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

[@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

[@李汶灏]

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.

[@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.

[@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

[@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 :)

[@jyotiagarwal6202]

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

[@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"

[@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

[@george4746]

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

[@RealEverythingComputers]

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

[@Salgandarin]

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

[@austins.219]

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

[@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?

[@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)

[@jrseinc]

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

[@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.

[@MominulKarim]

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

[@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...

[@akinaykenar1861]

Terrence Howard brought me here😁

[@oshoumap3947]

YOUR TEACHING IS JUST AWESOM .

[@lampochka3369]

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

[@Yournature108]

According to him - Assume = Asoom 😂 Contradiction = khnantadiction Co prime - Coh prime Multiply = Multhifly Two = theu K = khay 🤭 Times - thaimes

[@Homelander_21]

Best explanation ever.

[@vandomog1013]

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

[@rud1i3nfndn]

best explanation of this hands down.

[@janinapv9507]

Wow !! What a presentation!!😱

[@akuthota.madhusudhanarao7431]

Superb explanation I didn't see this type of explanation

[@nafeesathulkubra9709]

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

[@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.

[@abdullahsharaawy]

thank you very much

[@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

[@sankojuprithvi]

Marvellous video on mathematics

[@ericfricke4512]

Much better than that 20 minute video.

[@talhahasnain3936]

Just awesome Thank you so much

[@kornelijekovac9793]

But this doesn't prove it is irrational number. It just shows that assumption of irreducibility is wrong.

[@newmai5771]

Since it show our assumption that it is rational is wrong then which one will be right? Obviously irrational.

[@kornelijekovac9793]

@@newmai5771 But in the end he says that "a and b don't have other common factors other than 1". Than means it can be, for example (2*19) / (2*17 ) ? And then when you reduce them by 2, you get irreducible ratio? Maybe he just didn't finish "reducing numbers inside a letter", he purposely left them both multiplied by 2, and then presented it as an even numbers?

[@maximwynant1865]

I admire your ability to write with your mouse

[@diywithcaren1302]

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

[@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.

[@asterawoke-oc8hk]

thanks so much i will reinforce when you came to Ethiopia

[@Peter-bg1ku]

Thank you Sal

[@ahmedzamara7602]

You clutched up for me bro ❤️🙏

[@eleniidn757]

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

[@mathsmellow7516]

This is enough Thanks Khan

[@saifullahrahman]

thank you very much !

[@anonymous___4999]

Plz. Prove that a negative of an irrational no. Is an irrational no.?

[@MuffinsAPlenty]

Let x be an irrational number. If -x were rational, then -x = a/b for some integers a and b with b not 0. Multiply both sides by -1 to get x = -a/b is a rational number, which is a contradiction.

[@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.

[@HanSolo-dh4rn]

It confuses me how a video from a channel with over a million subscribers has about 15 comments and 6000 views?

[@greendaquil]

It's about math and it's a corporate channel.

[@happydays9306]

@@greendaquil wow u seem to be intelligent

[@aku7598]

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

[@anamikaahmed4887]

This was aweeesooommme!

[@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?

[@animore8626]

I'm confused. How can we assume that they are coprimes with mathematical notation? We just accept it, but couldn't it be possible for it to be reducible?

[@MuffinsAPlenty]

The point of a rational number is that it is a ratio of integers. So every rational number can be written in the form n/m where n and m are integers (and m is not 0). Every fraction like this _can_ be reduced. Just divide both the numerator and denominator by gcd(n,m). n/gcd(n,m) and m/gcd(n,m) are both integers, and they are coprime. And the result is an equivalent fraction to the one you started with.

[@animore8626]

That's perfectly fine, I understand that, but isn't the whole point of the proof showing that it's not fully reduced, even though we're assuming it's coprime?

[@MuffinsAPlenty]

You are right. That is the whole point of this particular proof. In logic, true assumptions can never lead to a false conclusion. And this is the point of proof by contradiction. If you want to show that something is false, you can assume that it is true and then show that your assumption leads to a false conclusion. (Since your assumption leads to a false conclusion, your assumption cannot be true, and as such, must be false.) In this case, we assume that sqrt(2) can be written as a fraction of integers. We are able to write it in reduced form (as I explained), and then this all leads to the fraction not being in reduced form. So if sqrt(2) is a rational number, then it can be written as a fraction which is fully reduced but can also be reduced more. This conclusion is nonsensical. It is definitely false. Therefore, our assumption that sqrt(2) is rational is false. But I would like to give you a slightly modified proof. A KZbinr by the name of Sci Twi has often pointed out that the proof given in this video isn't very elegant. There are proofs out there where you don't have to do anything with reducedness. So here's another argument: Suppose sqrt(2) = a/b where a and b are integers and b is not 0. Squaring both sides, we get 2 = a^2/b^2 Multiplying both sides by b^2, we get 2b^2 = a^2 Consider prime factorizations of 2b^2 and a^2. Since b^2 and a^2 are perfect squares, every prime factor appears an _even_ number of times in their prime factorizations. As such, 2b^2 has 2 appear an odd number of times, but a^2 has 2 appear an even number of times. Therefore, 2b^2 and a^2 have different prime factorizations. Since prime factorizations are unique, this means that 2b^2 and a^2 are different numbers. But our assumption that sqrt(2) = a/b leads us to the conclusion that 2b^2 = a^2, even though we know this is false. Therefore, our assumption that sqrt(2) = a/b is false. sqrt(2) is not a rational number.

