof that $log_2 3$ is irrational is by contradiction. For instance:




Assume the negation: that $log_2 3 = m/n$ for some integers $m$ and $n$. Then, by the property of logarithms, $2^{m/n} = 3$, which implies that $2^m = 3^n$. However, $2^m$ is even and $3^n$ is odd and an even number cannot be equal to an odd number. Therefore the assumption that $log_2 3$ is rational is wrong.




My understanding is that this form of proof by contradiction (assume the negation and arrive at a contradiction) is using the law of excluded middle (that proving $lnot lnot A$ is the same as proving $A$) and is therefore not a valid constructive proof.



So that leads to my two-part question:




  • Is the proof actually okay as a constructive proof (i.e., is my understanding wrong) and if so, why is it okay?

  • If it is not valid, what is a constructive proof that $log_2 3$ is irrational?






alternative-proof irrational-numbers constructive-mathematics intuitionistic-logic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What exactly do you mean by "constructively"?
    $endgroup$
    – Frpzzd
    Dec 21 '18 at 18:42










  • $begingroup$
    @Frpzzd - See en.wikipedia.org/wiki/Constructive_proof
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 18:43














2












2








2


1



$begingroup$


The usual proof that $log_2 3$ is irrational is by contradiction. For instance:




Assume the negation: that $log_2 3 = m/n$ for some integers $m$ and $n$. Then, by the property of logarithms, $2^{m/n} = 3$, which implies that $2^m = 3^n$. However, $2^m$ is even and $3^n$ is odd and an even number cannot be equal to an odd number. Therefore the assumption that $log_2 3$ is rational is wrong.




My understanding is that this form of proof by contradiction (assume the negation and arrive at a contradiction) is using the law of excluded middle (that proving $lnot lnot A$ is the same as proving $A$) and is therefore not a valid constructive proof.



So that leads to my two-part question:




  • Is the proof actually okay as a constructive proof (i.e., is my understanding wrong) and if so, why is it okay?

  • If it is not valid, what is a constructive proof that $log_2 3$ is irrational?






alternative-proof irrational-numbers constructive-mathematics intuitionistic-logic







share|cite|improve this question









$endgroup$




The usual proof that $log_2 3$ is irrational is by contradiction. For instance:




Assume the negation: that $log_2 3 = m/n$ for some integers $m$ and $n$. Then, by the property of logarithms, $2^{m/n} = 3$, which implies that $2^m = 3^n$. However, $2^m$ is even and $3^n$ is odd and an even number cannot be equal to an odd number. Therefore the assumption that $log_2 3$ is rational is wrong.




My understanding is that this form of proof by contradiction (assume the negation and arrive at a contradiction) is using the law of excluded middle (that proving $lnot lnot A$ is the same as proving $A$) and is therefore not a valid constructive proof.



So that leads to my two-part question:




  • Is the proof actually okay as a constructive proof (i.e., is my understanding wrong) and if so, why is it okay?

  • If it is not valid, what is a constructive proof that $log_2 3$ is irrational?






alternative-proof irrational-numbers constructive-mathematics intuitionistic-logic




alternative-proof irrational-numbers constructive-mathematics intuitionistic-logic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 18:38









Ted HoppTed Hopp

1628




1628








  • 1




    $begingroup$
    What exactly do you mean by "constructively"?
    $endgroup$
    – Frpzzd
    Dec 21 '18 at 18:42










  • $begingroup$
    @Frpzzd - See en.wikipedia.org/wiki/Constructive_proof
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 18:43














  • 1




    $begingroup$
    What exactly do you mean by "constructively"?
    $endgroup$
    – Frpzzd
    Dec 21 '18 at 18:42










  • $begingroup$
    @Frpzzd - See en.wikipedia.org/wiki/Constructive_proof
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 18:43








1




1




$begingroup$
What exactly do you mean by "constructively"?
$endgroup$
– Frpzzd
Dec 21 '18 at 18:42




$begingroup$
What exactly do you mean by "constructively"?
$endgroup$
– Frpzzd
Dec 21 '18 at 18:42












$begingroup$
@Frpzzd - See en.wikipedia.org/wiki/Constructive_proof
$endgroup$
– Ted Hopp
Dec 21 '18 at 18:43




$begingroup$
@Frpzzd - See en.wikipedia.org/wiki/Constructive_proof
$endgroup$
– Ted Hopp
Dec 21 '18 at 18:43










