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

You are watching: Product of rational and irrational numbers

Proof: evidence by contradiction, we assume the $qy$ 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 $y=fracax$. Because both $a$ and also $x$ space integers, $y$ is rational, resulting in a contradiction.



As I cite here frequently, this ubiquitous property is simply an instance of complementary view of the subgroup property, i.e.

THEOREM $ $ A nonempty subset $ m:S:$ that abelian group $ m:G:$ comprises a subgroup $ miff S + ar S = ar S $ wherein $ m: ar S:$ is the complement of $ m:S:$ in $ m:G$

Instances that this are ubiquitous in concrete number systems, e.g.




You can straight divide by $q$ presume the truth that $q eq 0$.

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

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

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


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

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

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

First of all, a minor picky 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 also $z$. For completeness, girlfriend should have said "Let $x$, $z$ be integers such the $q=fracxz$. Keep in mind that no $x$ no one $z$ is $0$." Basically, friend did no say what link $x/z$ had with $q$, though admittedly any kind of reasonable person would recognize what you meant. By the way, I more than likely would have actually chosen the letters $c$ and $d$ rather of $x$ and $z$.

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

I don"t think that correct. The seems like a good idea to show both x together an integer, and z together a non-zero integer. Climate you additionally want to "solve for" y, which as Eric point out out, friend didn"t rather do.

See more: Which Of The Following Occurs During The Light-Dependent Reactions Of Plants?

$$aincharline-picon.combbQ,bincharline-picon.combbRsetminuscharline-picon.combbQ,abincharline-picon.combbQimplies bincharline-picon.combbQimplies extContradiction herefore ab otincharline-picon.combbQ.$$

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

Q: does the multiplication of a and b result in a rational or irrational number?:


because b is rational: b = u/j whereby u and also j are integers

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

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

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