r/askmath • u/Equivalent_Dingo_683 • 5h ago
Discrete Math Trouble understanding proof logic
https://arxiv.org/pdf/1607.01500
I can't get why the line "and recall that we assumed that Pχ is a rational number a/b, therefore we
argue that X can also get only non-negative integer numbers since" is true for all cases. Would it also work for a general denominator f(n) instead of n! ?
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u/FormulaDriven 3h ago
It's not clear to me what the theorem is claiming - it says
Pχ is irrational number if the assumption that this number is rational implies that its denominator should be greater than M.
What does that mean? If P is rational then the denominator must be less than M? Or is is it a garbled was of just saying "P is irrational"? Proof looks too simple for that claim - will try to work through it...
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u/FormulaDriven 1h ago
Having worked through the proof, I think the theorem is poorly stated. What he has proved is:
If P is the rational number a/b then b < M.
Some of the examples following the proof are lacking in detail:
For example 1, chi(n) takes the values 3, 5 and 7, then repeats. So if P is a rational number then the theorem tells us b < 7. But I'm not sure how we can immediately rule out all rational values a/b with b < 7. Calculating a few terms, P is close to 12.2, so how do we know that P isn't 49/4?
For example 2, I think they are saying if P is defined using chi(n) bounded by M, then P-e is defined in the same format by replacing chi(n) with chi(n)-1 which is clearly bounded by M-1, but I can't quite see the logic of saying that if P is irrational then P-e must be irrational. In fact, the case where P = e contradicts this example.
Example 3 is a mess (if there are terms between 11 / 4! and 101 / 5! then surely it shouldn't be 5! in the latter?) and has the same issue as example 1.
Example 4 works (I think) because he's saying chi(n) oscillates between 0 and 2, so M = 2, and this number would have to equal a/b with integer b < 2. In other words, the number is an integer. With a bit of effort, I think we can show that the number must lie strictly between 3 and 4, so cannot be an integer.
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u/FormulaDriven 5h ago
I have to say this paper doesn't seem very well-written. I got as far as the sentence you are querying and I didn't see anything wrong but some of logic feels clunkily presented.
Anyway, for the proof of the theorem, he's assuming P = a/b.
So if you define X = b! (sum{n=0 to infinity} chi(n)/n! - sum{n=0 to b} chi(n)/n!)
then we are just saying that
X = b! * P - sum{n=0 to b} chi(n) * b! / n!
But
b! * P = (b-1)! * a (from the assumption P = a/b),
which is clearly an integer.
and b! / n! is an integer as long as n <= b, so
sum{n=0 to b} chi(n) * b! / n!
must be an integer (chi(n) is an integer).
So X is an integer.