If 0 < x < 1 , \(\lfloor x \rfloor \lceil x \rceil = 0\cdot1=0\) but \(0\nless\sqrt0\)
If 1 < x < 2 , \(\lfloor x \rfloor \lceil x \rceil = 1\cdot2=2\) and \(1<\sqrt2<2\)
So \(x=\sqrt2\) is a solution to \(\lfloor x \rfloor \lceil x \rceil = x^2\)
If 2 < x < 3 , \(\lfloor x \rfloor \lceil x \rceil = 2\cdot3=6\) and \(2<\sqrt6<3\)
So \(x=\sqrt6\) is a solution to \(\lfloor x \rfloor \lceil x \rceil = x^2\)
If 3 < x < 4 , \(\lfloor x \rfloor \lceil x \rceil = 3\cdot4=12\) and \(3<\sqrt{12}<4\)
So \(x=\sqrt{12}\) is a solution to \(\lfloor x \rfloor \lceil x \rceil = x^2\)
The product of the three smallest, positive, non-integer solutions = \(\sqrt2\cdot\sqrt6\cdot\sqrt{12}\,=\,\sqrt{144}\,=\,12\)
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