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.

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 $y = \fracab \cdot \fraczx$.

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.

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. Highly energetic question. Knife 10 call (not counting the combination bonus) in order to answer this question. The reputation necessity helps protect this inquiry from spam and non-answer activity. ## Not the prize you're looking for? Browse other questions tagged evaluation irrational-numbers or questioning your own question. Is over there a basic proof because that$\small 2\fracn3$is not an integer when$\fracn3$is no an integer? If$ab \equiv r \pmodp$, and$x^2 \equiv a \pmodp$climate$y^2 \equiv b \pmodp$for which problem of$r\$?
offered a rational number and also an irrational number, both higher than 0, prove that the product in between them is irrational.
Please aid me point out the error in my "proof" that the sum of 2 irrational numbers have to be irrational
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