Proof of Rational Root Theorem

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Prime Newtons

6 ай бұрын

In this video, I proved the rational root theorem using basic divisibility concepts

Пікірлер: 56

@BackflipsBen 6 ай бұрын

When I was 14-15 and doing quadratics in highschool, while everyone would go through the trouble of using the long quadratic formula, I would just factor the equations in my head. It was a good (and mostly correct) assumption at the time that the problems we were given had rational real solutions, so I saved myself a lot of time. I don't remember how I picked up the factoring trick, must have been one of my teachers back then, but this video definitely shined some light on why that trick worked.

@iMíccoli 6 ай бұрын

Yeah unfortunately I was one of those ppl that only used the quadratic formula but this year I'm avoiding it, I've learned some factoring tricks and it helps a lot 😅, it's more satisfying to solve equations with factoring instead of just plugging in a formula.

@MrDynamite110 6 ай бұрын

Very good proof. I always took the rational roots theorem for granted so it's nice seeing a proof for it. Even better when the proof is easy to understand. Good job, you're one of my favorite youtube math teachers

@eduardoteixeira869 6 ай бұрын

I knew this theorem, but I never have seen the proof of it. Thank you very much. Nice demonstration.

@souravpaul9021 4 ай бұрын

So nice presentation, take love from Bangladesh ❤️

@mazenzidieh 6 ай бұрын

Thank you... Nice presentation

@punditgi 6 ай бұрын

Prime Newtons is the Math Master! 🎉😊

@BabySeahorse420 6 ай бұрын

Ever since my freshman year of high school (when I took Algebra 2), I wondered why the p/q rule worked. This is the closest I’ve ever got to understanding it. Thank you!

@diegocabrales 6 ай бұрын

You forget to say that Cn, C0 ≠ 0 (since if they're equal to 0, then all its possible divisors are infinite, therefore having undetermined rational roots). By the way, you didn't formulate root rational theorem. It says that given a polynomial p(x) = Cn*x^n + Cn-1*x^(n - 1) + ... + C0 with all Ci being integers, where Cn, C0 ≠ 0, if it has rational roots, i.e., of the form x = a/b, where a, and b are non-zero integers, and relatively prime to each other [that means that gcd(a,b) = 1], then they satisfy that a | C0 (read it as "a divides C0"), and b | Cn (hence why they're non-zero integers). We search all possible rational roots of p(x) finding all possible divisors of C0, and Cn (i.e., all possible values of a, and b), and finding which rational numbers a/b make p(x) = 0 [the ones that make it are the rational roots of p(x)]. By the way again, it's not correct that given three integers a, b, and c, if a | bc, but a ∤ b (read it as "a does not divide b"), or a ∤ c, then a | c (for the first case), or a | b (for the second case). For example, we know that 6 = 2*3, so clearly 6 | 2*3. However, 6 ∤ 2, and 6 ∤ 3. What is true is that if a | bc, and gcd(a,b) = 1 (that means a, and b are relatively prime to each other), or gcd(a,c) = 1 (that means a, and c are relatively prime to each other), then a | c (for the first case), or a | b (for the second case). That's what's called Euclid's lemma, which we can prove by, e.g., Bézout's identity. Bézout's identity: Let a, and b be integers with gcd(a,b) = d. Then there exist integers x and y such that ax + by = d. Moreover, the integers of the form az + bt are exactly the multiples of d. You can search on Wikipedia about Euclid's lemma, and Bézout's identity for more information about them, if you're not familiar with them. So when we have Cn*a^n = -b[C0*b^(n - 1) + C1*a*b^(n - 2) + ... + Cn-1*a^(n - 1)] we can rewrite it as Cn*a^n = bk where k is an integer (since Ci, a, and b are also integers, and any integer combination of them in sums, substractions, and powers are also integers). That means that b | Cn*a^n. Now, because gcd(a,b) = 1, then it also follows that gcd(a^n,b) = 1, so by Euclid's lemma, b | Cn. Same thing happens with C0*b^n = -a[Cn*a^(n - 1) + Cn-1*a^(n - 2)*b + ... + C1*b^(n - 1)] which we can rewrite as C0*b^n = ak' where k' is an integer (by the same reason as with k), so a | C0*b^n. Now, because gcd(a,b) = 1, then it also follows that gcd(a,b^n) = 1, which means by Euclid's lemma that a | C0. Then we have just proved rational root theorem.

