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

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Proof: proof by contradiction, we assume that $qy$ is rational. As such $qy=fracab$ because that integers $a$, $b eq 0$. Since $q$ is rational, we have $fracxzy=fracab$ because that integers $x eq 0$, $z eq 0$. Therefore, $xy = a$, and $y=fracax$. Because both $a$ and also $x$ room integers, $y$ is rational, resulting in a contradiction.


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

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

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

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

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


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

In our case, we have actually the group $(BbbR^*,cdot)$ and its ideal subgroup $(BbbQ^*,cdot)$. By the arguments above $q in BbbQ^*$ and also $y in BbbRsetminus BbbQ$ suggests $qy in BbbRsetminus BbbQ$.


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

You have to solve $fracxzy = fracab$ for $y$. You gain $y = fracab cdot fraczx$.


Let"s see just how we have the right to modify your argument to make it perfect.

First of all, a minor picky point. Friend wrote$$qy=fracab qquad extwhere $a$ and also $b$ room integers, v $b e 0$$$

So far, fine.Then come her $x$ and $z$. For completeness, friend should have said "Let $x$, $z$ it is in integers such the $q=fracxz$. Note that neither $x$ no one $z$ is $0$." Basically, you did not say what connection $x/z$ had with $q$, despite admittedly any reasonable human being would know what girlfriend meant. By the way, I more than likely would have chosen the letter $c$ and $d$ rather of $x$ and $z$.

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


I don"t think that correct. It seems prefer a great idea to indicate both x together an integer, and also z as a non-zero integer. Climate you also want to "solve for" y, which as Eric points out, girlfriend didn"t rather do.

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$$aincg-tower.combbQ,bincg-tower.combbRsetminuscg-tower.combbQ,abincg-tower.combbQimplies bincg-tower.combbQimplies extContradiction herefore ab otincg-tower.combbQ.$$


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

Q: go the multiplication of a and also b result in a reasonable or irrational number?:

Proof:

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

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

before we declared a together irrational, but now that is rational; a contradiction. Therefore abdominal muscle must be irrational.


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