Let numbers are $ a,b,c$, then $ 2ab = k_1,2ac = k_2,2bc = k_3$, were $ k_1,k_2,k_3$ are odd integers. Therefore $ c = \sqrt\frac {k_2k_3}{2k_1},b = \sqrt {\frac {k_1k_3}{2k_2}},a = \sqrt {\frac {k_1k_2}{2k_3}}$ are irrationals.

Sorry for my stupid question, can someone explain why is $ \sqrt{\frac{k_2 k_3}{2 k_1}}$ irrational? Thx..

Alternately, if the three are $ a,b,c$, then $ (ab)(bc)(ca) = (abc)^2 = \frac{k}{8}$, for some $ k$ not divisible by $ 8$; then $ abc$ is irrational, since $ 8$ is not a square. Then $ a(bc)$ is irrational, while $ abc$ is rational; therefore $ a$ is irrational. Likewise $ b$ and $ c$ are irrational.

TFPNH wrote: Sorry for my stupid question, can someone explain why is $ \sqrt {\frac {k_2 k_3}{2 k_1}}$ irrational? Thx.. Assume that is rational: $ \sqrt {\frac {k_2 k_3}{2 k_1}}=\frac{x}{y}$ then $ 2k_1 x^2=k_2 k_3 y^2$ and remember that $ k_1,k_2,k_3$ are odd. then if $ 2^k | LHS$ then $ k$ is odd and if $ 2^k | RHS$ then $ k$ is even. Contradiction

mszew wrote: TFPNH wrote: Sorry for my stupid question, can someone explain why is $ \sqrt {\frac {k_2 k_3}{2 k_1}}$ irrational? Thx.. Assume that is rational: $ \sqrt {\frac {k_2 k_3}{2 k_1}} = \frac {x}{y}$ then $ 2k_1 x^2 = k_2 k_3 y^2$ and remember that $ k_1,k_2,k_3$ are odd. then if $ 2^k | LHS$ then $ k$ is odd and if $ 2^k | RHS$ then $ k$ is even. Contradiction Or just $ \sqrt {\frac {k_2 k_3}{2 k_1}}$ is $ \sqrt2$ times something which cannot cancel $ \sqrt2$, so it's irrational.

Thanks for the explanation, I got it now

Does this work? The part which I am unsure about is whether the following statement is true (I think it is). A real number $a$ is rational if and only if the product of the denominator and numerator in any representation of $a^2$ is a perfect square. Solution Let $a,b,c$ denote the three numbers. Then for some $x,y,z \in \mathbb{Z}$ we have that $2ab=2x+1$, $2bc=2y+1$ and $2ac=2z+1$. Solving the system of equations yields that, \[a^2 = \frac{(2x+1)(2z+1)}{2(2y+1)}\] If $a$ were rational, then it would follow that the product of the numerator and denominator in the above fraction would be a perfect square. However the largest power of $2$ that divides this product is $2$. Hence $a$ is irrational. Symmetry yields the same result for $b,c$.