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?:

Proof:

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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