1 Answer
1






active

oldest

votes


















6












$begingroup$

$x$ is irrational is defined as "$x$ is not rational", so a proof that shows that from the assumption that $x$ is rational we derive a contradiction is a valid constructive proof for
"$x$ is not rational", in fact it's the usual proof for such negative statements in e.g. intuitionistic logic.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In fact, it's not at all unusual to define $lnot phi$ as being "synactic sugar" for $phi rightarrow bot$; and in that case, a proof of this form would be a special case of the standard ${rightarrow} I$ rule.
    $endgroup$
    – Daniel Schepler
    Dec 21 '18 at 18:45












  • $begingroup$
    So let's see if I understand this. By definition, "$log_2 3$ is irrational" is by definition shorthand for "$log_2 3$ is not rational" and we're proving a negative. Furthermore, in constructivism, any theorem of the form "$lnot A$" can be validly proven by showing that $Avdashbot$? So even though the structure of the proof looks like we're proving $lnot lnot Avdashbot$, that's not what's going on?
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 20:29








  • 1




    $begingroup$
    @TedHopp indeed. A negative is proven by contradiction. How else could one prove a negative. But $lnotlnotphi$ does not show $phi$. The linked proof here that there are $a,b$ irrational such that $a^b$ is rational shows this by assuming 'tertium non datur' which is not allowed constructively.
    $endgroup$
    – Henno Brandsma
    Dec 21 '18 at 22:41













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









6












$begingroup$

$x$ is irrational is defined as "$x$ is not rational", so a proof that shows that from the assumption that $x$ is rational we derive a contradiction is a valid constructive proof for
"$x$ is not rational", in fact it's the usual proof for such negative statements in e.g. intuitionistic logic.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In fact, it's not at all unusual to define $lnot phi$ as being "synactic sugar" for $phi rightarrow bot$; and in that case, a proof of this form would be a special case of the standard ${rightarrow} I$ rule.
    $endgroup$
    – Daniel Schepler
    Dec 21 '18 at 18:45












  • $begingroup$
    So let's see if I understand this. By definition, "$log_2 3$ is irrational" is by definition shorthand for "$log_2 3$ is not rational" and we're proving a negative. Furthermore, in constructivism, any theorem of the form "$lnot A$" can be validly proven by showing that $Avdashbot$? So even though the structure of the proof looks like we're proving $lnot lnot Avdashbot$, that's not what's going on?
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 20:29








  • 1




    $begingroup$
    @TedHopp indeed. A negative is proven by contradiction. How else could one prove a negative. But $lnotlnotphi$ does not show $phi$. The linked proof here that there are $a,b$ irrational such that $a^b$ is rational shows this by assuming 'tertium non datur' which is not allowed constructively.
    $endgroup$
    – Henno Brandsma
    Dec 21 '18 at 22:41


















6












$begingroup$

$x$ is irrational is defined as "$x$ is not rational", so a proof that shows that from the assumption that $x$ is rational we derive a contradiction is a valid constructive proof for
"$x$ is not rational", in fact it's the usual proof for such negative statements in e.g. intuitionistic logic.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In fact, it's not at all unusual to define $lnot phi$ as being "synactic sugar" for $phi rightarrow bot$; and in that case, a proof of this form would be a special case of the standard ${rightarrow} I$ rule.
    $endgroup$
    – Daniel Schepler
    Dec 21 '18 at 18:45












  • $begingroup$
    So let's see if I understand this. By definition, "$log_2 3$ is irrational" is by definition shorthand for "$log_2 3$ is not rational" and we're proving a negative. Furthermore, in constructivism, any theorem of the form "$lnot A$" can be validly proven by showing that $Avdashbot$? So even though the structure of the proof looks like we're proving $lnot lnot Avdashbot$, that's not what's going on?
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 20:29








  • 1




    $begingroup$
    @TedHopp indeed. A negative is proven by contradiction. How else could one prove a negative. But $lnotlnotphi$ does not show $phi$. The linked proof here that there are $a,b$ irrational such that $a^b$ is rational shows this by assuming 'tertium non datur' which is not allowed constructively.
    $endgroup$
    – Henno Brandsma
    Dec 21 '18 at 22:41
















6












6








6





$begingroup$

$x$ is irrational is defined as "$x$ is not rational", so a proof that shows that from the assumption that $x$ is rational we derive a contradiction is a valid constructive proof for
"$x$ is not rational", in fact it's the usual proof for such negative statements in e.g. intuitionistic logic.






share|cite|improve this answer









$endgroup$



$x$ is irrational is defined as "$x$ is not rational", so a proof that shows that from the assumption that $x$ is rational we derive a contradiction is a valid constructive proof for
"$x$ is not rational", in fact it's the usual proof for such negative statements in e.g. intuitionistic logic.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 21 '18 at 18:43