@Maths_3.1415 6 ай бұрын

He also did the same thing bro

@diegocabrales 6 ай бұрын

@@Maths_3.1415No, he did not. He didn't say that Cn, and C0 cannot be zero because then rational roots would be undetermined. He didn't formulate rational root theorem but gave an attempt of solving it. He didn't say that a, and b are non-zero integers nor why they must be non-zero integers (but I've done so). He didn't explain well why a divides C0, and b divides Cn. Saying that as a | bc, but a ∤ b , then a | c is not a valid answer since that's not true in general, as I've shown with a counterexample (I'm generalizing to any similar case, and not giving only attention to the attempt of the video). Euclid's lemma needs to be well explained. I think I'm not missing any mistakes. If I am, I would thank the notification. A correct proof needs to be rigorous, writing/saying correctly the theorems/lemmas/etc., and without affirmations that are not true.

@Maths_3.1415 6 ай бұрын

@@diegocabrales Bro he clearly mentioned that the gcd(a,b)=1 by writing a,b are relatively prime. if gcd(a,b)=1 and a|(b^n)*Co then a|Co. But he didn't give a rigorous mathematical statement and proof because it's very obvious number theory result ○To Prove: if a|bc, gcd(a,b)=1 then a|c where (a,b,c)∈ℤ\{0} ○Proof:a|bc ⇒ bc=ak, k∈ℤ\{0}-----(1) gcd(a,b)=1 [given] ⇒gcd(ac,bc)=c ⇒gcd(ac,ak)=c [from (1)] ⇒gcd(c,k)=c/a Since gcd(c,k)∈ℕ⇒a|c ■ You can find rigorous proof of Rational Root Theorem on Math Stack Exchange. He just provided a simple proof without using mathematical symbols ●And at 14:11 he should have written it like this (a,b)∈ℤ\{0} where gcd(a,b)=1

@diegocabrales 6 ай бұрын

@@Maths_3.1415He mentioned that gcd(a,b) = 1, but he didn't use it in the proving attempt. He has just said that since a | bc, but a ∤ b, then a | c. Which is wrong, as I've just shown. By the way, you forgot to say that since gcd(a,b) = 1, then it also follows that gcd(a,b^n) = 1. This last statement plus this other one: a | C0*b^n, allow us to say that a | C0.

@Maths_3.1415 6 ай бұрын

@@diegocabrales now i understand what you were saying Yes you are right bro ;) Ya i forgot to mention that gcd(a,b)=gcd(a,b^n)=1 By the way what is your age? Are you preparing for any math competition? Which country are you from? The proof you wrote was nice and rigorous

@TranquilSeaOfMath 6 ай бұрын

Nice proof. You have a presentation that high school level and above should be able to follow. Congratulations on your subscriber achievement 🎉.

@iMíccoli 6 ай бұрын

Very nice and simple proof, now i can use this theorem while knowing why it works, good explanation teacher.👌

@axeitor 6 ай бұрын

Very interesting! I liked this proof

@kuber2455 6 ай бұрын

Nice explanation. Thank you :)

@dreamcastgh0st477 6 ай бұрын

good video! minor correction, though: you said that if a doesn't divide b, then a won't divide any powers of b. this isn't exactly true-- for example, take a=4 and b=6. a will divide b^2 = 36. you'll notice in my example though that a and b AREN'T coprime-- the reason this works is because they both share a prime factor, in this case 2. if the two numbers are coprime, like in this proof, then it's true that no power will ever be a multiple. just figured i should point this out so nobody picks up the wrong idea!

