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

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

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


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As I point out here frequently, this ubiquitous building is just an instance of complementary view 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 complement of $ m:S:$ in $ m:G$

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

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

Suppose $qy$ is reasonable then, you have $qy = fracmn$ for part $n eq 0$. This states that $y = fracmnq$ which states 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 then $h in H$ and $y in Gsetminus H$ suggests that $hy in Gsetminus H$. Proof: mean $hy in H$. You understand that $h^-1 in H$, and also therefore $y=h^-1(hy) in H$. Contradiction.

In ours case, we have the team $(BbbR^*,cdot)$ and also its ideal subgroup $(BbbQ^*,cdot)$. Through the arguments above $q in BbbQ^*$ and also $y in BbbRsetminus BbbQ$ implies $qy in BbbRsetminus BbbQ$.


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

You should solve $fracxzy = fracab$ because that $y$. You acquire $y = fracab cdot fraczx$.


Let"s see exactly how we deserve to modify your argument to make it perfect.

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

So far, fine.Then come your $x$ and also $z$. Because that completeness, you should have said "Let $x$, $z$ it is in integers such that $q=fracxz$. Keep in mind that no $x$ nor $z$ is $0$." Basically, friend did no say what connection $x/z$ had actually with $q$, despite admittedly any kind of reasonable person would understand what you meant. Through the way, I more than likely would have actually chosen the letter $c$ and $d$ instead of $x$ and $z$.

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


I don"t think the correct. It seems favor a an excellent idea to show both x together an integer, and z together a non-zero integer. Climate you additionally want come "solve for" y, which as Eric clues out, friend didn"t rather do.

See more: 2.8 Repeating As A Simplified Fraction Calculator, How Do You Convert 0


$$aingimpppa.orgbbQ,bingimpppa.orgbbRsetminusgimpppa.orgbbQ,abingimpppa.orgbbQimplies bingimpppa.orgbbQimplies extContradiction herefore ab otingimpppa.orgbbQ.$$


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

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

Proof:

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

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

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


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