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

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Proof: proof by contradiction, we assume that \$qy\$ is rational. As such \$qy=fracab\$ because that integers \$a\$, \$b eq 0\$. Since \$q\$ is rational, we have \$fracxzy=fracab\$ because that integers \$x eq 0\$, \$z eq 0\$. Therefore, \$xy = a\$, and \$y=fracax\$. Because both \$a\$ and also \$x\$ room integers, \$y\$ is rational, resulting in a contradiction.  As I point out here frequently, this ubiquitous property is merely an circumstances of complementary check out of the subgroup property, i.e.

THEOREM \$ \$ A nonempty subset \$ m:S:\$ of 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 the this room ubiquitous in concrete number systems, e.g.   You can directly divide by \$q\$ presume the reality that \$q eq 0\$.

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

A group theoretic proof: You understand that if \$G\$ is a group and \$H eq G\$ is among its subgroups climate \$h in H\$ and also \$y in Gsetminus H\$ implies that \$hy in Gsetminus H\$. Proof: mean \$hy in H\$. You know that \$h^-1 in H\$, and therefore \$y=h^-1(hy) in H\$. Contradiction.

In our case, we have actually the group \$(BbbR^*,cdot)\$ and its ideal subgroup \$(BbbQ^*,cdot)\$. By the arguments above \$q in BbbQ^*\$ and also \$y in BbbRsetminus BbbQ\$ suggests \$qy in BbbRsetminus BbbQ\$. It"s wrong. You wrote \$fracxzy = fracab\$. The is correct. Climate you stated "Therefore \$xy = a\$. The is wrong.

You have to solve \$fracxzy = fracab\$ for \$y\$. You gain \$y = fracab cdot fraczx\$.

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

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

So far, fine.Then come her \$x\$ and \$z\$. For completeness, friend should have said "Let \$x\$, \$z\$ it is in integers such the \$q=fracxz\$. Note that neither \$x\$ no one \$z\$ is \$0\$." Basically, you did not say what connection \$x/z\$ had with \$q\$, despite admittedly any reasonable human being would know what girlfriend meant. By the way, I more than likely would have chosen the letter \$c\$ and \$d\$ rather of \$x\$ and \$z\$.

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

I don"t think that correct. It seems prefer a great idea to indicate both x together an integer, and also z as a non-zero integer. Climate you also want to "solve for" y, which as Eric points out, girlfriend didn"t rather do.

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\$\$aincg-tower.combbQ,bincg-tower.combbRsetminuscg-tower.combbQ,abincg-tower.combbQimplies bincg-tower.combbQimplies extContradiction herefore ab otincg-tower.combbQ.\$\$

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

Q: go 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 also j space integers

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

before we declared a together irrational, but now that is rational; a contradiction. Therefore abdominal muscle must be irrational.

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