Proof: Square Root of 3 is Irrational

[@BenjiShock]

I read this proof in "Understanding Analysis" and he did not mention anything about Euclids Lemma so I was confused. Apparantly Euclids Lemma is something to be taken for granted for Mathematicians but for myself it was actually the most important part of the proof. Thanks so much for explaining it in full length! And I also learned about Euclids Lemma at the same time!

[@WrathofMath]

Oh that's funny, I was reading Understanding Analysis at a diner last night! Does it have a proof that sqrt(3) is irrational? I don't think I saw one in my copy, maybe it's a different edition. Anyhow, I believe I too have seen Euclid's lemma entirely ignored in explanations of this proof - I don't get it, I always try to be thorough! Glad it helped, and by "taken for granted" you mean it's taken for granted that the reader is familiar with it? That may be true, especially in an analysis text, but we can prove Euclid's lemma using the Euclid algorithm - so in that way we don't have to take it for granted. A number theory book would have such a proof, but not an analysis book.

[@BenjiShock]

@@WrathofMath Well I was trying to teach it to myself coming from an Engineering Degree just because I felt like I was losing my abilities in maths and I did not want to give that up and I saw the book recommended. Maybe I should start with a book that is slightly easier haha. Btw the proof was for sqrt(2) but the first Exercise in the book is actually about proving sqrt(3) and sqrt(6). There are solutions somewhere in the internet from Stephen Abbott as well so thats where I got the solution from, which I did not understand because he just skipped the explanation with Euclids Lemma. And thats what I meant with taken for granted just that he didnt mention it in the proof.

[@Pete-Prolly]

​@@BenjiShockah! That's the Springer book I'm currently using for ℝ Analysis! (I have the Springer book by Kenneth Ross too.) 🤓

[@AuburnKamstra]

I’m actually watching this video to see where I went wrong after accidentally proving that the square root of 4 was irrational, doing an exercise from the same book. I still don’t know where I went wrong, but at least I know I got the first question right

[@91JLovesDisney]

My math teacher is great, but she had about 30 seconds to explain this so your video came in very handy. Thank you, it's a neat proof

[@mikeychrisanthus9948]

So, this also proves that the square root of any integer that isn’t a perfect square is irrational. Wow.

[@eequaled]

Yeah. basically

[@lyingcat9022]

Not rigorously I don’t believe so. It seems you could use this to prove them individually. But to prove them all maybe you could use this in some mathematical induction proof in a more general form? One that excludes perfect squares.

[@eequaled]

@lyingcat9022 you can. Just put any integer "n" . n≠k². It will hold out

[@eequaled]

@@lyingcat9022 i used it and proved it so yeah

[@Ramkabharosa]

First observe that 1

[@jaya.5087]

The production quality is amazing and you explain so clearly, keep it up!

[@WrathofMath]

Thanks a lot Joseph! I will keep making the best math lessons I can, and occasionally I like to dabble in other sorts of math videos as well, check out my newest rap if you haven’t already! kzbin.info/www/bejne/iYPQZZydeNl9eck

[@jussef2057]

Right now was the perfect time for me to watch this video. Pre christmas vibes are kicking in and I have to do exactly this prove for my discrete mathematics class. Thanks a lot!

[@WrathofMath]

My pleasure, thanks for watching! I actually just put up my Christmas set for this year, so you can look forward to more beautiful Christmas lessons! Let me know if you ever have any video requests!

[@Xebtria]

Does that mean, that this proof by contradiction basically works for every square root of n if n is a prime number? Because we always will get to the the same exact point that a and b must be integer multiples of n, contradicting the original assumption of having a gcd of one, and therefore all square roots of primes must be irrational.

[@MarcusViniciusSilvaDaRosa]

Exactly

[@noobchickensupper6471]

Without assuming at first a and b are co prime, can we say at the end that since a and b are always in the form divisible by 3, they are not integers as the ratio of two integers can always be reduced to the ratio of two co prime numbers.

