Proof that dividing irrational number by an irrational number can result in an integer?

How can I prove that dividing an irrational number by an irrational number (besides itself) can result in an integer?

• @Alon: "proof" is a noun; the verb is "to prove", not "to proof". – Arturo Magidin Nov 24 '10 at 21:41
• @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. – Arturo Magidin Nov 24 '10 at 21:47
• I wonder if there are nontrivial examples... – J. M. is a poor mathematician Nov 24 '10 at 23:24
• Actually, 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! – Kevin Ventullo Nov 25 '10 at 0:04
• @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. – Arturo Magidin Nov 25 '10 at 1:00

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)