Proof that Square Root 2 is Irrational

[@frepi]

I love black boards. Lately, I attended classes where the teacher used a computer screen projection and this was so bad for learning. Black boards force the teacher to teach at the same pace that he or she writes. It allows the students to take notes and ask questions.

[@bobbysteakhouse7022]

What about white boards???

[@gold9994]

@@bobbysteakhouse7022 chalk is much better, it's satisfying

[@targetinstitute7175]

kzbin.info/www/bejne/pauplGuKrK2pZ80

[@johnbell3621]

At school the teacher would have 2 options if you were disruptive: 1. Throw the chalk at you. 2. Throw the larger chalk rubber at you.

[@DianeRyanONeill]

Yes , Absolutely agree

[@omargoodman2999]

To all the people criticizing the assumption of an irreducible fraction, it's a non-issue because the canonical form of a fraction is equal to any multiple thereof. So, even if we assumed that sqrt(2) is equal to some fraction, c/d, that isn't in lowest terms, c/d can be reduced to a/b anyway. Therefore, sqrt(2) is still equal to a/b where a and b share no common factors. This is basic part of the definition of a *rational number* that originated in ancient Greece during the time of Pythagoras in around 6 BCE. The "Pythagorean Order" believed that all numbers were perfect and divine and that any number could be expressed as a ratio two integers. Even an infinitely repeating decimal like 0.333... can be expressed as 1/3. If you couldn't calculate it down to a perfect ratio of just two integers, you just hadn't calculated enough. It was this very proof of sqrt(2) that demonstrated that there were, indeed, numbers that didn't follow this perfect structure; numbers that were irrational (cannot be expressed as a ratio of integers). And later, still, it was found that you can even go a step further. Even for irrational numbers like sqrt(2), they found that you could still describe them using an algebraic formula, using algebraic operations; exponents, addition, subtraction, multiplication, and division. These are now called Algebraic Irrational Numbers. But there are some that don't even obey that paradigm. Transcendental Irrational Numbers like _e_ and _pi_ won't even take an algebraic formula; the formula would keep going on and on with an infinite number of terms.

[@jeremystanger1711]

You don't even have to get that technical. By induction, if such a fraction were reducible, it would have to be infinitely reducible. This is obviously nonsense, so we still have a contradiction.

[@tahititoutou3802]

Anyways, even if a and b have common factors, the proof still holds!

[@Chris-5318]

+Jeremy, no rational can be infinitely irreducible. For a/b both a and b are natural numbers and are necessarily finite (as are all the real numbers). if g = gcd(a,b) then replace the a and b with A = a/g and B = b/g -NB you have effectively claimed that g can be infinite.- Oops, I had misunderstood Jeremy.

[@Chris-5318]

+Thititoutou Wrong. If for a/b both had a factor of 2, then there wouldn't be a contradiction at the end of the step where we deduce that b must be even too.

[@Chris-5318]

+Jeremy, Ooops, I see that I misunderstood you. I had a brain fart. I'm sorry about that.

[@idontknow-ms8mc]

Great explanation! I was watching a lecture for another class and the instructor mentioned this as an example of proving by contradiction and he definitely didn't spend 6 minutes talking it out, so much appreciated. I just subscribed, haha.

[@Awaneeshmaths]

kzbin.info

[@AmeeraAmir-jj7oj]

زبنت غايا

[@AmeeraAmir-jj7oj]

ظ رنم نننهنننننتفا اث

[@AmeeraAmir-jj7oj]

بثىق

[@rajendranchockalingam1079]

Good morning madam Super explanation and very simple way to understand I am from Tamil Nadu. Thanks

[@amitgupta-si4xw]

Excellent explanation i understood more than any other video I watched

[@org_central]

Bahoot tez ho rhe 😂😂😂😂 Naughty baccha

[@RDDance]

Even me!

[@OyeCBBA]

@@org_central 😀😂

[@aif22]

Same

[@MrFeatre]

If I had a teacher like her, I would go for Math classes everyday..

[@themotivator1214]

Why

[@mr.habibbi3158]

@@themotivator1214 coz he tryina smash

[@blessedlubasi6653]

Got confused at some point but I finally got it, good video.

