Sqrt of 2 is irrational[ Proof by Rational Root Theorem]

In this video, I showed that the square-root of 2 is irrational using the rational root theorem. This proof is easy to follow and uses a familiar concept in solving quadratic equations.

[@BartBuzz]

So much easier to prove rather than the usual proof by contradiction. Of course, now you should prove the rational root theorem 🤔

[@davidbrisbane7206]

By Descartes' Rule of Signs, then the equation x² - 2 = 0 has one positive real root and one negative real root 😀. The Rational Root theorem say that if x² - 2 = 0 has a rational root, then x = -1, -2, 1, or 2. On the other hand, we know from x² - 2 = 0, that x = √2 is a positive solution to this equation. But since x= √2 is not a possible rational solution to x² - 2 = 0, then we conclude that √2 is irrational.

[@mikevaldez7684]

Beautiful! Great exposition. One of the best mathematics instructors I've seen in all my years, both on the internet & all 6 colleges & University that I attended!!

[@StarThePony191]

Wonderful explanation! Always curious to find out why things are the way they are. Education, at least here, doesn't always do a good job at teaching us the why. But people like you truly show that, whatever happens, you never stop learning!

[@darma46]

your teaching way is very nice brother 🙂🙂

[@PrimeNewtons]

Thanks 😊

[@eduardoteixeira869]

Fantastic demonstration. I always have seen this demonstration by contradiction. Very interesting, thank you.

[@keithrobinson2941]

Nice proof. I guess the proof of contradiction that you briefly cited is more commonly used since it was the original proof, dating from the 5th or 6th century BC among the Pythagorean brotherhood. I don't think the Pythagoreans had access to the rational root theorem!

[@ahmedrafea8542]

Splendid. Thanks very much. Please never stop spreading mathematical wisdom all over those who are keen to learn math.

[@sajuvasu]

Wonder! Iam in high school and i have a pretty same thought as you sir..!! Teacher are saying 2 has no rational roots and we must call it irrational.. The proof(in the real analysis)is pretty simple.. But they aren't teaching it.... So happy to hear your thoughts... 🙂🙂😁

[@user-ti9xi1nt1i]

I sincerely wish you a happy new year sir

[@drChoosen]

Oh, very nice and clever. Thanks 👌

[@kuber2455]

Can you make a video that prove the rational root theorem?

[@f.r.y5857]

Rational : can be considered as p/q. eg 0.4444... = 4/9 0.232323... = 23/99. In calculator, the product of √2 is infinite unrepeated numbers, while 4/9 and 23/99 isn't.

[@user-dm5kh2mb2l]

nice😃 always love math!

[@FlexThoseMuscles]

thx for sharing!!

[@user-ds7po7kc7j]

i see a comment a buddy states that can you make a video about the prove of the rational root theorem, i agree with this budy. pls can you give a prove for this. thanks for this video.

[@lawrencejelsma8118]

Good proof. The rational root theorem for any degree polynomial were actually designed for other nth order integer or quotient polynomials but to find only if a positive integer or quotient p/q root exists. Then successive continued tests to find all rational number solutions. It doesn't hurt to look at all prime number representing factor quotients to see if they can find the root. However, the real solution in all ax^2 + bx + c = 0 integer or quotient answers come from ∆ defined as that inside the square root of the quadratic equation: ∆=b^2 - 4ac requirement that it forms a number of even power prime factors (not odf as you proved). That is, any positive ∆ = b^2-4ac is a (p1)^(2j)(p2)^(2k)(p3)^(2l) etc. prime number multiple of even powers. ... In your problem b^2 - 4ac was (0)^2 - 4(1)(-2) = 0 + 8 = 8. Good! A positive number result. But 8 = (2)(2)(2) = 2^3 and only prime number 2 to an odd power ... Not even ... Of 3 is what made this quadratic polynomial fail in the quadratic equation as being a possible integer or quotient root we'll be able to factor out in that quadratic equation example.

[@mathpro926]

Thank you

[@mkmathstutorials6645]

Thanks prime.

[@Tentin.Quarantino]

My first thought would be that this must work for all primes, meaning the square root of a prime is always irrational, right? Unless I’m missing something.

