Proof that square root of 2 is irrational | Algebra I | Khan Academy

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Algebra I on Khan Academy: Algebra is the language through which we describe patterns. Think of it as a shorthand, of sorts. As opposed to having to do something over and over again, algebra gives you a simple way to express that repetitive process. It's also seen as a "gatekeeper" subject. Once you achieve an understanding of algebra, the higher-level math subjects become accessible to you. Without it, it's impossible to move forward. It's used by people with lots of different jobs, like carpentry, engineering, and fashion design. In these tutorials, we'll cover a lot of ground. Some of the topics include linear equations, linear inequalities, linear functions, systems of equations, factoring expressions, quadratic expressions, exponents, functions, and ratios.
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[@Karansetiya16]

Proof by Khantradiction!

[@ratnatekchandani4421]

K Solo lol

[@jishnuvjn3378]

contradiction***

[@animore8626]

Jishnu, he was making a pun of the instructor's name.

[@alsilver7780]

W

[@alsilver7780]

jishnu vjn L

[@Skatinima]

The first guy who brought up the proof that sqrt(2) was irrational ended on a deserted island and was never heard of again.. Good times we live in. Sal didn't end up in a deserted island.

[@Upgradezz]

Seriously

[@enzoqueijao]

Or he did, and he was replaced by someone else

[@thedudethatneveruploads2617]

@HI3 They did; Hippasus was drowned at sea because he found the discovery on a ship while his shipmates threw him off because they refused to believe that a number couldn't be expressed as a/b with a and b being rational numbers.

[@obnoxiouslisper1548]

@@thedudethatneveruploads2617 i Guess they were acting in an irrational way

[@thedudethatneveruploads2617]

@@obnoxiouslisper1548 Much like the value of i, they needed to get real

[@DirtyToenailZ]

You're such a great teacher

[@frankkamiru7085]

exactly

[@cubingsphere3339]

Yupp.. u r right!!..👍🏻

[@shayorshayorshayor]

What do you do you do for a living now after 9 years?

[@rajsrivastava4896]

didn't understand after hours of practice...easily understood after watching this once.Thx man!!!

[@Kelorie]

What about a and b can be put in reducible.

[@soulimmortal8487]

@@Kelorie that means a/b are in lowest terms like 2/3 where it can't be divided further Not like 2/8 which becomes 1/4

[@Kelorie]

2/8 also a rational number bro

[@m4rquee11]

same, i couldn't understand all the steps but i immediately understood after watching this video

[@user-ks5ci6bs6x]

Y r u so stupid

[@cindywang9194]

There's probably a really simple answer to this, but I'm just gonna ask anyways: Why does a/b have to be irreducible? If it's reducible, couldn't you just simplify the answer?

[@corythecreeperplaysmc8464]

+Cindy Wang it keeps going on and on. Because NOW you assume r/t is an irreducible rational number, right? And then you go through that whole process again, and you find that r must be even, and t must also be even, therefore they can be reduced by 2. Which CONTRADICTS your assumption above. And then yes, you can go and say okay well the numbers are are the quotient of that reduction are now represented by x/y, and then you go through it again and find that x/y must also be two even numbers. And again, you find a contradiction.

[@silbid0]

Because if they are reducible, you just simply reduce them and end up with another fraction. You can then call it c/d and apply the exact sampe principle. So why don't just start with a/b beind irreducible and that's it? :D

[@real_Zuramaru]

At least give credit to DocWelcher's comment lol.

[@unflappabletoucan2558]

I guess the best way to put it is it doesn't mater what a and b are if the number is rational there are two numbers that make a ratio that = the number that you are finding the square root for (the square root of 25 can be expressed by some ratio) and the reason you assume it's irreducible is because if it wasn't then it's not complete 4/8 is really 1/2 for example just like 100/100 is 1 but don't get caught up on that because it doesn't make a difference if you don't reduce it because the fact is you should still get the same results because fractions that aren't reduced are actually just useless I could be wrong about this whole thing in which case rip me

[@Doivid_]

It's because that's how rational numbers are defined. Some n/m where n and m are coprime integers.

[@shainewinter8264]

Thank you! "No other common factors other than 1." I had a constant question about this where if we had 2/1, which is irreducible, then root-2 would be rational. But Khan to the rescue again =).

[@Zia568]

2/1= 2 and sqrt2 can't be equal to 2 hence it's obvious that for this case there is a contraction at very beginning of the proof hence sqrt2 is irrational. Therfore we must consider the case other than 2/1 by using our common sense

[@lampochka3369]

Thank you very much! I am really grateful for this lesson! You've helped me. Thanks again.