[@roninwarriorx4126]

I design the majority of my artwork in a root 2 rectangle. Phi is my favorite but 2 is easy.

[@Awaneeshmaths]

kzbin.info

[@sushmaverma6893]

You are from which country

[@donlodge1230]

This is the best description of this on KZbin. Thank you

[@samarjeetpal3869]

It was a very helpful video..I've been looking for explanations for this theorem, but I didn't understand any of them...thank u so much..

[@abdulhadi_abbasi7936]

Greatly explained mam .You highlighted each and every important point .Thank you very much .Your video widely helped me

[@donynam]

Let's use this method to prove that Square Root 3 is Irrational number.

[@niceguy4801]

What about if this logic applied in a rational number? Will it be true?

[@awaken6094]

I just tried it and it worked for me.. any help?

[@SARudra12]

this helped me a lot...she teaches really nicely...thank you Miss Tori Matta😊

[@alendebry3201]

¶

[@radiatingradian]

Thank you, I am just starting out with proofs at school and they are a bit confusing. This video explained it very well!

[@techosity]

You explain very smartly

[@gsssmustfapursurajsingh698]

EXCELLENT MATH TEACHER PRAISE WORTHY WORK

[@tejbahadur6616]

this just cant be true, so you are telling me that this is for college students ? i am an indian ,14 year old and this is in my first chapter of maths book

[@drexex0f]

😂

[@radhikasoni6231]

Best explaination. Being a ninth grader it's really helpful 👌👌

[@Jedimaster_is_CR]

Breh

[@himanabhdixit9747]

Hello madam You teach very well Love from India

[@quantumdevil5147]

Fantastic explaination 👍🏻👍🏻

[@ramya1758]

Very super mam your a good teacher of youtube and all videos super explanation is very good and this video is very useful of irrartion numbers thank u mam bye...

[@princendhlovu1874]

I now know how everything comes about, keep posting more

[@stephanund5206]

Let k=b/sqrt(2). In order to complete this process of proof successfully we HAVE TO assume that this expression for k is NOT an integer. Whether it is or not, we do not know at this point, and we won't find out beyond this point. To my opinion the here presented proof of sqrt(2) is rational did NOT fail. A vicious circle.

[@Surajgupta-dh3ri]

hlw

[@NASIR58able]

Well done Madam, Excellent methodology to explain. 🏅🏅🏅👌

[@amazingedits4980]

I am able to understand it more than anyone else

[@MamtaKumari-ct3kj]

Really it's so helpful....I can't expected that even I will understood your language or not but my expectation was wrong....😁 really it's so nice c video I have understood very well👍🏻 thnx so much

[@fasilmalik3027]

Brilliant explanation madam❤️👍👍

[@motopatalu2612]

You have solved √2 is an irrational number can not be written in form of rational number such as fraction form p/q well explained by you by contradiction method

[@awaken6094]

I tried this on √4 , and it still contradicted that the co primes have common factors

[@Kirsnkinder]

Root 4 is rational which can be reduced to 2x2 think

[@MuffinsAPlenty]

No, the argument falls apart for √4. If you arrived at a contradiction, you made a false claim in your argument somewhere. Likely, you claimed that if a^2 is divisible by 4, then a is divisible by 4. This is false.

[@humanrightsadvocate]

*3:10* Just because *(2n)² is even* doesn't mean that if *n² is even* than *n is even.* E.g. *n² = 2 (so, n² is even)* but then *n = √2 (therefore, n is not even)* Am I missing something here?

[@jcbcavalanche4558]

yes, n must be an integer - obviously. Otherwise you could sub in random decimals and ofc not come out with whole numbers let alone even whole numbers

[@fardeenbora8084]

Ma'am, instead using the argument that a and b must be even, I think it will be better if we use The fundamental theory of arithmetic which is applicable for all primes.

[@jelenajonjic]

Can u tell me how wolud that work? Tnq.

