On a straight line with marked origin 0 and unit 1, two points x and y are given. Is it possible, by any finite method, to geometrically define x^y if the given y is not a rational but an arbitrary point?
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$\begingroup$ So, is the question something like: Given three marks $x$, $y$ and $z$ where someone claims that $x^y=z$, is there a finite geometric construction (possibly involving other guessed points) they can use to prove they are right? $\endgroup$– hmakholm left over MonicaCommented Jun 1, 2016 at 12:59
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Answer to an earlier version of the question which asked to construct $x^y$ instead of merely defining it:
No, at least not if "any finite method" means compass and straightedge -- for example this is famously impossible when $x=2$ and $y=1/3$.
answered Jun 1, 2016 at 12:35
hmakholm left over Monicahmakholm left over Monica
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$\begingroup$ not necessarily straightedge and compass. If y is rational, x^y can be defined by a procedure involving similar triangles. The question is what happens if y is irrational. $\endgroup$– exp8jCommented Jun 1, 2016 at 12:43
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2$\begingroup$ @J.Avaris You should specify the rules of the game if you want us to play along! $\endgroup$– lynnCommented Jun 1, 2016 at 12:44
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1$\begingroup$ @ZubinMukerjee: The figure cannot be constructed just with compass and straightedge -- after drawing AB and the lines from B towards C and D, you need a "sliding ruler with marks" in order to find the line ACD. $\endgroup$ Commented Jun 1, 2016 at 12:48
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1$\begingroup$ @Zubin en.wikipedia.org/wiki/Doubling_the_cube $\endgroup$– lynnCommented Jun 1, 2016 at 12:51
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1$\begingroup$ @ZubinMukerjee: Doubling the cube is one of three famous ancient geometric construction problems that were only proved impossible (by compass and straightedge) in the 19th century. The other two ones are squaring the circle and trisecting an arbitrary angle. $\endgroup$ Commented Jun 1, 2016 at 12:51