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
Best Answer
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 $d=4\sqrt 2$.