Rational Roots Proof

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Күн бұрын

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

Пікірлер: 57

@CliffStamp 4 жыл бұрын

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.

@roberttelarket4934 4 жыл бұрын

Maybe nowadays. In the 1960's they never proved this in high school. You had to wait for undergraduate abstract algebra.

@hiwrenhere 3 жыл бұрын

@@roberttelarket4934 this is definitely not normal to see proved in high school

@roberttelarket4934 3 жыл бұрын

@@hiwrenhere: I love your name Integral Domain!!

@soupy5890 2 жыл бұрын

my school doesn't prove anything outside of Euclidean geometry

@garyhuntress6871 4 жыл бұрын

That was a rock star proof. I've never seen that before (in engineering) and I loved it!

@cendyywarlos 4 жыл бұрын

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

@rafadarkside9029 4 жыл бұрын

It's not a horrible proof. It's very elegant. This theorem is very useful when you combine it with Bézout's theorem.

@Illuminous_ 3 ай бұрын

he's so enthusiastic, this eases assimilation

@shayanmoosavi9139 4 жыл бұрын

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?

@eliyasne9695 4 жыл бұрын

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.

@drpeyam 4 жыл бұрын

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

@miro.s 4 жыл бұрын

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.

@forthrightgambitia1032 2 жыл бұрын

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.

@SpoonPhysics 4 жыл бұрын

Would love to see some examples of using a Green's function

@miro.s 4 жыл бұрын

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.

@yashuppot3214 4 жыл бұрын

At *5:30* you can also just look at that mod q and mod p and your are done.

@mtmpro7406 2 жыл бұрын

Your cheerfulness made my day ngl!

@utuberaj60 Жыл бұрын

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👋👌

@manuelfalzoialcantara92 3 жыл бұрын

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

@plaustrarius 4 жыл бұрын

Awesome so excited to watch thank you Dr. Peyam!!

@danielnelson6131 4 ай бұрын

Awesome video, thanks!

@Bedoroski 11 ай бұрын

Great explanation. Thx ❤

@thedeathofbirth0763 Жыл бұрын

Man, this guy is amazing! Thank you!

@daniellejdevlin8882 2 жыл бұрын

That was an amazing proof!! Thank you!

@6c15adamsconradwilliam3 4 жыл бұрын

Amazing! 👏🙂💪

@rajiv1625 Жыл бұрын

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?

@TheForever119 4 жыл бұрын

keep going Sir nice we are waitinf all amalyisis lectures

@vaginalarthritis1753 4 жыл бұрын

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

@drpeyam 4 жыл бұрын

Use the fact that [1/x] is between 1/x and (1/x)+1 and then use the squeeze theorem

@vaginalarthritis1753 4 жыл бұрын

@@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.

@sauravthegreat8533 4 жыл бұрын

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

@MrBeen992 3 жыл бұрын

Dr Peyam just did that proof this week.

@thomasborgsmidt9801 4 жыл бұрын

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"?

@jimnewton4534 9 ай бұрын

Does your proof suppose the leading term and constant term are relatively prime?

@drpeyam 9 ай бұрын

No because you can always reduce the fraction

@jimnewton4534 9 ай бұрын

@@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?

@omargaber3122 3 жыл бұрын

Thank you doctor ,make alot vedios like this please and about number theory

@aviraljain465 2 жыл бұрын

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

@thomasborgsmidt9801 4 жыл бұрын

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?