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?:

Proof:

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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