[@PrimeNewtons]

Correct

[@tomdekler9280]

It is a lovely video but I would have personally loved to see you first show when the rational root theorem _does_ work. Right now it feels like you have pulled out a theorem that is entirely designed to explain when and when something is rational, without explanation for how it was derived, and you've only shown that it disproves the most fundamentally basic example of something that isn't rational. I have no proof of this theorem or how I would satisfy it. Prove to me that sqrt(2.25) has rational roots.

[@tomctutor]

Very good, but RRTh can't be used to determine the irrationality of just any old number. Because there are an infinite set of real numbers that are not roots to any polynomial {P_n (x)} with integer (or rational) coefficients! e.g. let x = sin(1) or x = log(2) which are both definitely real numbers but don't zero any P(x) to my knowledge. So begs the question how can you determine the irrationality or otherwise of a random real number?

[@richardbraakman7469]

If you picked a real number at random then the probability of it being rational was 0 anyway :)

[@tomctutor]

@@richardbraakman7469 If we make a partition on the reals into rationals and irrationals, the by probability definitions P(irr)=1 - P(rat) = 1 or at least nearly 1? But it is a fact that between any two rationals we can find an irrational, and vica versa. So one might conclude there are as many rationals as irrationals between any two real numbers. Therefore I would claim P(irr) = P(rat) = 0.5 in any given interval on the reals.🤔

[@richardbraakman7469]

Picking at random from an infinite set might be an undefined operation anyway :) But I was basing my claim on the fact that the reals are a bigger infinity than the rationals, which makes them infinitely more dense on the number line.

[@xizar0rg]

@@tomctutor your assertion that "between any two irrationals there's a rational" suggests you're aware that Q is dense in R. The rest suggests you're unaware of measure theory. Given any compact interval in R, you can always find a countable set of intervals that contain all the rationals in that interval (trivially shown by noting that Q is countable). Thus Q has measure zero. Statistically speaking, there are no rational numbers. Why do you think we need infintesimals?

[@5Stars49]

r values also be 1/2 and -1/2 ???

[@dadoo6912]

no, there's only factors of leading coefficient in the denominator, which in this case is +-1

[@toastdog214]

well no, this equation doesn't even have rational roots. You can even just do difference of squares to find (x - √2)(x + √2) = 0

[@TeFurto777]

Cool

[@user-if5do8md6n]

this guy only need tow wear a eye patch he would be the lleader of the shield big fan sir

[@annaoaulinovna]

primes dont have rational number roots. they have single factors.

[@peterkiedron8949]

sqrt(2) is a meaningless symbol here. The RRT tells us that there are four possible rational roots: -1, +1, -2, +2. By plugging them in the equation we verify that neither of them is a root. Thus the root must be irrational according to RRT.

[@aurochrok634]

can be done without that theorem. let’s assume sqrt(2) = k/n, where k,n are mutually prime integers. squaring both sides and multiplying them by n^2, we will get 2*n^2 = k^2 , so k^2 is an even number. but then k is an even number, k = 2*l. so, 2*n^2 = 4*l^2 or n^2 = 2*l^2 and thus n must be an even number too. and that contradicts k and n being mutually prime numbers. so, sqrt(2) cannot be a rational number

[@dadoo6912]

everybody knows that, it's just a nice alternative method to prove that

[@random19911004]

Duh, that’s the standard method that everyone first sees

[@aurochrok634]

@@random19911004 so why involve the theorem at all? i just see it as an unnecessary complication . Occam’s razor etc

[@AurynBeorn]

It's just another way. By having it fresh in our minds, we may think of it to solve other problems.

[@aurochrok634]

@@AurynBeorn don’t get me wrong: i’m all for using the rational roots theorem (or any other theorem for that matter) when it provides the easiest, simplest etc solution to a problem . i just don’t think this is the case here, that’s all

[@harsh3198]

It’s false what if the roots are compex

[@GPSPYHGPSPYH-ds7gu]

Math is My Soul so I love Math PAZA M C69AoneA

[@mohamedelouajrachi66]

Folse

[@harsh3198]

Worst proof

[@artursowinski9402]

For me that is a very weak proof. With the ratio al root theorem you Can only show that a polynomial has no rational roots but that does not imply that this polynomial has irrational roots. You have to first show that a solution is possible and then use the ratio nal root theorem to conclude something. In this case its very easy to show that a solution exists Based on the darboux theorem.