Rational Roots Proof
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- čas přidán 8. 09. 2024
- Rational Roots Theorem Proof
In this video, I prove the rational roots theorem, which is a neat way of finding rational roots of polynomials. A little algebra delight to sweeten the day!
Rational Roots Example: • Rational Roots Theorem
Check out my Real Numbers Playlist: • Real Numbers
Wow, that is such an elegant proof. This is commonly taught in high school, but the proof (aside from a couple of points) is just very algebra.
Maybe nowadays. In the 1960's they never proved this in high school. You had to wait for undergraduate abstract algebra.
@@roberttelarket4934 this is definitely not normal to see proved in high school
@@hiwrenhere: I love your name Integral Domain!!
my school doesn't prove anything outside of Euclidean geometry
That was a rock star proof. I've never seen that before (in engineering) and I loved it!
I love the symmetry in this proof! I could immediately tell where we were going with the a_0q^n term after you'd finished the a_np^n segment
he's so enthusiastic, this eases assimilation
It's not a horrible proof. It's very elegant. This theorem is very useful when you combine it with Bézout's theorem.
In my opinion, teaching proofs should be mandatory.
I think that proofs are just a superior way of teaching concepts then unfulfilling memorization.
I personally find it much easier (and fun) to study the proof of something then memorising it.
The proof makes sense, if you forget a part of it you can just recreate it yourself. If you forget a part of a memorized formula you can't recreate it.
I think in early math classes only 3 minute proofs should be taught. I had a multivariable Calc teacher who would prove things like Stokes’ theorem, and barely explain how to use it, which just left me more confused
More important is focusing on developing problem solving and analytical thinking. It is much more better if students alone can discover some rule naturally during solving some problem.
There is a professor from MIT who did a bunch of MIT courses on AI where he says you need to make the equation 'sing to you'. I think sometimes that is visualisation, sometimes proof, sometimes application. It is surely more than just doing algorithmic computations that you forgot as soon as you finish the exams.
You can also get the second proof for divisor 'p' directly from the first form of equation after dividing both sides by 'q'. It can be more intuitive and natural way of thinking during problem solving to go in linear way than multipath way with returns back. Anyway, many matematical proofs follow the second possible path as you show.
Neat and simple proof, although I'm usually terrified by algebra this was easy to follow. I have a question though, do this theorem only work if a_i are rational? If so, is there a more generalized theorem for real or complex coefficients?
Would love to see some examples of using a Green's function
At *5:30* you can also just look at that mod q and mod p and your are done.
Dr Peyam, I look at many proofs of this, yours is the best, HIGH DIDACTIc quality!! thank you very much I think to subscribe to you
Superb Dr Peyam- or rather if may call you Dr.P (or Q)😊🙃-. I was also terrified as you about what this theorem was about- and wondered how prove it. You pulled out the solution- which is really elegant Dr.P, except, as the expression goes- you should mind your (p's) and q's"! That is you write your q's as the number "9" Nevertheless Dr.Peyam- your range of Math videos are amzing- and I am re-learning my math (I have a degree in Engineering- but never saw learning this fun).
Thanks a ton👋👌
Awesome so excited to watch thank you Dr. Peyam!!
Awesome video, thanks!
Your cheerfulness made my day ngl!
Great explanation. Thx ❤
Man, this guy is amazing! Thank you!
That was an amazing proof!! Thank you!
keep going Sir nice we are waitinf all amalyisis lectures
Thank you doctor ,make alot vedios like this please and about number theory
Amazing! 👏🙂💪
Great video 👍🏻👍🏻 😍😍. I believe you could’ve also elaborated on how if integer roots were mot produced then the answer had to be irrational. If q never divides An, then q has to divide p^n. But since p and q have to have no common factors, it follows that the answer will be an irrational number
Dr Peyam just did that proof this week.
Look the problem is this also proves that
AnP^n/Q^n......A1p/q is an integer and also this thing +A0 and -AnP^n/Q^n is an integer.
This implies An×root is an integer and the root is an integer not rational
This can be seen after we get all things to the right side just leaving AnP^n/Q^n on the left side.
Now multiply both side by 1/x^n and we get An = something
As An is an integer P and Q are also integer rest of the polynomial must be an integer.