@jeskov8tt Ай бұрын

Are 4 and 6 relatively prime?

@HuntDiscovers 6 ай бұрын

Good stuff!!!

@danielc.martin1574 5 ай бұрын

Just simple but beautiful theorem

@javaflux6059 5 ай бұрын

Excellent video

@ingiford175 6 ай бұрын

Good proof, I have seen it used a lot, but not seen it proven that much in comparison.

@Arkapravo 3 ай бұрын

WoW - I never knew of this theorem!

@davidbrisbane7206 6 ай бұрын

If you apply Decartes' Rule of Signs to 2x² - 3x +6 = 0, you discover that this equation has two real roots or two complex root and no negative roots, so this can be used to eliminate half the tests (i.e. all the negative possibilities) when applying the Rational Root theorem.

@ingiford175 6 ай бұрын

Yep, knowing the (max) count of the types of roots is always a good start. There is a more complex version which I only read about, but do not know how to use called Sturms Theorem to get good approximations of roots.

@davidbrisbane7206 6 ай бұрын

@@ingiford175 Strums theorem only confirms how many positive and negative roots a polynomial actually has. I guess if you use it to get bounds on where each of the roots might like then you could use the Newton-Raphson theorem to approximate each vale of the real roots

@ingiford175 6 ай бұрын

@@davidbrisbane7206 So it removes the complex possibilities from the count?

@davidbrisbane7206 6 ай бұрын

@@ingiford175 Yes. I only used the theorem once, but basically, you are looking at ranges of overlapping real numbers to determine that the real roots are in these ranges

@ingiford175 6 ай бұрын

@@davidbrisbane7206 Its like 4th on the list of things i am going over. picked p 2 books on spherical trig and that is my current reading. Then back to Jordan algerbras and such...

@zandergall9895 6 ай бұрын

I don’t think a and b being only natural numbers work as the root can be negative, however still a great proof

@Samir-zb3xk 6 ай бұрын

He couldve said a,b are integers but b≠0

@diegocabrales 6 ай бұрын

@@Samir-zb3xkBoth a, and b must be ≠ 0, for they are the divisors of C0, and Cn.

@Samir-zb3xk 6 ай бұрын

@@diegocabrales yea thats true

@WagesOfDestruction 3 ай бұрын

great job

@AzmiTabish 6 ай бұрын

Thank you, Sir

@Tenorsax333 6 ай бұрын

<a href="#" class="seekto" data-time="520">8:40</a>: You haven't even formulated the theorem yet! You just stated the requirements, and then you said that a rational number r is represented as a fraction of two coprime numbers a and b and should be a zero of the polynomial. But that's not what the rational roots theorem consists of. You have ignored the main thesis (which you then prove very nicely), namely: a divides the first coefficient C(0) and b divides the last coefficient C(n).

@calculus988 6 ай бұрын

Prime Newtons, you should do a video on the proof of the descartes rule of signs

@QuickStories_123 6 ай бұрын

Sir prove basic formula and make a video about it

@arnavnautiyal3039 6 ай бұрын

hey there i like your videos.could you please make a video on pentation just like you made one on tetration i would really appreciate it.

@jacobgoldman5780 6 ай бұрын

Technically a,b are Z/0 since we might have a rational root which is a negative number.

@PrimeNewtons 6 ай бұрын

That's correct

@Alians0108 6 ай бұрын

At the start, if you just make the x^2 coefficient to 1, does that decrease the number of possible roots by the rational room theorem? (assuming a and c aren't coprime) x^2+-3/2x+3=0 only has -1, 1, -3, 3 as possible roots, right? In this case, none of them are of course.

@davidbrisbane7206 6 ай бұрын

To apply the Rational Root theorem, you need integer coefficients.

@johnnolen8338 6 ай бұрын

I learned this as a rule tho not as a theorem while studying pre-calc in the 11th grade. Years later I took a course in Elementary Number Theory but still didn't learn the rational roots rule as a theorem. Today I learned something that I never knew before. 🤠 Thanks, Newton.