Is the sum of two irrational numbers almost always irrational?












9














Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$displaystylelim_{xtoinfty}dfrac{lambda(Pcap B(x))}{lambda(Rcap B(x))}=1$$ where $lambda$ is Lebesgue measure, $Rsubset mathbb{R}^2$ is the set of points with irrational coordinates and $Psubset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.






real-analysis number-theory probability-theory







share|cite|improve this question




















  • 1




    Related: Is the sum and difference of two irrationals always irrational?
    – mrtaurho
    yesterday










  • I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
    – Milo Brandt
    yesterday


















9














Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$displaystylelim_{xtoinfty}dfrac{lambda(Pcap B(x))}{lambda(Rcap B(x))}=1$$ where $lambda$ is Lebesgue measure, $Rsubset mathbb{R}^2$ is the set of points with irrational coordinates and $Psubset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.






real-analysis number-theory probability-theory







share|cite|improve this question




















  • 1




    Related: Is the sum and difference of two irrationals always irrational?
    – mrtaurho
    yesterday










  • I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
    – Milo Brandt
    yesterday
















9












9








9


4





Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$displaystylelim_{xtoinfty}dfrac{lambda(Pcap B(x))}{lambda(Rcap B(x))}=1$$ where $lambda$ is Lebesgue measure, $Rsubset mathbb{R}^2$ is the set of points with irrational coordinates and $Psubset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.






real-analysis number-theory probability-theory







share|cite|improve this question















Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$displaystylelim_{xtoinfty}dfrac{lambda(Pcap B(x))}{lambda(Rcap B(x))}=1$$ where $lambda$ is Lebesgue measure, $Rsubset mathbb{R}^2$ is the set of points with irrational coordinates and $Psubset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.






real-analysis number-theory probability-theory




real-analysis number-theory probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday







AlephNull

















asked yesterday









AlephNullAlephNull

2209




2209








  • 1




    Related: Is the sum and difference of two irrationals always irrational?
    – mrtaurho
    yesterday










  • I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
    – Milo Brandt
    yesterday
















  • 1




    Related: Is the sum and difference of two irrationals always irrational?
    – mrtaurho
    yesterday










  • I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
    – Milo Brandt
    yesterday










1




1




Related: Is the sum and difference of two irrationals always irrational?
– mrtaurho
yesterday




Related: Is the sum and difference of two irrationals always irrational?
– mrtaurho
yesterday












I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
– Milo Brandt
yesterday






I think you mean for your denominator to be the measure of $B(x)$ not of $Rcap B(x)$.
– Milo Brandt
yesterday












1 Answer
1






active

oldest

votes


















11














Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.



To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.






share|cite|improve this answer























  • Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
    – AlephNull
    yesterday








  • 1




    Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
    – Milo Brandt
    yesterday










  • I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
    – Stijn de Witt
    21 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%2f3063918%2fis-the-sum-of-two-irrational-numbers-almost-always-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









11














Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.



To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.






share|cite|improve this answer























  • Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
    – AlephNull
    yesterday








  • 1




    Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
    – Milo Brandt
    yesterday










  • I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
    – Stijn de Witt
    21 hours ago
















11














Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.



To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.






share|cite|improve this answer























  • Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
    – AlephNull
    yesterday








  • 1




    Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
    – Milo Brandt
    yesterday










  • I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
    – Stijn de Witt
    21 hours ago














11












11








11






Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.



To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.






share|cite|improve this answer














Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $lambda(NPcap B(x))=0$. In fact since $NPcap B(x)subset NP$, we just prove $lambda(NP)=0$ and we are done. Let $$NP_x={(x,y)|x+yin mathbb{Q}}$$
Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $lambda(NP)=0$.



To bring this back to the specific question you are asking, $NPcup P=mathbb{R}^2$, so for any open ball $B(x)$ $lambda(Pcap B(x))=1$.
On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $lambda(Rcap B(x))=1$. Hence you are taking the limit of $1/1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









Charlie FrohmanCharlie Frohman

1,468913




1,468913












  • Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
    – AlephNull
    yesterday








  • 1




    Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
    – Milo Brandt
    yesterday










  • I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
    – Stijn de Witt
    21 hours ago


















  • Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
    – AlephNull
    yesterday








  • 1




    Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
    – Milo Brandt
    yesterday










  • I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
    – Stijn de Witt
    21 hours ago
















Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday






Nice answer (I suppose you mean area $pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $overline{mathbb{Q}}$ is countable too.
– AlephNull
yesterday






1




1




Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday




Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero.
– Milo Brandt
yesterday












I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 hours ago




I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure)
– Stijn de Witt
21 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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3063918%2fis-the-sum-of-two-irrational-numbers-almost-always-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

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .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 ...
Read more

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .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 ...
Read more

Display a custom attribute below product name in the front-end Magento 1.9.3.8

Image
0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
Read more