Prove that there exist four distinct real numbers a, b, c, d such that exactly four of the numbers ab,ac,ad,bc,bd,cd are irrational So i am doing this example and i found a way to prove there exists that but it seems ugly. Can you guys help me find a better solution?
My proof:
a*b is irrational when one of them is irrational and the other is rational.
Assume one of the numbers is irrational (say "a"). We know that ab,ac,ad are irrational but the rest is not. So it is impossible with one irrational number.
Assume two of the numbers are irrational (say "a" and "b"). We know that ac,ad,bc,bd are irrational while ab may or may not be irrational. Since we want 4 integers to be irrational we want ab to be rational. So here i just come to a conclusion that a=b= irrational number that is a square root. Seems awful ^^. Any help would be greatly appriciated 
 A: You ask for four real numbers satisfying that; so we can give $\sqrt 2,2\sqrt 2,3\sqrt 3,4\sqrt 3$ for which we get their mutual multiplication as: $4,3\sqrt 6,4\sqrt 6,6\sqrt 6,8\sqrt 6,36$.

Let's see if we can construct an example by some arguments. If $a,b$ and $c$ are chosen as rational then $ab$, $ac$ and $bc$ are also rational and hence not desirable for us. So let's see what happens if $a$ and $b$ are rational; then $ab$ is rational. $c$ and $d$ are definitely irrational because otherwise according to what we discussed there will be more than 2 rational numbers. So $ac,bc,ad$ and $bd$ are irrational but $cd$ should be rational for our purpose. Hence you should choose $c$ and $d$ irrational so that their product is rational. Any idea about that? Yes, lets pick $a=1$, $b=2$, $c=3\sqrt 2$ and $d=4\sqrt 2$.
A: You don't need a proof, just a counter example.  You can always get two irrationals that are "completely unrelated" to each other (such as $\pi$ and $e$, or $\sqrt 2$ and $\sqrt 5)$.  Can set b "related" to a (such as b = rational/a) and the same for c and d.  Thus you'd get ab and cd are rational but ac, ad, bc, bd are not.  For example$a = \pi, b = 1/\pi, c= e, d=1/e$.  $ab = cd = 1$, but $\pi/e, \pi*e, e/\pi$ and $1/\pi*e$ are irrational.
Or if you want to be clever get three pairwise "related" irrationals where ab is rational and ac is rational but bc is not.  Easiest example is algebraic roots and inverses.  a = some n-th root, b= 1/a.  c = the n-th root to the n-1 power.  d can be pretty much anything you want as long as it is "unrelated" to a, b, c.
Example:  $a = 2^{\frac 1 3}, b = 2^{-\frac{1}{3}}, c = 2^{\frac 2 3}, d = 1 $ or $e$ or $\pi$ or $27/4$ or ....
A: One way to produce an example is to start by picking a pair of products that you'd like to make rational using distinct numbers.  The "nicest" pair is $ab$ and $cd$, because they use all four numbers you're trying to find. You can get these two products to be rational using the distinct numbers
$$a=-b=\sqrt2\qquad\text{and}\qquad c=-d=\sqrt3$$
The other four products are all $\pm\sqrt6$, which is irrational.
