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How to prove that $\forall [(a,b)]\in\mathbb{Q}$ with $a\neq0$ there must exist a $[\gamma]\in\mathbb{Q}$ such that $[\gamma]=[(\gamma_{1},\gamma_{2})]$ where : $$ [(a,b)]\cdot[\gamma]=1 $$
In what is known as the multiplicative inverse of $\mathbb{Q}$. My attempted proof is as follows : $$ $$ Attempted Proof : We have that $[(a,b)]\cdot[\gamma]\implies[(a,b)]\cdot[(\gamma_{1},\gamma_{2})]=1\implies [(a\gamma_{1},b\gamma_{2})]=[(1,1)]\implies a\gamma_{1}=1$ and $b\gamma_{2}=1$ which means $a\gamma_{1}=b\gamma_{2}$. $$ $$ I am stuck here and I am unable to proceed. I am told to consider two cases for $a$ that are $a>0$ and $a<0$

  • The statement "for all $[(a,b)] \in \mathbb Q$ there exists a $[\gamma] \in \mathbb Q$ such that $[(a,b)][\gamma] = 1$" is false: for $[(0,1)]$ we have that $[(0,1)][\gamma] = 0$ for all $[\gamma] \in \mathbb Q$. – azif00 Sep 24 '20 at 23:00
  • Thank you for letting me know I have fixed it –  Sep 24 '20 at 23:02
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    What if you look at $[(b,a)]$? Remember, the basic idea is that $[(a,b)]$ should turn out to behave like $\frac{a}b$. – Brian M. Scott Sep 24 '20 at 23:05
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    Hi Brain, I do notice that $[(a,b)]\cdot[(b,a)]$ should yield to $\frac{ab}{ba}=1$. However, I was thinking of reaching a conclusion in my proof like $\gamma_{1}=b$ and $\gamma_{2}=a$. Should I just directly compute $[(a,b)]\cdot[(b,a)]$ and prove that this operation is unique? @BrianM.Scott –  Sep 24 '20 at 23:12
  • @WiWo: I would: that’s the easiest way, once you hit upon the idea of using $[(b,a)]$. Proving that it’s unique is then just a matter of showing that if $[(\gamma_1,\gamma_2)]$ also works, then it’s equal to $[(a,b)]$. – Brian M. Scott Sep 24 '20 at 23:29
  • I unfortunately could not work this out. I have failed to see the main idea out of it. –  Sep 25 '20 at 00:08
  • The second to last line of your attempted proof is wrong; $[(a\gamma_1, b\gamma_2)]=[(1, 1)]$ does not imply $a\gamma_1=1$ or $b\gamma_2=1$. Recall what it means for two pairs to be equivalent in this construction... $(a, b)\sim (c, d)$ iff $ad=bc$, and in particular this condition does not require $a=c$ or $b=d$. – Atticus Stonestrom Sep 25 '20 at 00:31
  • I see, it was a stupid mistake. Now say I have that $a\gamma_{1}=b\gamma_{2}$ and as Brian implied $ab=ba$. How could I be able to show that $a\gamma_{1}=ab$ and $b\gamma_{2}=ab$? @AtticusStonestrom –  Sep 25 '20 at 00:34
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    Please give much more context, e.g. have you already proved that the fraction multiplication (and addition) operations are well-defined. – Bill Dubuque Sep 25 '20 at 00:37
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    Those equalities do not have to hold; in fact there are infinitely many valid choices for $\gamma_1$ and $\gamma_2$: for any non-zero natural number $n$, the assignment $\gamma_1=nb$ and $\gamma_2=na$ will work. However, you are asked only to show the existence of an inverse, so you need merely to give a single example of a pair $\gamma_1, \gamma_2$ that works – Atticus Stonestrom Sep 25 '20 at 00:37
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    (Bonus: can you prove that the choices $\gamma_1=nb$ and $\gamma_2=na$ for some non-zero natural number $n$ are the only possible choices? This will be enough to show that the inverse of $[(a, b)]$ is unique, since all of these choices lie in the same equivalence class) – Atticus Stonestrom Sep 25 '20 at 00:43
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    I believe you have answered my question. Thank you very much and thank you Brian. Dear Bill, I have defined everything and I am on my way to complete the construction of $\mathbb{Q}$ –  Sep 25 '20 at 00:45
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    I highly recommend that you peruse this and this question to be sure you are aware of all that needs to be done to obtain a complete, rigorous proof. It is more subtle than it may appear at first glance. It is impossible to infer from what little you wrote if you are on the correct path. It would also help if you state what your source is, e.g. cite the textbook you are using. – Bill Dubuque Sep 25 '20 at 00:47

1 Answers1

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(1)...$[(u,v)]=[(u',v')] \iff uv'=u'v.$

(2)... In particular with $u'=v'=1$ we have $[(u,v)]=[(1,1)] \iff u=v.$

(3)...$[(u,v)]=[(u',v')]$ does NOT imply $(u=u'\land v=v').$ E.g. $[(1,2)]=[(2,4)].$ In particular $[(a\gamma_1,b\gamma_2)]=[(1,1)]$ does not imply $a\gamma_1=1=b\gamma_2.$

(4)...If $a\ne 0$ and $b>0$ then let $\gamma_2=|a|$ and let $\gamma_1=b\cdot (|a|/a)\in \{b,-b\}.$ We have $\gamma_2>0$ and $b\gamma_2>0.$

And $a\gamma_1=b|a|=b\gamma_2$ so by (2) with $u=a\gamma_1$ and $v=b\gamma_2$ we have $[(a\gamma_1,b\gamma_2)]=[(1,1)].$

DanielWainfleet
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