[@animore8626]

Yeah, I definitely like that proof much more, it leads to much less confusion. Thanks for the answer!

[@eLiBiiLV]

Why do we assume that a/b is irreducible? Arent ratios of two integers still rational if they are reducible? Like 3/9 → 1/3; both rational. So why does a and b having a common factor of 2 make sqrt2 irrational? Im so confused at this part :(

[@TheCarpenterUnion]

I was very confused at first too. We set this up as if looking for a contradiction of the contradiction. By showing that a irreducable fraction is reducable, we have created a contradiction for the assumption that sqrt(2) is rational, thus it must not be rational. I was hung up on thinking that it had something to do with the properties of rationality or irrationality, but it doesn't. We are simply finding a negative for the negative to show that it is positive (in a manner of speaking)

[@eLiBiiLV]

thanks!

[@AbC-bj3gs]

that is the question where i thought, where i stuck. thanks

[@jonsentio3150]

@@TheCarpenterUnion you are really a savior even after nearly a decade! god bless you bro i was asking the same question

[@Escapist-qv8et]

You're a lifesaver!!!

[@mustafahassan6989]

Exceptional. Thank you for the help

[@amanisamitoama3001]

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

[@zoteha]

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

[@خوارزمة]

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.

[@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?

[@aparupanayak8760]

We can follow the same procedure to prove that root 4 is irrational

[@MuffinsAPlenty]

No, you cannot. When looking at the argument using sqrt(4), things fall apart at the step when you prove a and b are even. You can still prove that a is even, but you are unable to prove that b is even. Let's go through it. Assume sqrt(4) = a/b where a and b are integers, b is not 0, and the fraction is irreducible. Squaring both sides and multiplying by b^2, we get 4b^2 = a^2 Here, we get a^2 is divisible by 4. Hence, a must be divisible by 2. (You might be tempted here to think that a must be divisible by 4, but this isn't true! For example, if x = 2, then x^2 = 4 is divisible by 4, but x is not. Or if x = 6, then x^2 = 36 is divisible by 4, but x is not. The argument for a^2 being even so a must be even works because 2 is a prime.) So a = 2c for some positive integer c. Then 4b^2 = (2c)^2 = 4c^2 Dividing both sides by 4, we get b^2 = c^2 We cannot conclude that b^2 is even anymore. But let's see how far we go. If b^2 = c^2, then b = c or b = -c. Since we may as well assume everything is positive here, we have b = c. But remember that a = 2c, so since c = b, we have a = 2b. Then sqrt(4) = a/b = 2b/b = 2.

[@Jonitoni1234]

Cant this same proof applied tp sqrt of 4? And it is rational

[@sickboydroid]

I also think that

[@MuffinsAPlenty]

No, you cannot. When looking at the argument using sqrt(4), things fall apart at the step when you prove a and b are even. You can still prove that a is even, but you are unable to prove that b is even. Let's go through it. Assume sqrt(4) = a/b where a and b are integers, b is not 0, and the fraction is irreducible. Squaring both sides and multiplying by b^2, we get 4b^2 = a^2 Here, we get a^2 is divisible by 4. Hence, a must be divisible by 2. (You might be tempted here to think that a must be divisible by 4, but this isn't true! For example, if x = 2, then x^2 = 4 is divisible by 4, but x is not. Or if x = 6, then x^2 = 36 is divisible by 4, but x is not. The argument for a^2 being even so a must be even works because 2 is a prime.) So a = 2c for some positive integer c. Then 4b^2 = (2c)^2 = 4c^2 Dividing both sides by 4, we get b^2 = c^2 We cannot conclude that b^2 is even anymore. But let's see how far we go. If b^2 = c^2, then b = c or b = -c. Since we may as well assume everything is positive here, we have b = c. But remember that a = 2c, so since c = b, we have a = 2b. Then sqrt(4) = a/b = 2b/b = 2. So actually, it doesn't prove irrationality. Going through the argument shows that if sqrt(4) is rational, then it must be equal to 2. Isn't that cool?

[@cikambai]

THANKYOU SO MUCH.

[@malaikas389]

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

[@aligaming4652]

I AGREED that he is best teacher

[@redfirpineburngreen0]

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

[@Melomathics]

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.

[@danielmaxwell7536]

nice explanation ...very convincing

[@andreafillaine6594]

*THANKYOU!*

[@scottcollege213]

I don't get the reasoning. So what if it's reducible? 4/2 is reducible and it's a rational number.

