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

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


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

THEOREM $ $ A nonempty subset $ m:S:$ of abelian group $ 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 of this space ubiquitous in concrete number systems, e.g.

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You can straight divide through $q$ suspect the reality that $q eq 0$.

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


A group theoretic proof: You recognize that if $G$ is a group and $H eq G$ is one of its subgroups climate $h in H$ and $y in Gsetminus H$ implies that $hy in Gsetminus H$. Proof: suppose $hy in H$. You understand 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 its appropriate subgroup $(BbbQ^*,cdot)$. Through the arguments above $q in BbbQ^*$ and $y in BbbRsetminus BbbQ$ implies $qy in BbbRsetminus BbbQ$.


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

You must solve $fracxzy = fracab$ for $y$. You obtain $y = fracab cdot fraczx$.


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

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

So far, fine.Then come your $x$ and also $z$. Because that completeness, friend should have actually said "Let $x$, $z$ it is in integers such that $q=fracxz$. Note that no $x$ no one $z$ is $0$." Basically, friend did not say what connection $x/z$ had with $q$, though admittedly any kind of reasonable human would know what girlfriend meant. Through the way, I probably would have chosen the letters $c$ and also $d$ instead of $x$ and also $z$.

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


I don"t think that correct. It seems choose a good idea to suggest both x as an integer, and z together a non-zero integer. Then you additionally want to "solve for" y, which as Eric point out out, friend didn"t fairly do.

See more: The Rule Governing Exponents Is That An Exponent In The Denominator Is The Same As A


$$ainchathamtownfc.netbbQ,binchathamtownfc.netbbRsetminuschathamtownfc.netbbQ,abinchathamtownfc.netbbQimplies binchathamtownfc.netbbQimplies extContradiction herefore ab otinchathamtownfc.netbbQ.$$


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

Q: walk 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 j space integers

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

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


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