Theorem: If $q eq 0$ is rational and also $y$ is irrational, then $qy$ is irrational.

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Proof: proof by contradiction, we assume the $qy$ is rational. Therefore $qy=fracab$ for integers $a$, $b eq 0$. Because $q$ is rational, we have actually $fracxzy=fracab$ because that integers $x eq 0$, $z eq 0$. Therefore, $xy = a$, and $y=fracax$. Due to the fact that both $a$ and also $x$ space integers, $y$ is rational, leading to a contradiction.


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As I point out here frequently, this ubiquitous residential property is simply an instance of complementary see of the subgroup property, i.e.

THEOREM $ $ A nonempty subset $ m:S:$ the abelian group $ m:G:$ comprises a subgroup $ miff S + ar S = ar S $ whereby $ m: ar S:$ is the enhance of $ m:S:$ in $ m:G$

Instances the this are ubiquitous in concrete number systems, e.g.

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You can directly divide by $q$ suspect the reality that $q eq 0$.

Suppose $qy$ is rational then, you have $qy = fracmn$ for part $n eq 0$. This says that $y = fracmnq$ which says that $ exty is rational$ contradiction.


A team theoretic proof: You understand that if $G$ is a group and $H eq G$ is one of its subgroups then $h in H$ and also $y in Gsetminus H$ implies that $hy in Gsetminus H$. Proof: intend $hy in H$. You know that $h^-1 in H$, and also therefore $y=h^-1(hy) in H$. Contradiction.

In ours case, we have actually the group $(BbbR^*,cdot)$ and its proper subgroup $(BbbQ^*,cdot)$. By the arguments over $q in BbbQ^*$ and also $y in BbbRsetminus BbbQ$ indicates $qy in BbbRsetminus BbbQ$.


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It"s wrong. You created $fracxzy = fracab$. That is correct. Climate you claimed "Therefore $xy = a$. The is wrong.

You need to solve $fracxzy = fracab$ for $y$. You get $y = fracab cdot fraczx$.


Let"s see how we have the right to modify your discussion to do it perfect.

First the all, a minor picky point. Girlfriend wrote$$qy=fracab qquad extwhere $a$ and also $b$ space integers, v $b e 0$$$

So far, fine.Then come her $x$ and also $z$. For completeness, friend should have actually said "Let $x$, $z$ be integers such the $q=fracxz$. Note that neither $x$ no one $z$ is $0$." Basically, friend did no say what link $x/z$ had actually with $q$, despite admittedly any kind of reasonable human would recognize what you meant. Through the way, I probably would have chosen the letter $c$ and $d$ rather of $x$ and also $z$.

Now for the non-picky point. Girlfriend reached$$fracxzy=fracab$$From that you should have actually concluded straight that$$y=fraczaxb$$which ends things, because $za$ and also $xb$ are integers.


I don"t think it correct. That seems like a great idea to suggest both x as an integer, and also z together a non-zero integer. Then you additionally want come "solve for" y, which together Eric clues out, friend didn"t fairly do.

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$$ainaltoalsimce.orgbbQ,binaltoalsimce.orgbbRsetminusaltoalsimce.orgbbQ,abinaltoalsimce.orgbbQimplies binaltoalsimce.orgbbQimplies extContradiction herefore ab otinaltoalsimce.orgbbQ.$$


a is irrational, conversely, b is rational.(both > 0)

Q: walk the multiplication the a and also b an outcome in a reasonable or irrational number?:

Proof:

because b is rational: b = u/j wherein u and j space integers

Assume abdominal muscle is rational:ab = k/n, whereby k and n are integers.a = k/bna = k/(n(u/j))a = jk/un

before we claimed a as irrational, however now it is rational; a contradiction. Therefore abdominal must be irrational.


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