[@wotislaif1201]

Scott College the point is that assuming that it’s irreducible, we get that it’s reducible. Supposedly all rational numbers can be written through a irreducible fraction. If we assume that with sqrt(2) and get a contradiction, it means that it can not be true. Since there is only one other option, which is it being irrational, we’ve proved that the number is indeed so

[@anuj__ftw]

any one in 2021

[@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

[@yeshwantdeshmukh7768]

Just a great explanation 👍👍👍

[@senzubeats]

Can’t you just assume that either a or b is prime if it is not reducible, therefore either a or b must be odd. Even if you had a prime number of two, it can be reduced with an even number with it in a fraction, but if not you would have to have had an odd number with it in order to make it irreducible. Knowing this, it is impossible to yield an equivalent number in all scenarios. Lets say you had an even number as b and an odd number as a in the equation 2b^2=a^2 . Then following the rules (even*even=even , odd*odd=odd), it would be impossible to end up with any equivalency, thus the equation is not consistent and sqrt of 2 would be irrational. Not sure if this logically makes sense or is sufficient as a proof though.

[@senzubeats]

Actually, after some thought after posting this, I realized that this logic would mean that the square root of any number could be considered irrational. Will keep the comment up if someone wants to see my thought process lol.

[@killiancampion7320]

This video was amazing help thanks!

[@lakshyakhare6349]

Thank you

[@melonenlord2723]

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

[@nagamanisagi3977]

For even number only

[@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?

[@jdaregalario]

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

[@Hythloday71]

Have you shown that the square root operation is well defined ? If the square root of 2 is meaningless, or you fail to provide a reasonable definition of what irrationality is, it seems this is a faulty proof. Of course if you define irrationality simply as not rational, you haven't really said much beyond the vacuous. Also a/a or b/b or root2/root2 are all clearly drawn from the rationals. What grounds do you have for asserting root2 be equal to a/b with a not equal to b ? All integers are assignable to arbitrary intervals, clearly the constructible interval of root 2 is constructible. By what rational, say, are sides a, b of a unit right angle deemed special over the hypotenuse ? All this proves is that given a chosen measuring ruler system upon which a, b sides of unit triangle are defined, the same ruler system will not yield good answers for the other side, but its arbitrary, choose the hypotenuse as your, unit measure, and the other sides now are not exactly measureable, no big deal in a land of Gödel's incompleteness.

[@martinzone8153]

Thumb up for the effort!

[@TheCarpenterUnion]

This is how these are done. We assume well-defindedness.

[@marshallsmith3372]

Amazing explanation! Great job.

[@lolz0098]

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

[@angelatzw1503]

Doe this mean that irrational numbers can be written as a fraction if a and b ARE reducible?

[@GhostyOcean]

Irrational, by definition, cannot be written as a fraction of two integers, even if the ratio of those numbers is reducible. The reason we needed the assumption "a/b is irreducible" is for the contradiction to occur because we can show that a/b can be reduced.

[@lucasm4299]

I still hate this proof because what does irreducibility have to do with rationality?? 2/1 is rational, 4/2 is rational, etc. I was hoping for something like even=/ odd

[@eipiwau]

whats the problem if a and b would have factors in common?

[@codyw_]

+Simon Reeck Because we started with the idea that the two are irreducable, if they were both even, then they would have at the very least 2 in common, and would be divisible by 2. Which contradicts our original assumption that they were irreducable.

[@eipiwau]

+DocWelcher yes i've understood the assumption but i don't see a reason for assuming this. because lets a=2r and b=2t then sqrt(2)=2r/2t in other words sqrt(2)=r/t so i can still write this as a fraction...

[@codyw_]

+Simon Reeck But 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.

[@codyw_]

+Simon Reeck The whole thing is centered on that very assumption. If your assumption is that they were both reducible, then they would be two different numbers, therefore you wouldn't find the two numbers you're looking for. a and b are meant to represent THE numbers that equal sqrt(2).

[@eipiwau]

+DocWelcher now i understood ^^ Thank you!

[@ManishGupta-ce8nt]

Under root 4 me bhi ho jayega isi prakar se

[@fatimaalamien8786]

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

[@imuskansheikh]

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

[@akinaykenar1861]

Why does (a) have to be even? If it's odd squared its odd, if it's even squared its even.

[@mohanr.s1255]

Great sir🥰

[@aznstride4325]

I still don’t understand why it’s ok to assume that a and b are irreducible. To me, it’s analogous to trying to prove that something is a “living thing” (analogous to irreducible) So you assume that whatever you get should be a “living thing” but the proof ended up showing that the thing is a “Lion”. And then you say, well I got a “Lion”, and it breaks the assumption that I should get a “living thing” - proof. (Even though a lion is a living thing) Doesn’t make sense to me, why we can say an and b should be irreducible (living thing), when fractions don’t have to be irreducible (lion) to be a fraction (living thing)

[@groovysteroids7945]

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

[@thomasgale96]

sp helpful ty

[@nehaumbre6647]

Thanks 👍

[@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.

[@earlvalencia3610]

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

[@zowedyt641]

YO THANKS A LOT!!!!

[@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.

[@MsYoyojam]

Great contradiction example!

[@happydays9306]

ur welcome

[@qwertytrewq9870]

Very satisfying