[@nxtgamming3510]

Legend is learning 1 day before exam

[@WrathofMath]

Good luck!

[@rossk4770]

Hey, I like this explanation but it’s bugging me that it seems if you plug root 4 instead of root 3 at the beginning, you will end up at the same result; 4 divides a and divides b, so gcd assumption is violated, so root4 is irrational(which it’s clearly not). Am I messing up somewhere? Edit: I was missing Euclids lemma, which means you can’t get 4 divides a from 4 divides a^2 bc 4 isn’t prime(example a^2=36, a=6)

[@JaybeePenaflor]

Yup, and Euclid's Lemma is key here. Euclid's lemma states that if a prime p divides the product ab of two integers a and b, then p must divide at least one of those integers a or b. For the purposes of the video, 3 | ab implies that 3 | a or 3 | b or both. For example, 3 | 18 = (6)(3) and clearly 3 | 3 or 3 | 6. The or in this case is an inclusive OR. Or 3 | 18 = (9)(2) shows that 3 | 9. However, this is not the case for composite numbers. 6 | 18 = (6)(3). Here, 6 | 6 but 6 does not divide 3. Or if we express 18 = (2)(9), 6 does not divide both 9 and 2.

[@vinodb1008]

Kindly clarify if 3 divides p square then p divides p also

[@punditgi]

Very well explained! So, can you show us the proof that the square root of any prime number is irrational? We have to use the Fundamental Theorem of Arithmetic?

[@WrathofMath]

Thank you! And sure, thanks for the request, I'll try to do it soon :) We could use the fundamental theorem of arithmetic, but really all we need is Euclid's lemma again, which does not rely on the fundamental theorem of arithmetic.

[@punditgi]

@@WrathofMath Cool. I also saw a video from Mathologer that uses the Rational Roots Theorem to take whole radical expressions, turn them into equations, then determine if there are any valid rational roots. Very well done, but I'd love to see your explanation of that sort of approach. Thanks in advance and I love all your other videos! 💖👍

[@abdullahshaikh6623]

Sir! it was very well explained, we have a request can you explain that (2+2 root 5) is an irrational number

[@kenferman7602]

Thank you so much for this wonderful explanation

[@WrathofMath]

My pleasure, thanks for watching!

[@baked_potato_0-0]

Wrath of Math more like Wrath on me Math

[@ariananapal2959]

Best explanation so far

[@WrathofMath]

Glad it was clear, thanks for watching!

[@thehummelballkickers4485]

I just discovered your channel and I enjoy your explanations... Can you recommend any books for number theory and proofs?

[@WrathofMath]

I haven't read much from number theory texts specifically. I can say that I want An Introduction to the Theory of Numbers by GH Hardy, and have heard very good things about that book. For my own experience, I have read a good chunk of Number Theory by George Andrews, which is also good, although it is far from the easiest reading I have experienced in mathematics. It's a cheap textbook though, definitely worth it! I don't know if it's considered a classic in the field, but Andrews is a helluva mathematician! Oh and for proofs, I think Book of Proof by Richard Hammack is quickly becoming more of a standard text for the subject. If you look it up you can find it for free in PDF form, offered by official sources. It's physical edition is BEAUTIFUL though. Nice big print, and not too expensive. Also, Jay Cummings wrote "a long form mathematics textbook" on Real Analysis which is one of the finest textbooks I have ever read, and he is releasing a new textbook on Proof early next year, he claims in January, so keep your eyes peeled for that! I for one sure am excited!

[@kuei4604]

Thanks for the explanation, but I have a question.I think this proof is a little be weird since you can also prove that square root of 4 9 16 25... is irrational with the very same technique , which is obviously wrong.

[@WrathofMath]

Thanks for watching! In those cases, the proof would fall apart at the section where we conclude 3 divides a*a. We go from there to saying 3 must divide a, but we can only make this conclusion because 3 is prime. For example, if we knew 4 divided a*a, it doesn’t follow that 4 divides a, since a could have a factor of 2, and thus a*a has a factor of 4. Like, a = 6 for example.

