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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Is over there a simple proof because that $small 2fracn3$ is not an integer when $fracn3$ is not an integer?

If $ab equiv r pmodp$, and $x^2 equiv a pmodp$ climate $y^2 equiv b pmodp$ for which condition of $r$?

offered a reasonable number and also an irrational number, both greater than 0, prove that the product between them is irrational.

Please help me spot the error in mine "proof" that the amount of two irrational numbers have to be irrational

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