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

You are watching: The product of a rational and irrational number is rational.

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



As I cite here frequently, this ubiquitous residential or commercial property is merely an instance of complementary see of the subgroup property, i.e.

THEOREM $ $ A nonempty subset $ m:S:$ the 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 of this room ubiquitous in concrete number systems, e.g.




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

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

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

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


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

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

Let"s see just how we deserve to modify your discussion to do it perfect.

First of all, a minor stroller, stick point. Girlfriend wrote$$qy=fracab qquad extwhere $a$ and also $b$ room integers, through $b e 0$$$

So far, fine.Then come your $x$ and $z$. Because that completeness, you should have said "Let $x$, $z$ be integers such the $q=fracxz$. Note that neither $x$ nor $z$ is $0$." Basically, friend did no say what connection $x/z$ had with $q$, though admittedly any kind of reasonable human would know what you meant. Through the way, I probably would have chosen the letters $c$ and $d$ instead of $x$ and $z$.

Now for the non-picky point. You reached$$fracxzy=fracab$$From that you should have concluded straight that$$y=fraczaxb$$which ends things, because $za$ and $xb$ space integers.

I don"t think that correct. That seems favor a an excellent idea to show both x as an integer, and also z as a non-zero integer. Then you likewise want to "solve for" y, which as Eric clues out, you didn"t fairly do.

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$$ainptcouncil.netbbQ,binptcouncil.netbbRsetminusptcouncil.netbbQ,abinptcouncil.netbbQimplies binptcouncil.netbbQimplies extContradiction herefore ab otinptcouncil.netbbQ.$$

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

Q: walk the multiplication that a and b an outcome in a rational or irrational number?:


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

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

before we asserted a as irrational, but now that is rational; a contradiction. Therefore ab must it is in irrational.

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