[@kuei4604]

​@@WrathofMathYeah exactly. I figured it out soon afterwards XD.The different part is that p^2 has a factor : 4 does not imply p has a factor : 4, instead we can only be sure that p has a factor : 2 Just like your explanation above. Same for the 9 16 25 36... case. Thanks for the quick reply and great explanation again ❤.

[@trevtronix9338]

Thank you, great explanation!

[@WrathofMath]

Glad it helped, thanks for watching!

[@ChethanSG06]

Can you deduce sq root 9 and prove it as rational in the same way?

[@MarcusViniciusSilvaDaRosa]

sqrt(9) = 3, and 3 is a natural number. All natural numbers are also rational, so sqrt(9) is rational. Other way to look at it as reconize 3 as 3/1, a fraction between two integers, thus a rational number

[@migfed]

You are doing such a great videos, so interesting

[@WrathofMath]

Thanks a lot! Let me know if you ever have any requests!

[@muhammadumarsarfraz4705]

Thank you sooo much Sir 😊😊 Best lecture 😄

[@WrathofMath]

My pleasure! Thanks for watching!

[@nayefmsh8274]

This video actually felt fun listening to

[@WrathofMath]

Thank you!

[@jayceel251]

Thank you!

[@WrathofMath]

You're welcome!

[@mahmoudalbahar1641]

Thanks for this good video....please continue making such good videos.

[@WrathofMath]

You're welcome and thanks for watching! More are on their way!

[@gwenaellestelvana1397]

Thanks so much sir😢

[@ParsiaZahedi]

Could you please do the same as Square root of 8 is irrational first we assume that is rational and do the steps??

[@mikeychrisanthus9948]

You can do the exact same process that he did, just replace the 3’s with 8’s and carry through as shown. Alternatively, root(8) is root(4*2) which is 2 * root(2), which we already know is irrational.

[@timwhite7127]

You didn't need to go to all this trouble as I would have taken your word for it...

[@GomezUnplugged]

Man u r soo god I watched 7 videos but only now u made me understand thaks

[@WrathofMath]

So glad to help - thanks for watching!

[@aviationandminecraftpro6404]

thanks for this explantation :)

[@WrathofMath]

My pleasure, thanks for watching!

[@tewahedootube]

Damn u sound like Sheldon fr Good explanation tho

[@mudarisrinibas1470]

It would be easy enough if we use rational zeros law.

[@AMINE-dd1qy]

thanks! helped me alot

[@WrathofMath]

Glad to hear it!

[@munishsharma3801]

Bro ur the best teacher

[@WrathofMath]

Thank you!

[@sumitT-ti5if]

Thank you bro

[@WrathofMath]

Welcome!

[@lipichandra5438]

Could you explain polynomial and trigonometry topics as well? Pleaseeeeeeee 🥺🥺

[@madeleinezahn2870]

Great video! KZbin is what's carrying me through my discrete math class this quarter 😂

[@WrathofMath]

Thank you! Good luck with the rest of the course and let me know if you ever have any questions!

[@noobchickensupper6471]

Why do we need to assume that it is a fully reduced fraction? Rational numbers dont always appear as fully reduced fractions... Aah .. This proof bother me so much... It seems to have so many holes

[@WrathofMath]

Rational numbers don't always appear as fully reduced fractions, but they always can appear that way. There is no such thing as a fraction not in reduced form but that also cannot be put in reduced form, so we assume our fraction has been put in reduced form because it is a more useful form to have. Just as having the equation x - a = 0 assures us that x = a, having a rational number a/b assures us that the same number can also be expressed in a fully reduced form which may be a/b, but may be some different looking fraction c/d.

[@noobchickensupper6471]

@@WrathofMath thank you so much for replying... So by comcluding that our assumption was wrong we can say that √3 cannot be put as a ratio of coprime integers?