[@jyotiagarwal6202]

After listening to MIT professor I'm here. Understood really well now. Thanks

[@akinaykenar1861]

Terrence Howard brought me here😁

[@VoidHalo]

Thanks for this. I've taken up something of an obsession with root 2 lately, so I've been trying to find as many properties and neat tricks and such to do with it as I can. Really I'm watching this in preperation for the Mathologer video on this, which I assume goes more in depth, as they tend to do.

[@RealEverythingComputers]

Khan Academy to the rescue as always! Thanks so much!

[@Salgandarin]

Finally a video that explains every part of the answer. Thanks.

[@marshallsmith3372]

Amazing explanation! Great job.

[@anamikaahmed4887]

This was aweeesooommme!

[@killiancampion7320]

This video was amazing help thanks!

[@george4746]

Today I started learning about number theory and all of a sudden this proof makes more sense.

[@jrseinc]

much better than my highschool and coaching teacher. my eyes thanks Sal everyday for making the video with black background.

[@mustafahassan6989]

Exceptional. Thank you for the help

[@oshoumap3947]

YOUR TEACHING IS JUST AWESOM .

[@talhahasnain3936]

Just awesome Thank you so much

[@vandomog1013]

An excellent video, great introduction into proof by contradiction. Thanks very much

[@MominulKarim]

Thank You!! Thank You!! Thank You!! Thank You!! Thank You!! Thank You!!

[@astrobullivant5908]

This version is simpler than the version I usually use for the proof. I like it. I usually write that "if a^2 is even, a^2 must be a multiple of 4. If a^2 is a multiple of 4, then (a^2)/2 must be a multiple of two and an even number, which means that b must be an even number. If a and b must both be even, then the fraction can never be in lowest terms and therefore can't exist." I like your way of explaining it better.

[@squarerootof-1307]

tbh yours if much simpler to understand. thank you :)

[@squarerootof-1307]

wait but what is a?

[@squarerootof-1307]

tell me if im correct about your proof- let sqrt(2) = a/b (a/b)^2 = 2 2|a^2 -> 4|a 4|a -> 2|((a^2)/2) 2|((a^2)/2) -> 2|b 2|a & 2|b -> ¬(∃ (a/b)) ¬(∃ (a/b)) ⊢ sqrt(a) != a/b

[@astrobullivant5908]

@@squarerootof-1307 You have the gist of it down. Let sqrt(2) = a/b (a^2)/(b^2) = 2 a^2 = 2(b^2) , so a^2 must be even. Since a^2 must be even, it must also be divisible by 4, since all even perfect squares must be divisible by 4. We now continue to manipulate the expression to get: (a^2)/2 = b^2, Since a^2 is divisible by 4, and since any multiple of 4 divided by 2 is still an even number, b^2 must be an even number. Thus, a^2 and b^2 must be even numbers. Since the square-root of any even number must also be an even number, a and b must also both be even, which means that (a/b) can't be written in lowest terms, which is the contradiction.

[@squarerootof-1307]

@@astrobullivant5908 oh ok thanks :)

[@austins.219]

Woah we do proofs in algebra. Wish I had such an elite algebra class.

[@amissanime3007]

Here's a simpler way. ✓2 = p/q => 2 = (p/q)^2 => 2q = p^2/q Since 2q is natural And there's no common factor between p and q, Therefore p/q or p^2/q can't be natural. Hence, 2q = p^2/q can't be true.

[@malaikas389]

Thank you so much. You have absolutely no idea how much this helped me. I went through loads of videos b

[@isaac4727]

to make it clear for others struggling with why both numbers being even proves the irrationality of root 2, A and B are DEFINED to be coprime (all rationals can be reduced to this form) and by assuming the inverse statement (root 2 is rational) and working your way to both A and B being even, have now shown that A and B have a common factor of 2. This is against the proposition made of coprime rationality. The important part now is that we have made a contradiction, and it is a fact of logic that a contradiction can ONLY be formed if a previous step was incorrect, but in this case the only step that was wrong was the assumption of rationality, so root 2 MUST be irrational.

[@ZiyanShanavas]

Lyrics

[@akuthota.madhusudhanarao7431]

Superb explanation I didn't see this type of explanation

[@manvindersingh]

Thank you so much!

[@rud1i3nfndn]

best explanation of this hands down.

