4 ms·

The key idea is a bit obscured. A rational number is a finite computation of integers. It is finite but in magnitude and in level of detail. Thus, if you can

by lupire 2y ago

The key idea is a bit obscured. A rational number is a finite computation of integers. It is finite but in magnitude and in level of detail. Thus, if you can rewrite a computation as the same computation but with only a subset of the terms, or with smaller (absolute value) integer terms, in a repeatable way, then the computation must either collapse to 0, or be infinite in either number of terms or size of a term. That is, not rational. https://en.m.wikipedia.org/wiki/Proof_by_infinite_descent https://en.m.wikipedia.org/wiki/Proof_by_infinite_descent Any finite set of rational numbers is equivalent to a set of integers.

potbelly83 2y ago
It's not immediately clear (though I'm sure it is true) from reading the article that F(\pi) + F(0) != 0 (assuming \pi = \frac{a}{b}). Given that F(x) is an alternating series, maybe F(\pi) = F(0) = 0. (again I'm sure this is not the case, but it's not discussed in the proof)
- ykonstant 2y ago
  The article shows in (1) that F(\pi) + F(0) is the integral of f(x)sin(x) from 0 to pi; the integrand is non-negative there and thus so is the integral.
  - red_trumpet 2y ago
    Even stronger, the integrand is positive, hence the integral is positive as well, so it cannot be zero.
- csense 2y ago
  F(\pi) + F(0) is the integral of f(x) sin(x) from 0 to \pi. f(x) sin(x) is the product of four factors (not necessarily integers): f(x) sin(x) = (1/n!) (x^n) (a-bx)^n sin(x) Examine the signs of these terms on the interval 0 to \pi. All four factors are non-negative everywhere, so the integral has to be non-negative. Could the integral be zero? 1/n! is a nonzero constant. x^n = 0 only when x = 0. (a-bx)^n = 0 only when x = a/b = \pi. And sin(x) = 0 only when x = 0 or x = \pi. The four factors are all positive inside the interval, and nice continuous functions, so there's definitely a rectangle of positive area underneath their product (and the product can never be negative so you can never get a negative area to get the integral back down to 0).
alimw 2y ago
Are you talking about this proof in particular? I see no connection.
CaptainNegative 2y ago
I don't think that follows at all. Any series of rational numbers that converges to a rational (such as any appropriate geometric series) will still converge to a rational if you remove any finite or co-finite subset of terms.

by lupire 2y ago

potbelly83 2y ago

It's not immediately clear (though I'm sure it is true) from reading the article that F(\pi) + F(0) != 0 (assuming \pi = \frac{a}{b}). Given that F(x) is an alternating series, maybe F(\pi) = F(0) = 0. (again I'm sure this is not the case, but it's not discussed in the proof)

ykonstant 2y ago
The article shows in (1) that F(\pi) + F(0) is the integral of f(x)sin(x) from 0 to pi; the integrand is non-negative there and thus so is the integral.
- red_trumpet 2y ago
  Even stronger, the integrand is positive, hence the integral is positive as well, so it cannot be zero.
csense 2y ago
F(\pi) + F(0) is the integral of f(x) sin(x) from 0 to \pi. f(x) sin(x) is the product of four factors (not necessarily integers): f(x) sin(x) = (1/n!) (x^n) (a-bx)^n sin(x) Examine the signs of these terms on the interval 0 to \pi. All four factors are non-negative everywhere, so the integral has to be non-negative. Could the integral be zero? 1/n! is a nonzero constant. x^n = 0 only when x = 0. (a-bx)^n = 0 only when x = a/b = \pi. And sin(x) = 0 only when x = 0 or x = \pi. The four factors are all positive inside the interval, and nice continuous functions, so there's definitely a rectangle of positive area underneath their product (and the product can never be negative so you can never get a negative area to get the integral back down to 0).

alimw 2y ago

Are you talking about this proof in particular? I see no connection.

CaptainNegative 2y ago

I don't think that follows at all. Any series of rational numbers that converges to a rational (such as any appropriate geometric series) will still converge to a rational if you remove any finite or co-finite subset of terms.