[@Ramesh-k4g]

use the concept of co-primes

[@fardeenbora8084]

@@Ramesh-k4g thanks

[@friedshrimpsareusuallybland]

@@jelenajonjic im late but there are two theorems. Theorem 1: If a is a natural number and p is a prime number, then if p divides a^2 then p also divides a. Theorem 2: If a and b are two natural numbers and p is a prime number, then if p divides ab then p divides a or p divides b or p divides both. You can apply it in the equation so as to prove that a/b indeed has a common factor other than 1, hence proving its not a rational number

[@loicboucher-dubuc4563]

then can it be written as a reducible fraction...?

[@GurshaanGaming]

I was not able to sleep so that’s why I am watching this vedio 🤭🤭🤭🤭😴 But now I am going to sleep

[@suavecreators1843]

😂😂

[@zanyzara8788]

Lol

[@blueworld1696]

But why 🤣?

[@IDMYM8]

*vedio*

[@yusratariq2186]

HAHAHA GO SLEEP DUDE

[@hkayakh]

Now explain why sqrt of 3 is irrational

[@yamkelamampoza7510]

It's been two year I am also waiting

[@ryanbutton8718]

Well put and easy to follow. Thank you.

[@sushmaverma6893]

You are from which country???

[@jlinkels]

Nice presentation. But I have to recommend to the lady that she unlearns to write the "2" as a delta

[@MrJason005]

An actual delta looks like this: δ I think you are referring to the partial derivative symbol, because, her 2, if looked at a certain way, does remind someone of ∂

[@AnasKhan-ff4yo]

Thanks mam this helps me lot in my examination

[@hqs9585]

The "no common factor" statement was presented as a given, however it should be explained or proof why is that the case , then the proof will follow.

[@magicbaboon6333]

What is the definition of rational. It is that it can be expressed as a ratio. Which is a/b in this case. If they are co prime then we can simplify until they are not

[@Grizzly01]

@@magicbaboon6333 Your last sentence has got the definition the wrong way around. Should read 'If they are _not_ coprime, then we can simplify until they are.'

[@rikkardo9359]

Why exactly can't a and b have any common factors? I tried the proof with n instead of 2 and it seems to work, "proving" that there are no rational numbers at all... My math is most certainly wrong, but please tell me how

[@MuffinsAPlenty]

Every rational number can be expressed in the form a/b where a and b have no common factors. You can always divide both the numerator and denominator by gcd(a,b), and you will have an equivalent fraction to the one you started with where the numerator and denominator are integers with no common factors. So while rational numbers don't _have to_ be written in reduced form, they always _can_ be. Starting with the reduced form makes the argument cleaner. Let's work through the argument with sqrt(n). Say sqrt(n) = a/b where a and b are positive integers which have no common factors. Squaring both sides and multiplying by b^2 gives us: nb^2 = a^2. Now here's where the argument breaks down for general n. Yes, a^2 is a multiple of n. However, this does not mean that a is a multiple of n. For example, let's say n = 4. It's possible for 4 to divide a square number without dividing the square root of that square number. For instance, 4 divides 36 = 6^2, but 4 does not divide 6. The argument works fine for n = 2. The argument actually works just fine for n being _prime_ since prime numbers have the property that if they divide a product of integers, they must divide at least one of the factors. But not every number has that property.

[@rikkardo9359]

@@MuffinsAPlenty Great explanation, thanks

[@arshia6619]

can we prove that √4 isn't an irrational number with this method?

[@ErNaveenKumarOfficial]

Can you explain root 16 is not a irrational no by this contradiction method ?

[@wantedgamer1972]

your explanation is awesome

[@noobchickensupper6471]

Why do thry need to be irreducible? Please can you explain

[@lorenzopombowulfes3903]

I hate that number 2. It looks like a comic sans font of Windows 95 AND can be easily confused with symbols like alpha or the curved 'd' of the partial derivative. How would she solve the negative gradient of a potential [alpha]/(r^2)?? ^.^ => -(2/2x * 2/r^2) -(2/2y * 2/r^2) -(2/2z * 2/r^2)

[@Chris-5318]

A small slip would make it look like a 3.

[@wanwisawonguparat6372]

I don't understand something in your proof. Why a and b haven't common factors? (I'm not good at English . Sorry about it.)