Thus the thing from the second term to A0 must be an integer
Neat proof.
Now any natural number (n) may be written as the product a limited number of prime numbers, but includes ALL prime numbers though the vast majority of the powers is raising a particular number to 0'th power.
example n=9:
9 = 2^0 * 3^2 * 5^0 * 7^0 .......
Now as p/q has no common factor: Both can be written in the form
p = (r0^a0) * (r1^a1) * (r2^a2) *.....
Where the series of a0, a1, a2 ... an is the series of the prime numbers.
All the a1....an are whole positive numbers.
Now as p and q do not have a common factor m = p/q might be written as:
n=p/q = (r0 ^a0) * (r1^a1) * (r2^a2) * (r3^a3) * ....
where every a is a whole number, but may be negative. Each prime number appear once, and only once. In case of an ax appearing twice, they can be consolidated by addition of the powers - if 5^2 is a common factor for p and q then n will have 5^2 * 5^-2 = 5^(2-2) = 5^0 = 1.
If n has one and only one nonzero ax then n^(1/ax) is a prime number. If n can be written with only one ax (rest of a's being = 0) = 1, then n is prime.
I don't know if that makes sense to You or is relevant?
Thank you very much sir❤
I have a question that if a number is written in p/q form and p divides a polynomial's constant term and q divides its leading content then is it its rational root?
Hmm..... very inspirational!
Now any prime numbered root of a rational number is irrational - as far as I recall.
2^(½) is irrational.
But what about a polynomial that has (x^2 - 2) as a "root"?
Good day Dr. Peynam, I was wondering if you take certain math questions from viewers or commenters? I'm a math undergrad, and i'm currently trying to work out a limit question. Its for my own practice, not a project or assignment, and I can't seem to find any like it online.
It asks to the determine the limit of f(x) = x*[1/x], where [ ] is the greatest integer function ( floor function ), as x approaches 0.
The only aid it gives us is to examine the values of f(x) on the intervals of the form
1/(n+1) < x < 1/n where n is an integer.
For reference:
This question is taken from *Pearson International Edition CALCULUS EARLY TRANSCENDENTALS 7th EDITION*
Page 89 Q73
By Edwards & Penny
Use the fact that [1/x] is between 1/x and (1/x)+1 and then use the squeeze theorem
@@drpeyam Oh wow! Thank you so much! This notification came just as I figured out my own solution. Though yours is more straightforward and simple.
In my approach, I noted that:
if 1/(n+1) < x < 1/n..........(1)
then
(n+1) > 1/x > n..................(2)
And as x approaches 0/ |x| INFINITY
By definition, n+1 > [1/x] >=n, for some integer n.
(multiply by x throughout)
x(n+1)>= x[1/x] >= xn...........(3)
from the inequalities (1)and (2), our (3) becomes
(n+1)/n >= x(n+1)>= x[1/x] >= xn >= n/(n+1)........(4)
But again, from inequality (1) and (2), n < 1/x and 1/(n+1) < x.
Therefore, (n+1)/n = 1/( n(1/n+1) ) = 1/(1/x)x = 1.
So the extremes of (4) can be simplified to
1 >= x[1/x] >= 1
Applying the squeeze theorem yields the limit as 1 as x approaches 0.
.....
Really long and complicated I know. It just shows I still have a long way to go :P
Thanks again for the response Dr. Peynam, and have a nice day.
Elegant
thanks
best explanation, keep the good work
Beautiful !
Does your proof suppose the leading term and constant term are relatively prime?
No because you can always reduce the fraction
@@drpeyam you can reduce the entire polynomial by multiplying each term by a scalar, but you can't simply reduce the leading and trailing terms. Or is there something I'm not seeing?
Hermosos, maravilloso...
Sounds sufficient 👌
Where my irrational root theorem at
Dr Peyam can you please send the notes which you use for these proofs? Thank you
sites.uci.edu/ptabrizi/math140asp20/
@@drpeyam Thanks a lot!!
@@drpeyam Thanks !
salamatt
( Mona e to the x Lisa ) I m missing it
I'v been told if you comment this video early, Dr Peyam says hi
Uploaded 1 minute ago but comments from a month ago
It's gauss theorem
yayyy