How can I prove that dividing an irrational number by an irrational number (besides itself) can result in an integer?
Proof that dividing irrational number by an irrational number can result in an integer?
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2@Alon: "proof" is a noun; the verb is "to prove", not "to proof". – 2010-11-24
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0@Alon: not really (number-theory), which usually refers to a slightly different field of study. I re-tagged it as (real-number-computation), which seems better. – 2010-11-24
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0I wonder if there are nontrivial examples... – 2010-11-24
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0Actually, you can say something a bit stronger. If $\frac{a}{b}$ is an integer, and one of $a$ or $b$ is irrational, then the other must be irrational! – 2010-11-25
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0@J.M., nontrivial examples? I think not, since it is in general hard to prove the arithmetic nature of the constants one finds in Analysis. For instance $\dfrac{\pi^2}{\zeta(2)}=6$, but there is no known similar relation between $\pi^k$ and $\zeta(3)$. – 2010-11-25
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0@Américo: I'm pessimistic about it too, but then again I am pessimistic about a lot of stuff... – 2010-11-25
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5@J.M.. In a sense, there can be no "nontrivial examples": If $a/b = n$, with $n$ an integer, then it means that $a=nb$, that is, we are just taking an irrational, multiplying it by an integer, and dividing by the original irrational. Presumably you mean two irrationals that we obtain via independent processes, that don't *seem* to be related, but in fact turn out to be related, like $\pi^2$ and $\zeta(2)$ above. – 2010-11-25
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0On the other hand, if we replace "irrational" with "transcendental" in the OP's question, the question is now much more intricate, yes? (Schnauel comes to mind.) – 2010-11-25
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0@J.M.: How so? If $b$ is transcendental, then so is $nb$ for any nonzero integer $n$, so we still have the same examples: just take $nb/b$ for any transcendental $b$. And if $a$ and $b$ are transcendental and $a/b=m$ an integer, then $a=bm$ just as before. I don't see it; what am I missing? – 2010-11-25
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0@Arturo: Sorry, I'm baked today; I seem to have conflated with the fact that things like $\frac{\pi}{e}$ are not known to be rational or not... – 2010-11-25
9 Answers
By giving an example. For instance, $2\sqrt{2}$ divided by $\sqrt{2}$?
Note well: To prove that something can happen, all you need to do is exhibit a single instance of it happening. To prove that something must happen you need to show that it happens always. So to show that dividing an irrational number by an irrational number other than itself can be an integer, all you need to do is cook up a single example of it happening. If you wanted to prove that, say, a rational divided by rational other than zero must be a rational, an example would not suffice; you would need to show that for any rational $q$ and any rational $r\neq 0$, you would have $\frac{q}{r}$ a rational.
The basic idea works much more generally: it amounts to how a subgroup acts on its complement.
$\rm If\:\ \ a,b \in A \subset B\ \ \: then\ \ ab \in A \iff b \in A\ \ $ if $\rm\ \ a^{-1}\in A,\ $ and $\rm\ A,\:B\ $ are monoids.
Thus $\rm\ \ b\not\in A\ \ \Rightarrow\ \ ab\not\in A\ \ and\ \ (ab)/b = a\in A\ $ as you desire.
Your case is simply $\rm\ \ A = \mathbb Q,\ \ B = \mathbb R,\ \ a\in \mathbb Z$
The key property above is essentially this "complementary" view of a subgroup:
THEOREM $\ \ $ A nonempty subset $\rm\:A\:$ of abelian group $\rm\:B\:$ comprises a subgroup $\rm\iff\ A\ \bar A\ = \bar A\ $ where $\rm\: \bar A\:$ is the complement of $\rm\:A\:$ in $\rm\:B$
Instances of this are ubiquitous in concrete number systems, e.g.
transcendental algebraic * nonalgebraic = nonalgebraic if nonzero rational * irrrational = irrational if nonzero real * nonreal = nonreal if nonzero even + odd = odd additive example integer + noninteger = noninteger
A trivial example would be $2\pi/\pi=2$. In fact every example would be of this form (one irrational would be an integer multiple of the other), for if $a,b$ are distinct irrational numbers such that $\frac{a}{b}=c$ where $c$ is an integer, then, $a=bc$
An example would suffice to prove this claim. If you don't want to divide an irrational number by itself, how about $2\sqrt{2}/\sqrt{2}$. You should be able to prove that if a number is irrational, then twice that number is also irrational.
If x is an irrational then x/-x works and is the canonical example that satisfies the OP's rules
Start by assuming that you have a case where it does result in an integer, and see what that says about the irrational numbers. From there you should be able to give examples. That is, if $x/y = n$ for some integer $n$, what does that say about the relationship between $x$ and $y$? Choosing a specific integer $n$ and a specific $y$ should lead you to an example.
(While writing this, many answers were posted. I'm adding this anyway to give a naive approach to use if you don't already see how it will come out.)
An example from Euler together with the proof of the irrationality of $\pi^2$:
$$\dfrac{\pi^2}{\zeta(2)}=6,$$
where
$$\zeta(2)=\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^2}.$$
if a,b and c are integers, and $$a^b=c$$ then $$(\ln \ c)/(\ln \ a)=b$$ and the natural log of all integers $>1$ is irrational.
Root 32 (irrational)/Root 2 (irrational)= Root 16= 4( Rational)