[@bashirahmadwani6501]

Easily explained mam. Thank you so much

[@ronnietoyco4421]

Nice Explanation. BTW if to Squared equivalence a squared & b squared equations ... could be 4b = a & 4k = b where; a/b = 1,,, 🤔

[@ratulchoudhury9144]

Ma'am why a and b are taken as coprime ? Please reply

[@demolition-man729]

Why can't you reduce a/b when it's rational

[@Anubhuti_Atmachintan]

Nice explanation I like your way to teach.

[@nicoleleung3177]

when u assume root(2) is a rational number, why a/b must be irreducible? is this part of the definition of rational number?

[@vishnurahul3378]

Yes the definition of a rational number is a fraction in its simplest terms. Even if this wasn't the case consider getting the simplest form of the fraction a/b here. Through the same process used in the video it can be shown that the simplest fraction of a/b can be divided further which is clearly absurd and not possible

[@highvoltage1393]

how come just cause a and b are even they are irrational? 4/2 is rational?

[@haileesteinfeld9996]

I think it's because in the first she assumed a and b do not have any common factors (relatively prime),and in your case the common factor is 2

[@pattyrick5479]

how would the result be any different if you were to put a perfect square under the radical, because then it is rational and if you were to continue the proof there would still be a contradiction saying it couldnt be rational

[@MuffinsAPlenty]

Let's go through the argument with 4 instead of 2. Suppose √4 is rational. Then √4 = a/b where a and b are integers and b is not 0. Square both sides to get 4 = a^2/b^2. Multiply both sides by b^2 to get 4b^2 = a^2 Now, the left hand side is divisible by 4. So the right hand side must also be divisible by 4. This means a^2 is divisible by 4. *Here's where things are different: we **_cannot_** conclude that a is divisible by 4 - the best we can do is conclude that a is divisible by 2* (I will explain why later) So a = 2c for some integer c. Then 4b^2 = (2c)^2 = 4c^2 Dividing both sides by 4, we get b^2 = c^2. Since b and c are both positive, we get b = c. So √4 = a/b = (2c)/b = (2b)/b = 2. And we get the actual answer instead of a contradiction. Now, why can we not conclude that a^2 is divisible by 4? Well, let's look at some examples. Suppose a = 2. Then a^2 = 4 is divisible by 4, but a = 2 is not divisible by 4. Or suppose a = 6. Then a^2 = 36 is divisible by 4, but a = 6 is not divisible by 4. So why does it work for 2 when it doesn't work in general? One way is to notice that 2 is a prime number. If n^2 is a perfect square which is divisible by a prime number p, then n must be divisible by p as well. You can see this by taking a prime factorization of n, and then squaring all of the factors to obtain a prime factorization of n^2. Since p is a prime dividing n^2, it follows that one of the prime factors in the prime factorization of n^2 is p. But the prime factors in the prime factorization of n^2 are the same (but appearing twice as many times) as the prime factors in the prime factorization of n. Therefore, p is a prime factor in the prime factorization of n. So n is divisible by p. More generally, by slightly modifying this argument, if m is a number which is not divisible by the square of any prime number and if n^2 is divisible by m, then n must also be divisible by m. So since 2 is a prime (more specifically, since 2 is not divisible by the square of any prime), we know that if 2 divides a^2, 2 must also divide a. The same thing is not true for 4 since 4 _is_ divisible by the square of a prime - namely 2.

[@habeebhussain3300]

Mam please solve 1+root 3

[@suba8936]

Excellent Teaching

[@rittenbrake1613]

I enjoy her voice

[@vaibhavmahore7246]

I also

[@aarshtiwari9889]

AWESOME EXPLANATION.... ❤

[@lifetimephysics8308]

Which standard maths are u teaching here¿

[@kusumpatel1981]

1.41 what did I hear

[@kiip7579]

Why a/b has no common factors? For a rational number, "no common factors" is not a compulsory condition. 4/6 is a rational number but it has a common factor of 2.

