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

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Proof: evidence by contradiction, we assume the $qy$ is rational. Thus $qy=fracab$ because that integers $a$, $b eq 0$. Since $q$ is rational, we have actually $fracxzy=fracab$ because that integers $x eq 0$, $z eq 0$. Therefore, $xy = a$, and $y=fracax$. Since both $a$ and also $x$ are integers, $y$ is rational, causing a contradiction.


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As I point out here frequently, this ubiquitous residential property is merely an circumstances of complementary watch of the subgroup property, i.e.

THEOREM $ $ A nonempty subset $ m:S:$ that abelian team $ m:G:$ comprises a subgroup $ miff S + ar S = ar S $ wherein $ m: ar S:$ is the match of $ m:S:$ in $ m:G$

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

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You can directly divide by $q$ presume the fact that $q eq 0$.

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


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

In our case, we have actually the team $(BbbR^*,cdot)$ and also its appropriate subgroup $(BbbQ^*,cdot)$. By the arguments over $q in BbbQ^*$ and $y in BbbRsetminus BbbQ$ implies $qy in BbbRsetminus BbbQ$.


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It"s wrong. You wrote $fracxzy = fracab$. That is correct. Climate you said "Therefore $xy = a$. That is wrong.

You should solve $fracxzy = fracab$ for $y$. You acquire $y = fracab cdot fraczx$.


Let"s see exactly how we deserve to modify your dispute to make it perfect.

First the all, a minor stroller, stick point. Friend wrote$$qy=fracab qquad extwhere $a$ and $b$ room integers, with $b e 0$$$

So far, fine.Then come your $x$ and $z$. For completeness, girlfriend should have actually said "Let $x$, $z$ it is in integers such the $q=fracxz$. Keep in mind that no $x$ nor $z$ is $0$." Basically, girlfriend did not say what connection $x/z$ had actually with $q$, though admittedly any reasonable person would know what friend meant. By the way, I most likely would have actually chosen the letters $c$ and also $d$ rather of $x$ and $z$.

Now because that the non-picky point. You reached$$fracxzy=fracab$$From the you should have actually concluded straight that$$y=fraczaxb$$which end things, due to the fact that $za$ and also $xb$ space integers.


I don"t think the correct. The seems favor a great idea to suggest both x as an integer, and z as a non-zero integer. Then you also want come "solve for" y, which together Eric points out, you didn"t quite do.

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$$ainstclairdrake.netbbQ,binstclairdrake.netbbRsetminusstclairdrake.netbbQ,abinstclairdrake.netbbQimplies binstclairdrake.netbbQimplies extContradiction herefore ab otinstclairdrake.netbbQ.$$


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

Q: does the multiplication of a and b result in a rational or irrational number?:

Proof:

because b is rational: b = u/j where u and j room integers

Assume ab is rational:ab = k/n, where k and n space integers.a = k/bna = k/(n(u/j))a = jk/un

before we asserted a as irrational, yet now that is rational; a contradiction. Therefore abdominal muscle must it is in irrational.


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