Theorem: If $q \neq 0$ is rational and $y$ is irrational, then $qy$ is irrational.

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


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It"s wrong. You composed $\fracxzy = \fracab$. The is correct. Climate you said "Therefore $xy = a$. The is wrong.

You must solve $\fracxzy = \fracab$ because that $y$. You gain $y = \fracab \cdot \fraczx$.


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As I cite here frequently, this ubiquitous building is merely an instance of complementary watch of the subgroup property, i.e.

THEOREM $\ $ A nonempty subset $\rm\:S\:$ the abelian group $\rm\:G\:$ comprises a subgroup $\rm\iff\ S\ + \ \bar S\ =\ \bar S\ $ where $\rm\: \bar S\:$ is the complement of $\rm\:S\:$ in $\rm\:G$

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

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You can directly divide through $q$ assuming the truth that $q \neq 0$.

Suppose $qy$ is rational then, you have $qy = \fracmn$ for part $n \neq 0$. This says that $y = \fracmnq$ which claims that $\texty is rational$ contradiction.


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

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


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

First of all, a minor stroller, stick point. You wrote$$qy=\fracab \qquad\textwhere $a$ and $b$ are integers, v $b \ne 0$$$

So far, fine.Then come your $x$ and also $z$. For completeness, friend should have said "Let $x$, $z$ be integers such that $q=\fracxz$. Keep in mind that no $x$ no one $z$ is $0$." Basically, you did no say what link $x/z$ had with $q$, despite admittedly any reasonable person would know what you meant. By the way, I more than likely would have actually chosen the letters $c$ and also $d$ rather of $x$ and $z$.

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 $xb$ space integers.


I don"t think the correct. That seems prefer a great idea to indicate both x as an integer, and z as a non-zero integer. Then you also want to "solve for" y, which as Eric point out out, you didn"t fairly do.

See more: What Is The Square Root Of 864 In Simplest Radical Form? Square Root Of 864 Simplified


$$a\in\piersonforcongress.combbQ,b\in\piersonforcongress.combbR\setminus\piersonforcongress.combbQ,ab\in\piersonforcongress.combbQ\implies b\in\piersonforcongress.combbQ\implies\textContradiction\therefore ab\not\in\piersonforcongress.combbQ.$$


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

Q: does the multiplication the a and also b an outcome in a rational or irrational number?:

Proof:

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

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

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


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