my favorite type of equation
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- Опубликовано: 15 сен 2024
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11:11 Good Place To 🗿
cant we just use rational root theorem on f(epsilon)=0 and the fact that 0
basically what penn did tbf
A 5 minute video would have been too short 😂
if you just clear the denominators by multiplying thru by b^3, you can conclude that b divides a^3.
Quite a nice feature of the solution is that it turns out that ε(n) associated to a given n is always greater than ε(n+1).
Floor functions - a MP Favorite
Hi,
11:12 : missing "to stop".
@ 9:39 Should be -nab^2.
Masterfully done, Sir! The substitutions are downright brilliant, without them I think the proof would have been MUCH more complicated. 👍
"All solutions are irrational" part is not correct.
Correct statement should be "All solutions except x=0 is irrational"
If you take natural numbers to exclude 0, which I believe he typically does, it works.
Actually, this brought me to a bit of a thought: He writes r=n+a/b, but then doesn't really justify why a is not 0. In the n=0 case, of course it *is* 0, but maybe there should have been a step to show that it isn't otherwise.
At 7:38, why not multiply both sides by b^2? Then after observing that all the terms except for a^3/b are integers. So a^3/b = m, some integer. With (a,b)=1, then b must equal 1 contradicting the b>1 assumption.
Faster: Noting (x^3 + x)/(x^2 + 1) = x, rewrite as x = floor(x) + x/(x^2+1) -> frac(x) = x/(x^2+1) (*), which solves part a.
Part b: If x = p/q with p = aq+b, gcd(p, q) = 1, plug into (*) to get b(p^2+q^2) = pq^2 -> p, q^2 | b. Set b = pq^2 n -> n(p^2+q^2) = 1 -> n = 1, {p, q} = {0, 1} -> only solution is x = 0.
A good place to start with the first step of the floor function!
Nice video as always. But at around 9:30 I had to take a nab...
Sometimes I wish you could give pointers why such seemly mundane equation could appear in advanced math, which is why we are following in the first place.
at ~ 4:57 you say epsilon cant be 0, but shouldnt it be 0 for n=0 since thats an obvious solution to the equation?
EDIT: okay i think 0 is not included in the natural numbers here by the fact that 0 is indeed rational ^^
Yeah, he didn't explicitly say 0 isn't included in his definition of the Natural Numbers in this problem but the irrationality statement fails for x in [0,1)
Excellent! Math is always easy
x = 0 is a solution though!
n is a natural number, not an integer
@@fredfred9847 natural/integer aside (for me 0 is a natural number), by the requirement that b > a >= 1 we still exclude integers. Not every solution is irrational, because x=0 is a rational solution
@@skylardeslypere9909 Positive integers are not natural numbers. Eg. first natural number is 0. There might be a bizzare Peono arithmetic with axiom that 1 is a natural number, but I've never seen one. It's always 0 is a natural number.
@@topilinkala1594 Wait what are you saying? That 1,2,3,... are not natural numbers? Only 0? The natural numbers are always either the set {0,1,...} or the set {1,2,...} but never just the number 0. Am I misinterpreting your comment?
@@topilinkala1594the natural numbers are just a convenient definition for certain numbers. There's no 'axiom' defining a number to be natural. Definitions and axioms are different things.
Nice problem, but I was disappointed by number of things. In the first part, Michael didn't properly justify the existence of a zero of f in [0,1[ (f is continuous as it is a polynomial and so the zero exists by the intermediate value theorem). In the second part, the question is not clear: it should say "Show that there are no rational solutions for x≥1" (this is how Michael interpreted the question, and this must be correct as x=0 is a rational solution). Also Michael should have mentioned that we can exclude the case a/b=0, from work done in the first part. Finally, it would have been much clearer and cleaner to quote the rational root theorem, instead of giving a rather sketchy proof of it. At the very least, he should have mentioned the theorem and that he was proving it in this particular case.
he proved the theorem for (0,1), and assumed by extension can be proved for for (n,n+1), maybe as exercise!
nice proof...