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.

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.

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\$.

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.

Highly active question. Knife 10 call (not count the association bonus) in order come answer this question. The reputation necessity helps safeguard this inquiry from spam and also non-answer activity.

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