A hole in one?

Number Theory Level pending

Consider a unit circle centered at the origin.

Remove all the points that have rational y y values.

Then remove all the points that have irrational x x values.

What is the area, A A , of the union of the remaining points?

π 2 < A < 9 π 10 \dfrac{\pi}{2} < A < \dfrac{9\pi}{10} 0 0 π \pi π 10 < A < π 2 \dfrac{\pi}{10} < A < \dfrac{\pi}{2} π 2 \dfrac{\pi}{2}

Geoff Pilling by Geoff Pilling , 53, USA

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1 solution

Geoff Pilling
Apr 4, 2017

Since there are infinitely many more irrational numbers than rational numbers, once you remove all the points with irrational x x values, although there are still some points remaining, in fact an infinite number, they make up 0% of the total area.

So, the area of the left over points is 0 \boxed0

One of the oddities of different types of infinities....

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I guess we're dealing with Lebesgue measures and integration here, (at least that is what I based my answer on). So does the set of all points ( x , y ) , x , y ∈ ( R − Q ) (x,y), x,y \in \mathbb{(R - Q)} , on the unit circle have area π \pi , since the Lebesgue measure of the irrationals on [ − 1 , 1 ] [-1,1] is 1 1 ? It always seems weird that even though the rationals are dense on the reals, they contribute nothing to any measure, even when paired up with irrationals as is the case here ....

Brian Charlesworth - 4 years, 2 months ago

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