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.
Albert1
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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 $\sqrt{2}^{\log_29}=3$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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