[@yeshwantdeshmukh7768]

Just a great explanation 👍👍👍

[@danielmaxwell7536]

nice explanation ...very convincing

[@NoahAndABadger]

People in the comments either don't understand or are some kind of math major and can spot flaws in the proof 😓

[@texchip977]

J there are no flaws in the proof...

[@janinapv9507]

Wow !! What a presentation!!😱

[@nafeesathulkubra9709]

Superb Explanation ......Tnx alot sir .....

[@Escapist-qv8et]

You're a lifesaver!!!

[@eleniidn757]

thank u so much i have an exam in a few days and you explain things so well :)

[@thomasgale96]

sp helpful ty

[@sankojuprithvi]

Marvellous video on mathematics

[@bujzi_bhairava]

Best explanation ever.

[@abdullahsharaawy]

thank you very much

[@Peter-bg1ku]

Thank you Sal

[@RedShiftedDollar]

Is it ok to restate the initial question of is sqrt2 rational in a different form such as this: x^2=2. Is x rational?

[@saifullahrahman]

thank you very much !

[@earlvalencia3610]

Thanks for the video. Our discussion is kinda hard to understand.

[@ibrahimshoukat2703]

Thank ya :)

[@HariCharann]

Hello sir I love the way you explained it. But I still have a doubt : if root 2 is expressed in decimal and a decimal number can be expressed in a fraction. Fractions are expressed in p/q form the how can we say root 2 is irrational number.

[@veerdotmp3]

+Hari Charann Dude, if a fraction can be denoted in a the form of its decimal expansion, it doesn't mean that it straight away is a rational number. If the decimal expansion is terminating or repeating, only then the number can be said to be rational. If it is non terminating and doesn't repeat itself (i.e 0.1011011101111..., or even Pi.) then it is irrational. √2 is non recurring and non terminating, hence it is irrational.

[@HariCharann]

+VeBz You took some time to go through my question and answered it. Thank you

[@Kelorie]

@@veerdotmp3 what about..root 2= a reducible fraction..this proof is wrong?

[@Justhritik]

@@Kelorie well, its not wrong basically its correct coz u get the value of root 2 as 1.414 and it doesn't end there, it is 1.41421356237309504880168872420969807856967187537694 and continues, now if u think about it, you get such a decimal value only if the fraction is reducible over infinite times

[@jukit3906]

@@Kelorie the thing is that if a number is rational then its necessary that there exists a irreductible fraction which is equal to the number. If you want to have fun you can try to prove this by yourself for instance lol (should be straightforward, just use the definition of a rational number and extrapolate using basic arithmetic)

[@qwertytrewq9870]

Very satisfying

[@diywithcaren1302]

Thankuuu smm! I know i will get a solution to my problem when i come to Khan Academy and I am right!

[@lakshyakhare6349]

Thank you

[@nehaumbre6647]

Thanks 👍

[@andreafillaine6594]

*THANKYOU!*

[@asithyaasithy1257]

i saw a video from ding and was confused there because he didnt say that a/b assumption is irreducable now im here and very thankful to u luv from india

[@Neroner94]

Excellent!!

[@aarushaeron]

first comment

[@mathsmellow7516]

This is enough Thanks Khan

[@ahmedzamara7602]

You clutched up for me bro ❤️🙏

[@user-pc3ii1ez3l]

can we generalize it for all √3 √5 ...... for all

[@MuffinsAPlenty]

For any prime number p, sqrt(p) is irrational, and the proof generalizes _very_ nicely. The only thing that you really have to change is the whole idea of a^2 being even, so a is even argument. Instead, you use the following proof about prime numbers: For any prime number p, if p divides a product of integers, then it divides one of the factors of that product. (In other words, if a and b are integers and p divides ab, then p divides a or p divided b or both.) In particular, if p divides a^2, then p divides one of the factors, but there is only one factor: a. The rest of the proof follows identically to what is given in the video. You can actually generalize the exact same result a little bit further: let n be any positive integer having the property that there exists a prime p dividing n but that p^2 does not divide n. Then sqrt(n) is irrational. The proof is still almost identical: First step: Since p divides n but p^2 does divide n, n = p·m, where p does not divide m. Assume sqrt(n) = a/b in lowest terms Then n = a^2/b^2 So n·b^2 = a^2 So p·m·b^2 = a^2 (replacing n with p·m) Now a^2 is divisible by p, so a is divisible by p: let a = p·c. Then p·m·b^2 = (p·c)^2 = p^2 · c^2 Divide both sides by p to get m·b^2 = p·c^2 So now the left hand side is divisible by p. Since whenever a prime divides a product of integers it must divide one of the factors, we know that p must divide m or b^2. But p doesn't divide m, so p divides b^2. Using the same fact, p divides b. Now both a and b are divisible by the prime p, so a/b is not in lowest terms, a contradiction. Using the above fact (that if a prime but not its square divides a positive integer, then the square root of that number is irrational), you can prove sqrt(n) is irrational for any positive integer n which is not a perfect square. But it doesn't follow the same argument. It uses the fact I just proved.

