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

You are watching: Product of rational and irrational number

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.



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.




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 $qy in BbbRsetminus BbbQ$.


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?:


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.

Highly energetic question. Earn 10 call (not counting the association bonus) in order to answer this question. The reputation requirement helps defend this inquiry from spam and non-answer activity.

Not the answer you're feather for? Browse various other questions tagged evaluation irrational-numbers or questioning your very own question.

Is there a basic proof for $small 2fracn3$ is no an integer when $fracn3$ is no an integer?
If $ab equiv r pmodp$, and also $x^2 equiv a pmodp$ climate $y^2 equiv b pmodp$ for which condition of $r$?
provided a rational number and an irrational number, both greater than 0, prove that the product between them is irrational.
Please help me spot the error in my "proof" the the sum of 2 irrational numbers should be irrational
site design / logo design © 2021 stack Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.11.16.40760

her privacy

By click “Accept every cookies”, friend agree stack Exchange have the right to store cookie on your maker and disclose information in accordance with our Cookie Policy.