[@noobchickensupper6471]

@@WrathofMath by the way can we show taht it is irrational without first assuming that they are coprime... Then we can say that as a and b or the numerator and denominator will always have a common factor 3 √3 is therefore irrational? Like how vsauce has done it with his proof of √2 being irrational?

[@noobchickensupper6471]

@@WrathofMath and just because we assume that does not make it fully reduced fraction right? So if at the end we find out that they still have common factors that does not make it irrational... It is just that assumption was wrong and maybe we need to reduced it further right?

[@Chimangonyirenda-o5s]

Clarify where you explained about 3k for some integer k

[@MarcusViniciusSilvaDaRosa]

If the the numbers is a multiple of 3, then you kust be able to write it as 3 times some smaller integer. Just that

[@noobchickensupper6471]

Can we proof that √4 is a rational number... Using the same method?

[@WrathofMath]

Thanks for watching, I'm not sure what you mean. Showing sqrt(4) is rational is just a computation. We know sqrt(4) = sqrt(2^2) = 2. Since 2 is rational, sqrt(4) is rational.

[@noobchickensupper6471]

@@WrathofMath Thank you for replying. What I mean is can we use the same method to show that √4 is a rational number? Or there is no need to prove?

[@noobchickensupper6471]

In other words.. Will this way of proving holds true for the fact that there exists a rational number whose square is 4

[@noobchickensupper6471]

So why do we need to prove √3 is irrational? According to you we can also say that √3 = 1.732050807.....which is non terminating and not repetitive and cannot be expressed in the form a/b where both b are....... And so on... Couldn't we just write like this as well then... ?

[@acosusnjar4397]

@@noobchickensupper6471 We cant since Euclids theorem wont work since 4 and 9 are not prime numbers

[@bhaitlauncherwala8770]

The wrath of math saved me from the mercy of the liberal arts (Better to reign in hell, than serve in heaven)

[@Anshika-ij7ii]

This video is really nic. Bt it didn't get those many views ❤️

[@WrathofMath]

Thank you! If you know anyone who would benefit from it, sharing it is a huge help! Let me know if you ever have any video requests!

[@Anshika-ij7ii]

Sure ❤️

[@paulduncan3214]

What would happen if you tried putting 4 instead of 2,3, etc .

[@Kittyyatansakul]

the square root of 4 is already rational, both 2 and 3 are irrational, the question knows that these numbers are irrational but wants us to prove that root of 2 or root of 3 is irrational You cannot take the root of 4 to prove it is irrational because it's not irrational

[@AuburnKamstra]

Would you happen to have a video showing where this proof fails in the face of a perfect square? I’m very very curious

[@WrathofMath]

sure! kzbin.info/www/bejne/gGbRqJ2qYq1nkMU

[@chandan1452]

Why do I think I know it from grade9

[@WrathofMath]

Thanks for watching and do you mean you learned this in 9th grade?

[@chandan1452]

Yup from my IIT text

[@solore7399]

Lol Indian education.....i hve to study this for 10th grade.

[@sareem7]

@@solore7399 Actually ,if you study IIT syllabus, you have to learn it in class 9 ✌🏻.

[@Alone_boy_36]

Bekar

[@cheatyhotbeef2636]

Well I just can't seem to get past the language barrier.

[@WrathofMath]

English isn't your first language?

[@mahmoudalbahar1641]

Cheaty Hotbeef My advice is you should immerse your self in English to be better by (listening+reading+talking )very much..... equivalently practice the language.......that of course needs persistence.

[@cheatyhotbeef2636]

@@WrathofMath It is, but math isn't D:

[@WrathofMath]

Boy! Maybe I should hire you to write my jokes 😂

[@Michael-yy2vl]

neat

[@Inerself2222]

Fun fact: if you think your first well refresh your page you will find your self the last😂😂

[@WrathofMath]

Is this comment addressed to people who comment "First!"?

:)))

[@WrathofMath]

Thanks for watching!