MHB We can find two irrational numbers x and y to make xy rational,true or false

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It is true that two irrational numbers x and y can be found such that x^y is rational. A well-known example is when x = y = √2; if √2^√2 is rational, the proof is complete. If it is irrational, then setting x = √2^√2 and y = √2 leads to x^y = 2, which is rational. Another example provided is √2^(log₂9) = 3, further supporting the claim. Thus, the statement holds true with valid examples demonstrating the existence of such irrational numbers.
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we can find two irrational numbers $x$ and $y$
to make $x^y$ rational,true or false statement?
if true then find else prove it .
 
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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 .

what is a reasonable number ?
 
kaliprasad said:
what is a reasonable number ?
sorry it should be edited as :
$x,y $ irrational numbers
$ x^y$ rational number
 
Here's a famous example. Let x= y= \sqrt{2}. Either x^y= \sqrt{2}^\sqrt{2} is irrational or it is rational. If it is rational we are done. If it is irrational, let x= \sqrt{2}^\sqrt{2} and y= \sqrt{2}. Then x^y= (\sqrt{2}^\sqrt{2})^\sqrt{2}= \sqrt{2}^{(\sqrt{2}\sqrt{2})}= \sqrt{2}^2= 2. In either case, there exist two irrational numbers, x and y, such that x^y is rational.
 
Another example is $\sqrt{2}^{\log_29}=3$.
 
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