Right. If you want to be fancy about it, consider a polynomial mod p of degree p+1. It has at least one root in the integers mod p, because the “complex roots” (those that live in field extensions) come in orbits of the Frobenius automorphism (the analog of taking the complex conjugate) of size p. Of course, one could also say that using the opposite signs of f(-large) and f(+large) is a parity-based proof, since it shows that we have to cross the x-axis an odd number of times :-) (counting multiple roots properly of course) C
On Jan 30, 2020, at 5:41 PM, Bernie Cosell <bernie@fantasyfarm.com> wrote:
On 31 Jan 2020 at 0:18, Gareth McCaughan wrote:
On 30/01/2020 23:16, Cris Moore via math-fun wrote:
This is too easy, but one could also consider the fact that any polynomial of odd degree (and real coefficients) has at least one real root...
Is there a parity-based proof of that that isn't strictly inferior to "f(-large) and f(+large) are large and of opposite sign, so by the intermediate value theorem there's a root somewhere between"?
Is it not sufficient to show that complex roots always come in pairs, and so when you've exhausted all the complex roots there must be [at least] one more?
/Bernie\
Bernie Cosell bernie@fantasyfarm.com -- Too many people; too few sheep --
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Cris Moore moore@santafe.edu "It is bound to be very imperfect. But I think it possible that I have got my statues against the sky.” — Virginia Woolf