Theorem: If $q
eq 0$ is rational and also $y$ is irrational, then $qy$ is irrational.
You are watching: The product of a rational and irrational number is rational.
Proof: proof by contradiction, us assume that $qy$ is rational. Because of this $qy=fracab$ for integers $a$, $b eq 0$. Since $q$ is rational, we have actually $fracxzy=fracab$ for integers $x eq 0$, $z eq 0$. Therefore, $xy = a$, and $y=fracax$. Due to the fact that both $a$ and $x$ are integers, $y$ is rational, bring about a contradiction.
As I cite here frequently, this ubiquitous residential or commercial property is merely an instance of complementary see of the subgroup property, i.e.
THEOREM $ $ A nonempty subset $ m:S:$ the 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 of this room ubiquitous in concrete number systems, e.g.
You can directly divide by $q$ suspect the reality that $q eq 0$.
Suppose $qy$ is reasonable then, you have $qy = fracmn$ for part $n eq 0$. This claims 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 understand that $h^-1 in H$, and therefore $y=h^-1(hy) in H$. Contradiction.
In our case, we have the group $(BbbR^*,cdot)$ and also its proper subgroup $(BbbQ^*,cdot)$. Through the arguments above $q in BbbQ^*$ and also $y in BbbRsetminus BbbQ$ means $qy in BbbRsetminus BbbQ$.
It"s wrong. You composed $fracxzy = fracab$. That is correct. Climate you stated "Therefore $xy = a$. The is wrong.
You have to solve $fracxzy = fracab$ because that $y$. You gain $y = fracab cdot fraczx$.
Let"s see just how we deserve to modify your discussion to do it perfect.
First of all, a minor stroller, stick point. Girlfriend wrote$$qy=fracab qquad extwhere $a$ and also $b$ room integers, through $b e 0$$$
So far, fine.Then come your $x$ and $z$. Because that completeness, you should have said "Let $x$, $z$ be integers such the $q=fracxz$. Note that neither $x$ nor $z$ is $0$." Basically, friend did no say what connection $x/z$ had with $q$, though admittedly any kind of reasonable human would know what you meant. Through the way, I probably would have chosen the letters $c$ and $d$ instead of $x$ and $z$.
Now for the non-picky point. You reached$$fracxzy=fracab$$From that you should have concluded straight that$$y=fraczaxb$$which ends things, because $za$ and $xb$ space integers.
I don"t think that correct. That seems favor a an excellent idea to show both x as an integer, and also z as a non-zero integer. Then you likewise want to "solve for" y, which as Eric clues out, you didn"t fairly do.
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$$ainptcouncil.netbbQ,binptcouncil.netbbRsetminusptcouncil.netbbQ,abinptcouncil.netbbQimplies binptcouncil.netbbQimplies extContradiction herefore ab otinptcouncil.netbbQ.$$
a is irrational, vice versa, b is rational.(both > 0)
Q: walk the multiplication that a and b an outcome in a rational or irrational number?:
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 asserted a as irrational, but now that is rational; a contradiction. Therefore ab must it is in irrational.
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given a reasonable number and also an irrational number, both better than 0, prove the the product in between them is irrational.
Please help me spot the error in my "proof" that the sum of 2 irrational numbers need to be irrational
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