Hello ! Here is a little problem : Proove that there exists two irrationnals a & b such as $\displaystyle a^b$ is a rationnal. The proof is very easy, but you have to know it
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I know what you are looking for. But I have a different solution. $\displaystyle e^{\ln 2} = 2$.
This is good too Erm...is ln(2) irrationnal ? I love the demo i read, because it's like...juggling with maths hihi
$\displaystyle \left( 2^\pi\right)^\frac{1}{\pi}$
Yes, this is the kind of demo it was given ^^ $\displaystyle ((\sqrt{2})^{\sqrt{2}})^{\sqrt{2}}$
Originally Posted by Moo This is good too Erm...is ln(2) irrationnal ? I love the demo i read, because it's like...juggling with maths hihi Suppose otherwise, then there exist integers $\displaystyle a$ and $\displaystyle b$ such that $\displaystyle e^{a/b}=2$. Then: $\displaystyle e=\root a \of{2^b}$ Now the left hand side is transcendental and the right hand side algebraic - a contradiction. RonL
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