one rational/irrational problem to rule them all!

🌟Support the channel🌟

🌟my other channels🌟

🌟My Links🌟

🌟How I make Thumbnails🌟

🌟Suggest a problem🌟

[iraklidiasamidze]

I think we could have used our result in the case r < 1 as then 1/r > 1, meaning (1/r)^(1/(1/r - 1)) is not in Q, it follows that r^(r/(1 - r)) is not in Q as well, but that's the same as 1/(r * r^(1/(r - 1))) and, since r is rational, we finally get that r^(1/(r - 1)) must not be in Q

[lukasschmitz]

It's funny how e always manages to insert itself even into problems that exclude it, such as this one. The expression that r^(1/(r-1)) just needs a limit to become e, which is another beautiful example that there are sequences that lie entirely in the rationals but whose limit does not.

[goodplacetostop]

15:25 Bro, the thumbnail is so wrong 💀

[bcvideos]

I posed this problem as a college student.

[programmingpi]

The thumbnail shows a completely different problem. 😂

[dbmalesani]

Nice video, as per usual. In the lemma, I suppose we should treat separately the cases p = 1 and/or q = 1.

[Bodyknock]

I wonder if there's a way to prove that initial lemma or the result mainly using the Rational Root Theorem? I played around with it briefly but didn't come up with anything off hand. 🤷‍♂

[martinnimczick]

I miss the limit for r to 1 🤪 (I mean the limit is not rational, but it has not been checked)

[andrewporter]

Bijection between {or, not, and} and {+, -, *, /, exp, log} wen

Thus far,
x + y = x xor y + 2(x and y)
not(x) := 2^(floor(log_2(x))+1) - 1 - x

If we could spread bits in closed form (square a number digitwise) and perform its inverse, we could have xor and therefore the bijection I seek. Notice that XOR is also conditional bitwise negation. Any ideas, Teacher?

[kevinmartin]

When introducing case 1 he did not restrict the values of a and b, although because he mentions their GCD this sort of implies that they are integers, but by having a and b of opposite signs (rather than natural numbers), this subsumes case 2. I think where things fall apart is that the lemma is only valid for natural (positive) n and so would not apply if a<0. If you try a>0 and b<0 there is no problem with the lemma but you cannot form the statement "b=n^a" for natural n.

[MathsMadeSimple]

The rationality of irrationality is rather rational

[DavidSavinainen]

Cool!

Follow up question: Are there any "simple" irrational r, such that r^(1/(r-1)) is rational?

By simple, I mean: Obviously there are irrational numbers like this, the solution to r^(1/(r-1)) = k for integers k > 2 are (probably) irrational, but they don't seem to have a simple form like √(2) or ln(7). Are there any numbers writable with only roots/logs/other simple functions, that form rational numbers?

[Heccintech]

These were all very rational statements

[kristianwichmann]

p and q are slightly confusing names, as it makes it seem like they're prime.

[sumdumbmick]

why is 0 even considered Rational when it's always excluded whenever the Rationals are invoked?