Timeline for Theorems that are 'obvious' but hard to prove
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2011 at 19:32 | comment | added | David E Speyer | Have you seen the proof which looks at Re(f) and Im(f) separately and studies the combinatorics of how they cross? I think a high school student could understand this. I like this exposition arxiv.org/abs/math/0511248 | |
| Jan 12, 2011 at 14:34 | comment | added | Qiaochu Yuan | @Thierry: agreed. Anything Euler attempted unsuccessfully to prove can't be all that easy. I think your comments apply to comments on several other answers as well in relation to how rooted algebraic topology is in many mathematicians' educations these days. | |
| Jan 11, 2011 at 14:43 | comment | added | Thierry Zell | If I'm following this correctly, some comments say the example is unsuitable because proving the theorem is actually easy, while the oldest comment says it's unsuitable because it's not obvious. What a mess! Keeping in mind how long it took between the result being conjectured and the first actual proof, I think we are too far removed in time from the result to truly appreciate it from a historical perspective, and the FTA is too fundamentally rooted in students educations to imagine how hard it would be for someone who was a blank slate. | |
| Jan 11, 2011 at 4:52 | comment | added | roy smith | Doug's argument is made completely explicit in the book of Steenrod and Chinn, aimed precisely at high school students. | |
| Jan 11, 2011 at 1:11 | comment | added | Douglas Zare | Here is a proof whose main idea is understandable by many high school students. The winding number of the image of the circle of radius $r$ changes from $0$ at $r=0$ to the degree of the polynomial for $r$ large, and it can only change when there is a $0$ of the polynomial. | |
| Jan 11, 2011 at 0:43 | comment | added | roy smith | The division theorem implies a non constant polynomial defines an open map from the sphere to itself. Since any such map is also continuous, hence closed, it is surjective. that's the best I can do. It seems conceivable to make that seem plausible to someone with a little intuition, if not obvious. | |
| Jan 10, 2011 at 23:30 | comment | added | Andrés E. Caicedo | I do not know why this is plausible. What would be an "obvious" reason to expect that a degree 6 polynomial with real coefficients has a complex root? | |
| Jan 10, 2011 at 22:45 | history | answered | Eric Hsu | CC BY-SA 2.5 |