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

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Proof: evidence by contradiction, we assume the \$qy\$ is rational. Thus \$qy=fracab\$ because that integers \$a\$, \$b eq 0\$. Since \$q\$ is rational, we have actually \$fracxzy=fracab\$ because that integers \$x eq 0\$, \$z eq 0\$. Therefore, \$xy = a\$, and \$y=fracax\$. Since both \$a\$ and also \$x\$ are integers, \$y\$ is rational, causing a contradiction.  As I point out here frequently, this ubiquitous residential property is merely an circumstances of complementary watch of the subgroup property, i.e.

THEOREM \$ \$ A nonempty subset \$ m:S:\$ that abelian team \$ m:G:\$ comprises a subgroup \$ miff S + ar S = ar S \$ wherein \$ m: ar S:\$ is the match of \$ m:S:\$ in \$ m:G\$

Instances of this are ubiquitous in concrete number systems, e.g.   You can directly divide by \$q\$ presume the fact that \$q eq 0\$.

Suppose \$qy\$ is reasonable then, you have actually \$qy = fracmn\$ for part \$n eq 0\$. This claims that \$y = fracmnq\$ which states that \$ exty is rational\$ contradiction.

A team theoretic proof: You recognize that if \$G\$ is a group and also \$H eq G\$ is just one of its subgroups climate \$h in H\$ and also \$y in Gsetminus H\$ suggests that \$hy in Gsetminus H\$. Proof: suppose \$hy in H\$. You know 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 also its appropriate subgroup \$(BbbQ^*,cdot)\$. By the arguments over \$q in BbbQ^*\$ and \$y in BbbRsetminus BbbQ\$ implies \$qy in BbbRsetminus BbbQ\$. It"s wrong. You wrote \$fracxzy = fracab\$. That is correct. Climate you said "Therefore \$xy = a\$. That is wrong.

You should solve \$fracxzy = fracab\$ for \$y\$. You acquire \$y = fracab cdot fraczx\$.

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

First the all, a minor stroller, stick point. Friend wrote\$\$qy=fracab qquad extwhere \$a\$ and \$b\$ room integers, with \$b e 0\$\$\$

So far, fine.Then come your \$x\$ and \$z\$. For completeness, girlfriend should have actually said "Let \$x\$, \$z\$ it is in integers such the \$q=fracxz\$. Keep in mind that no \$x\$ nor \$z\$ is \$0\$." Basically, girlfriend did not say what connection \$x/z\$ had actually with \$q\$, though admittedly any reasonable person would know what friend meant. By the way, I most likely would have actually chosen the letters \$c\$ and also \$d\$ rather of \$x\$ and \$z\$.

Now because that the non-picky point. You reached\$\$fracxzy=fracab\$\$From the you should have actually concluded straight that\$\$y=fraczaxb\$\$which end things, due to the fact that \$za\$ and also \$xb\$ space integers.

I don"t think the correct. The seems favor a great idea to suggest both x as an integer, and z as a non-zero integer. Then you also want come "solve for" y, which together Eric points out, you didn"t quite do.

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\$\$ainstclairdrake.netbbQ,binstclairdrake.netbbRsetminusstclairdrake.netbbQ,abinstclairdrake.netbbQimplies binstclairdrake.netbbQimplies extContradiction herefore ab otinstclairdrake.netbbQ.\$\$

a is irrational, conversely, 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 where u and j room integers

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

before we asserted a as irrational, yet now that is rational; a contradiction. Therefore abdominal muscle must it is in irrational.

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