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

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Proof: proof by contradiction, we assume the \$qy\$ is rational. Therefore \$qy=fracab\$ for integers \$a\$, \$b eq 0\$. Because \$q\$ is rational, we have actually \$fracxzy=fracab\$ because that integers \$x eq 0\$, \$z eq 0\$. Therefore, \$xy = a\$, and \$y=fracax\$. Due to the fact that both \$a\$ and also \$x\$ space integers, \$y\$ is rational, leading to a contradiction.  As I point out here frequently, this ubiquitous residential property is simply an instance of complementary see of the subgroup property, i.e.

THEOREM \$ \$ A nonempty subset \$ m:S:\$ the abelian group \$ m:G:\$ comprises a subgroup \$ miff S + ar S = ar S \$ whereby \$ m: ar S:\$ is the enhance of \$ m:S:\$ in \$ m:G\$

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

Suppose \$qy\$ is rational then, you have \$qy = fracmn\$ for part \$n eq 0\$. This says 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 know that \$h^-1 in H\$, and also therefore \$y=h^-1(hy) in H\$. Contradiction.

In ours case, we have actually the group \$(BbbR^*,cdot)\$ and its proper subgroup \$(BbbQ^*,cdot)\$. By the arguments over \$q in BbbQ^*\$ and also \$y in BbbRsetminus BbbQ\$ indicates \$qy in BbbRsetminus BbbQ\$. It"s wrong. You created \$fracxzy = fracab\$. That is correct. Climate you claimed "Therefore \$xy = a\$. The is wrong.

You need to solve \$fracxzy = fracab\$ for \$y\$. You get \$y = fracab cdot fraczx\$.

Let"s see how we have the right to modify your discussion to do it perfect.

First the all, a minor picky point. Girlfriend wrote\$\$qy=fracab qquad extwhere \$a\$ and also \$b\$ space integers, v \$b e 0\$\$\$

So far, fine.Then come her \$x\$ and also \$z\$. For completeness, friend should have actually said "Let \$x\$, \$z\$ be integers such the \$q=fracxz\$. Note that neither \$x\$ no one \$z\$ is \$0\$." Basically, friend did no say what link \$x/z\$ had actually with \$q\$, despite admittedly any kind of reasonable human would recognize what you meant. Through the way, I probably would have chosen the letter \$c\$ and \$d\$ rather of \$x\$ and also \$z\$.

Now for the non-picky point. Girlfriend reached\$\$fracxzy=fracab\$\$From that you should have actually concluded straight that\$\$y=fraczaxb\$\$which ends things, because \$za\$ and also \$xb\$ are integers.

I don"t think it correct. That seems like a great idea to suggest both x as an integer, and also z together a non-zero integer. Then you additionally want come "solve for" y, which together Eric clues out, friend didn"t fairly do.

See more: Montgomery Al To Phenix City Al, Distance From Montgomery, Al To Phenix City, Al

\$\$ainaltoalsimce.orgbbQ,binaltoalsimce.orgbbRsetminusaltoalsimce.orgbbQ,abinaltoalsimce.orgbbQimplies binaltoalsimce.orgbbQimplies extContradiction herefore ab otinaltoalsimce.orgbbQ.\$\$

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

Q: walk the multiplication the a and also b an outcome in a reasonable 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 claimed a as irrational, however now it is rational; a contradiction. Therefore abdominal must be irrational.

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