Contrapositive proof: If $H$ and $K$ are nontrivial subgroups of $Bbb Q$, then $Hcap K$ is also nontrivial.












-1












$begingroup$


I'm reading "Contemporary Abstract Algebra," by Gallian.



This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.



I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.




$mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $Bbb Q$ under addition. Show that $H cap K$ is a nontrivial subgroup of $Bbb Q$.




The contrapositive problem can be stated as follows:




$mathcal{C}$: Let $H$ and $K$ be subgroups of $Bbb Q$. Suppose $Hcap K={0}$ is trivial. Show that either $H$ or $K$ is trivial.





Thoughts:



To me, it seems $mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.



My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $Hcap K$ is nontrivial, contradicting the hypothesis that $Hcap K={0}$. That's just silly.




A lovely proof of the original statement $mathcal{O}$ is given here, with variations on its theme here and here.



Please help :)






group-theory alternative-proof rational-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
    $endgroup$
    – Shaun
    Jan 24 at 11:44


















-1












$begingroup$


I'm reading "Contemporary Abstract Algebra," by Gallian.



This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.



I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.




$mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $Bbb Q$ under addition. Show that $H cap K$ is a nontrivial subgroup of $Bbb Q$.




The contrapositive problem can be stated as follows:




$mathcal{C}$: Let $H$ and $K$ be subgroups of $Bbb Q$. Suppose $Hcap K={0}$ is trivial. Show that either $H$ or $K$ is trivial.





Thoughts:



To me, it seems $mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.



My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $Hcap K$ is nontrivial, contradicting the hypothesis that $Hcap K={0}$. That's just silly.




A lovely proof of the original statement $mathcal{O}$ is given here, with variations on its theme here and here.



Please help :)






group-theory alternative-proof rational-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
    $endgroup$
    – Shaun
    Jan 24 at 11:44
















-1












-1








-1





$begingroup$


I'm reading "Contemporary Abstract Algebra," by Gallian.



This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.



I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.




$mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $Bbb Q$ under addition. Show that $H cap K$ is a nontrivial subgroup of $Bbb Q$.




The contrapositive problem can be stated as follows:




$mathcal{C}$: Let $H$ and $K$ be subgroups of $Bbb Q$. Suppose $Hcap K={0}$ is trivial. Show that either $H$ or $K$ is trivial.





Thoughts:



To me, it seems $mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.



My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $Hcap K$ is nontrivial, contradicting the hypothesis that $Hcap K={0}$. That's just silly.




A lovely proof of the original statement $mathcal{O}$ is given here, with variations on its theme here and here.



Please help :)






group-theory alternative-proof rational-numbers







share|cite|improve this question









$endgroup$




I'm reading "Contemporary Abstract Algebra," by Gallian.



This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.



I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.




$mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $Bbb Q$ under addition. Show that $H cap K$ is a nontrivial subgroup of $Bbb Q$.




The contrapositive problem can be stated as follows:




$mathcal{C}$: Let $H$ and $K$ be subgroups of $Bbb Q$. Suppose $Hcap K={0}$ is trivial. Show that either $H$ or $K$ is trivial.





Thoughts:



To me, it seems $mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.



My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $Hcap K$ is nontrivial, contradicting the hypothesis that $Hcap K={0}$. That's just silly.




A lovely proof of the original statement $mathcal{O}$ is given here, with variations on its theme here and here.



Please help :)






group-theory alternative-proof rational-numbers




group-theory alternative-proof rational-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 11:37









ShaunShaun

9,380113684




9,380113684












  • $begingroup$
    This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
    $endgroup$
    – Shaun
    Jan 24 at 11:44




















  • $begingroup$
    This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
    $endgroup$
    – Shaun
    Jan 24 at 11:44


















$begingroup$
This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
$endgroup$
– Shaun
Jan 24 at 11:44






$begingroup$
This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
$endgroup$
– Shaun
Jan 24 at 11:44












2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.