[@MuffinsAPlenty]

While it is not a compulsory condition, every rational number can be expressed in the form a/b where a and b have no common factors. You can always divide both the numerator and denominator by gcd(a,b), and you will have an equivalent fraction to the one you started with where the numerator and denominator are integers with no common factors. So while rational numbers don't have to be written in reduced form, they always _can_ be.

[@Grizzly01]

4/6 is not that fraction in its simplest (aka irreducible) form, which you should always strive for. Divide both the numerator and denominator by 2, and what do you get? 4/6 = 2/3 2 and 3 are coprime, so the fraction is now irreducible.

[@Grizzly01]

@@MuffinsAPlenty "So while rational numbers don't have to be written in reduced form, they always can be." I would say that they always _should_ be, unless there is a very compelling reason not to do so.

[@rkumaresh]

Good explanation.

[@ratulbanerjee8456]

Isn’t the proof hold for any irrational number which are square root of somthing

[@devanandshaji6573]

Thanks,hope this helps me 😊

[@misan2002]

So 3/6 is irrational? But Google says it isn't. Please help

[@Grizzly01]

Why would you think that 3/6 is irrational? It is both rational (by definition) and reducible (3/6 = 1/3)

[@misan2002]

@@Grizzly01 becase 3/6 isn't in its simplest form

[@Grizzly01]

@@misan2002 You are getting 'irreducible' and 'irrational' mixed up, aren't you?

[@misan2002]

@@Grizzly01 yes, maybe.

[@leosousa7404]

There are mistakes. sqrt(2) = a/b 2 = a^2/b^2 2*b^2 = a^2 if a^2 is a multiple of 2, that is, an odd number, and a is an intenger, then a is multiple of root of 2, which is rational like you assumed: a^2 = 2c a = sqrt(2)sqrt(c) however you can see here that a is not an odd number, since it's a multiple of root of 2, and not 2 is a an integer? 2b^2 = (sqrt(2)*sqrt(c))^2 2b^2 = (a/b *sqrt(c))^2 since 2*b^2 = a^2 then a^2 = (a/b * sqrt(c))^2 thus sqrt(c) = b so yes, a is an integer. Thus you haven't proven sqrt(2) is irrational. You cant assume a^2 is a multiple of 4 just because it is a multiple of 2. If you say an odd*odd = odd you are suddenly assuming a wrong statement without any explanation, invalidating the contradiction. Further on, if you assume 2n multiple for any multiple you can contradict Yourself with this result: say that a and b Are multiples of 2, they can't be equal each other otherwise a/b = 1, so you can still divide them by 2 until one of them has no common factor with the other, leading to a rational number that corresponds to the root of 2.

[@aaronbernal3189]

Proof: If a is a multiple of 2 then there exists k such that 2k=a, then squaring both sides 4k^2=a^2, then there exists k' (i.e k^2) such that 4k'=a^2 then, a^2 is a multiple of 4

[@Chris-5318]

+Leo You used a circular argument. There is no a and b such that √2 = a/b. Pretending that it is true for the purpose of argument doesn't make it actually be true. The proof really say that "IF √2 = a/b THEN ... contradiction". I could pretend that your name is Fred Smith, that doesn't mean that your name actually is Fred Smith (it doesn't mean that it isn't Fred Smith either).

[@Chris-5318]

+Leo. Euclid's lemma en.wikipedia.org/wiki/Euclid%27s_lemma says that if p is a prime and p|ab (p divides a times b) then p|a and/or p|b. Now, 2 is a prime and because 2b^2 = a^2, we have 2|a^2 and so 2|a. So if √2 = a/b, we also have that 2|a must also be true.

[@shalinishandilya7245]

Mam I have a problem _ what is the difference between a rational number and fraction? Please answer me mam

[@centerofmath]

Hi Shalini, A rational number is a type of fraction, although fractions also can describe things which are not rational numbers. Check out en.wikipedia.org/wiki/Fraction_(mathematics) for more information.

[@shalinishandilya7245]

***** thank you very much mam😃😃😃😃

[@silverruv6220]

It will come in board exam 100% garentee

[@kasperjoonatan6014]

what about square root of 3 ? how to prove that?