[@ronitsinghpatel5319]

Good Job

[@redfirpineburngreen0]

thank you. followed it quite well. there is hope for me!

[@firdausspusma8476]

THANKYOU SO MUCH.

[@groovysteroids7945]

It’s all mathematics... you passed that elevator. Thanks Khan

[@ericfricke4512]

Much better than that 20 minute video.

[@marcosgutierrez9100]

To everyone saying this proof is wrong because you can apply this logic to any other non-prime number and still get out a contradiction, I think you missed the step where he squared everything in the equation and got out 2 on the left side of the equation. If it was, say, 4, then it could be reduced in half twice which changes the variables on the right.

[@imuskansheikh]

Thank uh....after a long time I m able to do this question....🙄😅

[@mohanr.s1255]

Great sir🥰

[@jcruyff_14]

Sir, I understand that assuming a and b are coprime is crucial for our proof, but I'm uncertain about their coprimality since they could be any integers. You mentioned reducing them to coprime if they're reducible. However, what if they're reducible but we choose not to reduce them due to their unknown specifics?

[@anuj__ftw]

any one in 2021

[@NeemeVaino]

At 2:45 it is clear already, no further argument necessary: Since the numerator and denominator do not have common divisors, squaring them does not make them have any either, so they don't cancel out, so the ratio of squares cannot be an integer. Period. Say, A and B are are products of a list of primes, not sharing any of them in common: A=p1×p3×p5×...×pn and B=p2×p4×p6×...×pm. The squares are respectively A² = p1²×p3²×p5²×...×pn² and B²=p2²×p4²×p6²×...×pm². Example: 257²×1409²×2448769²/(32²×3²×17²×577²×665857²) ≠ 2

[@o.m9434]

Neeme Vaino yea but just cos the ratio of squares isn’t an integer doesn’t mean it’s irrational

[@o.m9434]

Saikishore Gowrishankar that’s fine but the comment says now that you know the ratio isn’t an integer you’ve shown it’s irrational Not necessarily I’m not saying the proof is wrong I’m just saying the ratio not being an integer isn’t the reason why it’s a proof

[@Upgradezz]

You are correct dear, very insightful

[@huckthatdish]

They can if b is 1. Take any perfect square and this works fine. Which makes sense. It would be concerning if you could prove the square root of 9 is irrational. 3^2 / 1^2 = 9. And 3/1 is in fact the irreducible rational representation of root 9, so all is well in the world. You’ve got a great start to a proof that the root of any non perfect square integer is irrational though. Because if b isn’t 1 you’re argument holds. And b = 1 is equivalent to the number inside the radical being a perfect square.

[@NeemeVaino]

1 is not a product of a list of primes (does it need to say that empty list should be excluded from this proof?) or, does it need to say that "while B ≠ 1"

[@jdaregalario]

Could've said gcd(a,b) = 1

[@forrestberg591]

Someone get Terrence Howard to watch this video haha

[@ganixxmniga1594]

Understand able

[@biogrisha4433]

Cool, finally decided to check the proof of why that ubiquitous sqrt 2 is irrational. I wasn't disappointed.

[@amanisamitoama3001]

I like you man 🤩, I understanded after seeing your video 🌷🌷🌷

[@O_P_Shorts]

Thanksssssss

[@zowedyt641]

YO THANKS A LOT!!!!

[@fatimaalamien8786]

I really love this channel but please don’t make it in Hindi cause I can’t understand

[@maximwynant1865]

I admire your ability to write with your mouse

[@asterawoke-oc8hk]

thanks so much i will reinforce when you came to Ethiopia

[@aligaming4652]

I AGREED that he is best teacher

[@mainulislam6765]

Ok help me out on this one. Couldn’t it just be said that since p/q has no common factor, then the division of their squares should yield no integer but we see that their division yields 2 and that in itself is a contradiction?

[@lolz0098]

I still dont understand if a and b is reducible it mean that sqrt of 2 s irrational?

[@skebess]

Is this the only non "overly complicated" proof? I keep seeing it everywhere.

