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

offered a rational 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" the the amount of two irrational numbers need to be irrational

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