The statement that two irrational numbers \(x\) and \(y\) can be found such that \(x^y\) is rational is true. A well-known example is when both \(x\) and \(y\) are set to \(\sqrt{2}\). In this case, \(x^y = \sqrt{2}^{\sqrt{2}}\) may be either rational or irrational. If it is irrational, setting \(x = \sqrt{2}^{\sqrt{2}}\) and \(y = \sqrt{2}\) leads to \(x^y = 2\), which is rational. Another example includes \(\sqrt{2}^{\log_2 9} = 3\), further confirming the existence of such irrational pairs.
PREREQUISITESMathematicians, educators, students studying number theory, and anyone interested in the properties of irrational numbers and their relationships with rational numbers.
Albert said:we can find two unreasonable numbers $x$ and $y$
to make $x^y$ reasonable,true or false statement?
if true then find else prove it .
kaliprasad said:what is a reasonable number ?