1: If all $G_i$ have the property $P$, then so does



$$lim_{longrightarrow}G_i .$$



2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,



begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}



Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$



3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is beautiful. Thank you! :)
    $endgroup$
    – Shaun
    Jan 25 at 10:21



















4












$begingroup$

This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$



It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
    $endgroup$
    – Shaun
    Jan 24 at 15:55










  • $begingroup$
    The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:09










  • $begingroup$
    I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
    $endgroup$
    – Shaun
    Jan 24 at 16:17










  • $begingroup$
    Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:27








  • 1




    $begingroup$
    OK, I stand corrected
    $endgroup$
    – Reiner Martin
    Jan 25 at 10:58











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%2f3085770%2fcontrapositive-proof-if-h-and-k-are-nontrivial-subgroups-of-bbb-q-then%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.



1: If all $G_i$ have the property $P$, then so does



$$lim_{longrightarrow}G_i .$$



2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,



begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}



Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$



3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is beautiful. Thank you! :)
    $endgroup$
    – Shaun
    Jan 25 at 10:21
















1












$begingroup$

Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.



1: If all $G_i$ have the property $P$, then so does



$$lim_{longrightarrow}G_i .$$



2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,



begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}



Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$



3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is beautiful. Thank you! :)
    $endgroup$
    – Shaun
    Jan 25 at 10:21














1












1








1





$begingroup$

Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.



1: If all $G_i$ have the property $P$, then so does



$$lim_{longrightarrow}G_i .$$



2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,



begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}



Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$



3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.






share|cite|improve this answer











$endgroup$



Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.



1: If all $G_i$ have the property $P$, then so does



$$lim_{longrightarrow}G_i .$$



2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,



begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}



Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$



3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 25 at 10:26









Shaun

9,380113684




9,380113684










answered Jan 25 at 10:09









Hiro WatHiro Wat

1238




1238












  • $begingroup$
    This is beautiful. Thank you! :)
    $endgroup$
    – Shaun
    Jan 25 at 10:21


















  • $begingroup$
    This is beautiful. Thank you! :)
    $endgroup$
    – Shaun
    Jan 25 at 10:21
















$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21




$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21











4












$begingroup$

This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$



It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
    $endgroup$
    – Shaun
    Jan 24 at 15:55










  • $begingroup$
    The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:09










  • $begingroup$
    I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
    $endgroup$
    – Shaun
    Jan 24 at 16:17










  • $begingroup$
    Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:27








  • 1




    $begingroup$
    OK, I stand corrected
    $endgroup$
    – Reiner Martin
    Jan 25 at 10:58
















4












$begingroup$

This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$



It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
    $endgroup$
    – Shaun
    Jan 24 at 15:55










  • $begingroup$
    The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:09










  • $begingroup$
    I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
    $endgroup$
    – Shaun
    Jan 24 at 16:17










  • $begingroup$
    Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:27








  • 1




    $begingroup$
    OK, I stand corrected
    $endgroup$
    – Reiner Martin
    Jan 25 at 10:58














4












4








4





$begingroup$

This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$



It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.






share|cite|improve this answer









$endgroup$



This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$



It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 12:08









Reiner MartinReiner Martin

3,509414




3,509414












  • $begingroup$
    This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
    $endgroup$
    – Shaun
    Jan 24 at 15:55










  • $begingroup$
    The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:09










  • $begingroup$
    I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
    $endgroup$
    – Shaun
    Jan 24 at 16:17










  • $begingroup$
    Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:27








  • 1




    $begingroup$
    OK, I stand corrected
    $endgroup$
    – Reiner Martin
    Jan 25 at 10:58


















  • $begingroup$
    This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
    $endgroup$
    – Shaun
    Jan 24 at 15:55










  • $begingroup$
    The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:09










  • $begingroup$
    I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
    $endgroup$
    – Shaun
    Jan 24 at 16:17










  • $begingroup$
    Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
    $endgroup$
    – Reiner Martin
    Jan 24 at 16:27








  • 1




    $begingroup$
    OK, I stand corrected
    $endgroup$
    – Reiner Martin
    Jan 25 at 10:58
















$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55




$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55












$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09




$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09












$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17




$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17












$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27






$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27






1




1




$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58




$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58


















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%2f3085770%2fcontrapositive-proof-if-h-and-k-are-nontrivial-subgroups-of-bbb-q-then%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

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

The Binding of Isaac: Rebirth/Afterbirth

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
Read more