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

1607




1607








  • 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













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%2f3048782%2fprove-constructively-that-log-2-3-is-irrational%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









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

108k347114




108k347114








  • 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




















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%2f3048782%2fprove-constructively-that-log-2-3-is-irrational%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

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more