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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Is there a straightforward proof because that $small 2fracn3$ is no an integer when $fracn3$ is not an integer?

If $ab equiv r pmodp$, and $x^2 equiv a pmodp$ then $y^2 equiv b pmodp$ because that which problem of $r$?

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

Please aid me clues the error in my "proof" that the sum of 2 irrational numbers must be irrational

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