[@MuffinsAPlenty]

You can prove it in the same way :) The only difference is that instead of saying "a^2 must be even so a must also be even" you say that "a^2 must be divisible by 3 so a must also be divisible by 3." Since 3 is a prime, we know this is true. You can check by taking a prime factorization of a, and then squaring each of the factors to get a prime factorization of a^2. Since 3 is a prime dividing a^2, it must be a prime factor of a^2. But the primes appearing in the prime factorization of a^2 are precisely the primes appearing in the prime factorization of a (since you squared a prime factorization of a to get a prime factorization of a^2). Therefore, 3 must be a prime factor of a. So a is divisible by 3.

[@ayan701]

Thanks mam i was also finding this that why a is even but other tutors were explaing only by prime factors vision.

[@cagataytekin6372]

hocam buyukluk saka mi

[@tinula]

Thanks Madam. very useful

[@praneethasajja3457]

What is your hairstyle name Ma'am? It is awesome

[@xuan8641]

It's a bob cut

[@abdulhameedafridi9524]

That's a really good Explanation.

[@singhbalmiki6157]

Ma,am from where you are

[@MedYasserLarousi]

So clear explanation...

[@sandeshthapa2003]

Thank you very much ma'am

[@aravindan5091]

Excellent teaching

[@sureshsah6241]

Your video is really helpful for me

[@patrickwilliams7411]

Thanks for the video.

[@mnmathclasses8447]

Very nice Hi Iam maths(lect) from India

[@rickideemus]

OK, but what if it just *_isn't always possible_* to reduce a fraction to lowest terms? I mean, I'm "pretty sure" that's always possible, but that doesn't PROVE it! I guess we have to look elsewhere for that part of the proof?

[@Grizzly01]

It is always possible to reduce any fraction to simpler terms, unless it's there already. And if it's there already, it doesn't need reducing, does it?

[@Islandnomad1.-._]

Still the principle of contradiction not satisfying!!! But it opens the way for more possibilities... Irrationality is an ambiguity

[@hassizahananahala7356]

you make irrational eveytime you turn ...

[@madhukarvishwas1749]

Thank you tari matta g for this video ,overwhelming pretty video

[@Awaneeshmaths]

kzbin.info

[@Leberteich]

Nice. You use one fact though that you do not proof, which is that odd x odd cannot be even. I know it's true but to be bulletproof you should have shown why/how.

[@code4baiano646]

what are ratoinal numbers

[@centerofmath]

a rational number is any number that can be expressed as a fraction (p/q) of two integers, a numerator p and a non-zero denominator q. Since q may be equal to 1, every integer is a rational number by definition.

[@officialwork9581]

Beautiful maths N ma'am🥰 💕

[@prospermaaweh9203]

Made it simple to understand....thumbs up 😃😃

[@johndaniel3040]

I am not a math person. This is an audio visual sleeping pill for my low IQ self. I admire you're abilties greatly.

[@Surajgupta-dh3ri]

hlw bro

[@ChandraMathematicsClasses]

Proved beautifully I have also proved it but in another way

[@ChiangKaiSheksBrother]

So convoluted.

[@PritishTechDuniya]

Mam sorry to say they have only 1 is common factor

[@rmnalpha3751]

I enjoy her voice you also enjoy it 😂😂 👇 Like whose enjoy her voice

[@baoskai6738]

-10^333222,10^333222

[@sumbalkhan9902]

Great explanation

[@bwalyahenschel403]

prove that ⅓√2+√5 is irrational

[@babitasingh2991]

a and b have no common factor except 1

[@lidoo86]

Hmmm...I think there is a mistake in the vid at 2:06...The square of square root 2 is not 2...It is +-2

[@JohnJoe131]

She meant root 2 X root 2 which gives 2

[@wadehosking4446]

The principle root dude.

[@sighisoaraa]

She squared both sides of the equation. So sqrt(2)^2 = 2

[@drumetul_dacic]

There is no mistake. If we assume: sqrt(2) = {-x, x}, then (-x)*(-x) = x*x.

[@razielkeren6480]

why a and b can't have a common factor?

[@amdadullahshohel1908]

Thank you somuch sisters