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Topic: Proof involving a polynomial and pi, e
Replies: 3   Last Post: Aug 24, 2013 11:52 AM

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William Elliot

Posts: 1,490
Registered: 1/8/12
Re: Proof involving a polynomial and pi, e
Posted: Aug 24, 2013 5:04 AM
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On Sat, 24 Aug 2013, lookskywalker wrote:

> How would one prove the below?
>
> If p(x) is a polynomial with integer coefficients, then the curve y=p(x)
> cannot cross the x-axis at pi, pi + 1, pi^2, e, e+1 or e^2.
>
> Seems kind of unbelievable, because these curves hit all kinds of real
> numbers, since they are continuous lines.


There are only countably many solutions to all of the polynomials in Z[x] (the
algebraic numbers) while there are uncountable many real numbers. That is
most of the numbers can't be algebraic or as it's said, the real numbers are
almost everywhere transcendental (is not algebraic) or the real numbers are
almost nowhere algebraic. On the other hand, unlike the integers, both the
rationals and the algebraic are dense subsets of the reals. That's why you
find infinitely many of them in every tiny interval.

However, proving a particular number, such as e or pi, is transcendental is a
difficult matter.



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