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

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Proof: evidence by contradiction, us assume that \$qy\$ is rational. Thus \$qy=fracab\$ because that integers \$a\$, \$b eq 0\$. Since \$q\$ is rational, we have \$fracxzy=fracab\$ for integers \$x eq 0\$, \$z eq 0\$. Therefore, \$xy = a\$, and \$y=fracax\$. Due to the fact that both \$a\$ and also \$x\$ are integers, \$y\$ is rational, bring about a contradiction.  As I mention here frequently, this ubiquitous home is merely an instance of complementary watch of the subgroup property, i.e.

THEOREM \$ \$ A nonempty subset \$ m:S:\$ of abelian group \$ 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 space ubiquitous in concrete number systems, e.g.   You can straight divide through \$q\$ suspect the reality that \$q eq 0\$.

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

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

You must solve \$fracxzy = fracab\$ for \$y\$. You obtain \$y = fracab cdot fraczx\$.

Let"s see exactly how we have the right to modify your argument to make it perfect.

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

So far, fine.Then come your \$x\$ and also \$z\$. Because that completeness, friend should have actually said "Let \$x\$, \$z\$ it is in integers such that \$q=fracxz\$. Note that no \$x\$ no one \$z\$ is \$0\$." Basically, friend did not say what connection \$x/z\$ had with \$q\$, though admittedly any kind of reasonable human would know what girlfriend meant. Through the way, I probably would have chosen the letters \$c\$ and also \$d\$ instead of \$x\$ and also \$z\$.

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

I don"t think that correct. It seems choose a good idea to suggest both x as an integer, and z together a non-zero integer. Then you additionally want to "solve for" y, which as Eric point out out, friend didn"t fairly do.

See more: The Rule Governing Exponents Is That An Exponent In The Denominator Is The Same As A

\$\$ainchathamtownfc.netbbQ,binchathamtownfc.netbbRsetminuschathamtownfc.netbbQ,abinchathamtownfc.netbbQimplies binchathamtownfc.netbbQimplies extContradiction herefore ab otinchathamtownfc.netbbQ.\$\$

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

Q: walk the multiplication of a and also b result in a reasonable or irrational number?:

Proof:

because b is rational: b = u/j wherein u and j space integers

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

before we claimed a together irrational, however now the is rational; a contradiction. Therefore abdominal muscle must it is in irrational.

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