Henno BrandsmaHenno Brandsma

114k348123




114k348123








  • 1




    $begingroup$
    In fact, it's not at all unusual to define $lnot phi$ as being "synactic sugar" for $phi rightarrow bot$; and in that case, a proof of this form would be a special case of the standard ${rightarrow} I$ rule.
    $endgroup$
    – Daniel Schepler
    Dec 21 '18 at 18:45












  • $begingroup$
    So let's see if I understand this. By definition, "$log_2 3$ is irrational" is by definition shorthand for "$log_2 3$ is not rational" and we're proving a negative. Furthermore, in constructivism, any theorem of the form "$lnot A$" can be validly proven by showing that $Avdashbot$? So even though the structure of the proof looks like we're proving $lnot lnot Avdashbot$, that's not what's going on?
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 20:29








  • 1




    $begingroup$
    @TedHopp indeed. A negative is proven by contradiction. How else could one prove a negative. But $lnotlnotphi$ does not show $phi$. The linked proof here that there are $a,b$ irrational such that $a^b$ is rational shows this by assuming 'tertium non datur' which is not allowed constructively.
    $endgroup$
    – Henno Brandsma
    Dec 21 '18 at 22:41
















  • 1




    $begingroup$
    In fact, it's not at all unusual to define $lnot phi$ as being "synactic sugar" for $phi rightarrow bot$; and in that case, a proof of this form would be a special case of the standard ${rightarrow} I$ rule.
    $endgroup$
    – Daniel Schepler
    Dec 21 '18 at 18:45












  • $begingroup$
    So let's see if I understand this. By definition, "$log_2 3$ is irrational" is by definition shorthand for "$log_2 3$ is not rational" and we're proving a negative. Furthermore, in constructivism, any theorem of the form "$lnot A$" can be validly proven by showing that $Avdashbot$? So even though the structure of the proof looks like we're proving $lnot lnot Avdashbot$, that's not what's going on?
    $endgroup$
    – Ted Hopp
    Dec 21 '18 at 20:29








  • 1




    $begingroup$
    @TedHopp indeed. A negative is proven by contradiction. How else could one prove a negative. But $lnotlnotphi$ does not show $phi$. The linked proof here that there are $a,b$ irrational such that $a^b$ is rational shows this by assuming 'tertium non datur' which is not allowed constructively.
    $endgroup$
    – Henno Brandsma
    Dec 21 '18 at 22:41










1




1




$begingroup$
In fact, it's not at all unusual to define $lnot phi$ as being "synactic sugar" for $phi rightarrow bot$; and in that case, a proof of this form would be a special case of the standard ${rightarrow} I$ rule.
$endgroup$
– Daniel Schepler
Dec 21 '18 at 18:45






$begingroup$
In fact, it's not at all unusual to define $lnot phi$ as being "synactic sugar" for $phi rightarrow bot$; and in that case, a proof of this form would be a special case of the standard ${rightarrow} I$ rule.
$endgroup$
– Daniel Schepler
Dec 21 '18 at 18:45














$begingroup$
So let's see if I understand this. By definition, "$log_2 3$ is irrational" is by definition shorthand for "$log_2 3$ is not rational" and we're proving a negative. Furthermore, in constructivism, any theorem of the form "$lnot A$" can be validly proven by showing that $Avdashbot$? So even though the structure of the proof looks like we're proving $lnot lnot Avdashbot$, that's not what's going on?
$endgroup$
– Ted Hopp
Dec 21 '18 at 20:29






$begingroup$
So let's see if I understand this. By definition, "$log_2 3$ is irrational" is by definition shorthand for "$log_2 3$ is not rational" and we're proving a negative. Furthermore, in constructivism, any theorem of the form "$lnot A$" can be validly proven by showing that $Avdashbot$? So even though the structure of the proof looks like we're proving $lnot lnot Avdashbot$, that's not what's going on?
$endgroup$
– Ted Hopp
Dec 21 '18 at 20:29






1




1




$begingroup$
@TedHopp indeed. A negative is proven by contradiction. How else could one prove a negative. But $lnotlnotphi$ does not show $phi$. The linked proof here that there are $a,b$ irrational such that $a^b$ is rational shows this by assuming 'tertium non datur' which is not allowed constructively.
$endgroup$
– Henno Brandsma
Dec 21 '18 at 22:41






$begingroup$
@TedHopp indeed. A negative is proven by contradiction. How else could one prove a negative. But $lnotlnotphi$ does not show $phi$. The linked proof here that there are $a,b$ irrational such that $a^b$ is rational shows this by assuming 'tertium non datur' which is not allowed constructively.
$endgroup$
– Henno Brandsma
Dec 21 '18 at 22:41




















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