[@MuffinsAPlenty]

It's the simplest proof in that it doesn't rely on "heavy-duty" mathematical machinery. For example, here's another nice proof. sqrt(2) is a root of the polynomial x^2−2. Since all of the coefficients of the polynomial are integers, the Rational Root Theorem states that the only possible rational roots are 1, −1, 2, and −2. Plugging each of these in, we see that none of these are roots of x^2−2. Therefore, x^2−2 has no rational number roots. Since sqrt(2) is a root, sqrt(2) is not rational. Of course, this proof relies on the machinery of polynomials, and, in particular, on the Rational Root Theorem. So it is a valid proof, but if you don't understand why the Rational Root Theorem is true, then you may not find this to be a solidly convincing proof. (And the most accessible proof of the Rational Root Theorem is fairly similar to the proof given in this video.)

[@swim_ad]

@@MuffinsAPlenty Since squirt(2) is a root, squirt(2) is rational.

[@rancorjoy5412]

Mark Allen roots do not have to be rational, for example - X-pi = 0 has a root of X = Pi Pi is irrational Lazy example but it is a proof by contradiction that your assertion is not correct.

[@MsYoyojam]

Great contradiction example!

[@happydays9306]

ur welcome

[@dragonfly3139]

I dont understand what if i plug in 5/3 in this case a is 5, 5 squared is 25 that is not an odd number how does that hold up against the proof

[@MuffinsAPlenty]

25 _is_ an odd number.

[@ChandraMathematicsClasses]

You proved it beautifully I have also proved it but in another way

[@franciscomerlos8899]

can you show how you proved it in another way?

[@EPC-ue2ci]

isnt this true for sqrt of 3 as well?

[@ViolinistJeff]

Ok, how do we express with mathematical symbols that a/b is an irreducible fraction??

[@aku7598]

At present, the root 2 value is computed to 10 trillion digits. Maybe little digression inserted, odd x odd = odd

[@AceOfHearts001]

What I dont understand is why it is significant to say it is even... it is the factor of two which is preventing a over b to be a reduced fraction. 'Even' is just a property of having a factor of 2.

[@yxtee]

But doesn't this only show that sqrt(2) cannot be an irreducible rational number, not that it cannot be a reducible one(any integer)?

[@superroydude]

yxtee All rational numbers can be reduced to a point. For example you can write 2 like: (2/1) or 8/4 or 1600/ 800, but the most reduced version is 2/1 and this is the case for all numbers. If a number can't be reduced to a smallest ratio, ( keep in mind this could be anything e.i 3554/678) then it was never expressable as a ratio to begin with.

[@prav8141]

√2 is not an integer.

[@melonenlord2723]

But does this also work for sqrt(3) or does this need another proof?

[@nagamanisagi3977]

For even number only

[@mithunkrishna8880]

i can't hear anything. What should i do to get the sound right

[@AlAnsikk]

wow Khan sounds just like DJ Vlad!

[@HariOmSinghRawat]

Thanks Sir But Which Application Do You Use For Writing

[@happydays9306]

idk

[@patrickjane3315]

Rational: Numbers that can be represented as the ratio between two integers. Based on that, we can expect that any method used to provesqrt(2) to be irrational, must show us that it can not be represented by such ratio. However, this "proof by contradiction" only allows us to conclude that both terms are even, and that is it. I mean, they are even but they still composes a ratio that we assumed to be equal to sqrt(2). Maybe this confuses me because I can't understand why wr assumed that a and b must be co-primes. It is explained that we can reduce ratio between two integers to the lowest terms. And that idea is used to justify our assumption, but I don't understand why is that.

[@TomasAragorn]

If you can write sqrt(2) as a fraction of two even numbers, then you can also write it as the ratio of two integers where they are not both even. Just divide the even numbers in the ratio by 2 until one is odd

[@ggnore9089]

How is the principle root of 2^2 = 2 ? Can anyone explain, I'm sorry if this this dumb, I just don't understand how sqrt of 2^2 is 2

[@nataliachira1031]

this is a little late but i hope you still find it useful, whenever you have a square root and you square it, the square root cancels out and you are left with whatever is under it.

[@onetapper3270]

But this works for everything that has a root. For example if you use √25, you will get 25 as a common factor of a and b. That must mean √25 = ±5 is also irrational?

[@ManishGupta-ce8nt]

Under root 4 me bhi ho jayega isi prakar se

[@priya200328]

His voice reminds me of the rock (Dwayne Johnson)

[@happydays9306]

no

[@niceguy4801]

What about if this logic applied in a rational number? Will it be also true or not?

[@happydays9306]

idk