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.
@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.
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?
+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.
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
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
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 =).
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
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.
Sal showed a direct contradiction with evidence. But for your one question raises why p^2/q can't be natural. Can you prove it? Sal's method is called contradiction of constructivism but your one is just a normal contradiction with assumptions.
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 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.
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.
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
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
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.
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"
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.
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
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.
+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.
@@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)
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.
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?
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.
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.
Isn't the last line you wrote in 3:01 enough for the proof ? If you multiply a perfect square by another number ( which is not a perfect square ) , youll never get a perfect square . Hence 2 * b^2 = a^2 is impossible .
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.
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
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.
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.
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 :(
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)
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?
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.
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?
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 TH-camr 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.
Wouldn't it be easier and quicker to stop the proof at 3:00: 2b^2 = a^2? Since 2b^2 is necessarily even, but a^2 is not necessarily even (e.g. 3x3), you have exceptions to the equation, wherein the numbers could be different by virtue of one being even and the other being odd, and thus the equation is already invalid... right?
You can stop around there, yes, but not quite for the reason you stated, as Gaben pointed out. But you are on to something! If you make a slight change to your argument, it works! The left hand side has an odd number of 2's in its prime factorization, but the right hand side has an even number of 2's in its prime factorization. Since prime factorizations are unique, it is impossible for the left- and right-hand sides to be equal, which is a contradiction.
Why is it that the left hand side has an odd number of 2's in its prime factorization, but the right hand side has an even number of 2's in its prime factorization.
Hi, Tony Liu. For some reason, I didn't get notified about your response, and I stumbled upon this again, so sorry for being 8 months late on this! Suppose you have a positive integer n, which has a prime factorization. You can get a prime factorization for n^2 by squaring the prime factorization for n. But squaring the prime factorization for n is the same thing as repeating each prime an additional time. So each prime appears _twice_ in the prime factorization of n^2 for each time it appears in the prime factorization of n. So no matter how many times 2 appears in the prime factorization of n, it appears twice as many times in the prime factorization of n^2. Hence, it appears an _even_ number of times in the prime factorization of n^2. (For example, 30 = 2∙3∙5, so 30^2 = (2∙3∙5)^2 = 2^2∙3^2∙5^2; another example is 12 = 2^2∙3, so 12^2 = (2^2)^2∙3^2 = 2^4∙3^2) But n could be any positive integer. In particular, the result must be true for both a and b. Since a and b are positive integers, 2 must appear an even number of times in the prime factorization of a^2 and in the prime factorization of b^2. Therefore, 2b^2 has 2 appear one more time than an even number of times, so it appears an odd number of times. But 2 can only appear an even number of times for a^2.
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.
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.
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.
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.
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?
+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.
+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...
+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.
+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).
@@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?
Proof by Khantradiction!
K Solo lol
contradiction***
Jishnu, he was making a pun of the instructor's name.
W
jishnu vjn L
didn't understand after hours of practice...easily understood after watching this once.Thx man!!!
What about a and b can be put in reducible.
@@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
2/8 also a rational number bro
same, i couldn't understand all the steps but i immediately understood after watching this video
Y r u so stupid
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.
Seriously
Or he did, and he was replaced by someone else
@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.
@@thedudethatneveruploads2617 i Guess they were acting in an irrational way
@@obnoxiouslisper1548 Much like the value of i, they needed to get real
You're such a great teacher
exactly
Yupp.. u r right!!..👍🏻
What do you do you do for a living now after 9 years?
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?
+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.
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
At least give credit to DocWelcher's comment lol.
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
It's because that's how rational numbers are defined. Some n/m where n and m are coprime integers.
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 =).
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
After listening to MIT professor I'm here. Understood really well now. Thanks
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.
Sal showed a direct contradiction with evidence. But for your one question raises why p^2/q can't be natural. Can you prove it? Sal's method is called contradiction of constructivism but your one is just a normal contradiction with assumptions.
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.
tbh yours if much simpler to understand. thank you :)
wait but what is a?
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
@@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.
@@astrobullivant5908 oh ok thanks :)
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.
Finally a video that explains every part of the answer. Thanks.
Thank you very much! I am really grateful for this lesson! You've helped me. Thanks again.
much better than my highschool and coaching teacher. my eyes thanks Sal everyday for making the video with black background.
YOUR TEACHING IS JUST AWESOM .
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
Neeme Vaino yea but just cos the ratio of squares isn’t an integer doesn’t mean it’s irrational
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
You are correct dear, very insightful
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.
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"
Today I started learning about number theory and all of a sudden this proof makes more sense.
Woah we do proofs in algebra. Wish I had such an elite algebra class.
Thank You!! Thank You!! Thank You!! Thank You!! Thank You!! Thank You!!
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.
Lyrics
Best explanation ever.
Khan Academy to the rescue as always! Thanks so much!
An excellent video, great introduction into proof by contradiction.
Thanks very much
thank you very much
best explanation of this hands down.
Wow !! What a presentation!!😱
Superb explanation I didn't see this type of explanation
According to him -
Assume = Asoom 😂
Contradiction = khnantadiction
Co prime - Coh prime
Multiply = Multhifly
Two = theu
K = khay 🤭
Times - thaimes
Superb Explanation ......Tnx alot sir .....
Just awesome
Thank you so much
Marvellous video on mathematics
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
Thank you Sal
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.
+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.
+VeBz You took some time to go through my question and answered it. Thank you
@@veerdotmp3 what about..root 2= a reducible fraction..this proof is wrong?
@@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)
You clutched up for me bro ❤️🙏
Terrence Howard brought me here😁
People in the comments either don't understand or are some kind of math major and can spot flaws in the proof 😓
J there are no flaws in the proof...
I AGREED that he is best teacher
Thankuuu smm! I know i will get a solution to my problem when i come to Khan Academy and I am right!
Much better than that 20 minute video.
thank u so much i have an exam in a few days and you explain things so well :)
thank you very much !
This is enough
Thanks Khan
This was aweeesooommme!
Exceptional. Thank you for the help
You're a lifesaver!!!
Just a great explanation 👍👍👍
nice explanation ...very convincing
sp helpful ty
thanks so much i will reinforce when you came to Ethiopia
*THANKYOU!*
thank you. followed it quite well. there is hope for me!
Great sir🥰
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.
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?
Thank you so much. You have absolutely no idea how much this helped me. I went through loads of videos b
You proved it beautifully I have also proved it but in another way
can you show how you proved it in another way?
THANKYOU SO MUCH.
I like you man 🤩, I understanded after seeing your video 🌷🌷🌷
Cool, finally decided to check the proof of why that ubiquitous sqrt 2 is irrational. I wasn't disappointed.
Thank you
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.
Thanks 👍
Understand able
This video was amazing help thanks!
Thanksssssss
I admire your ability to write with your mouse
YO THANKS A LOT!!!!
can we generalize it for all √3 √5 ...... for all
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.
Isn't the last line you wrote in 3:01 enough for the proof ? If you multiply a perfect square by another number ( which is not a perfect square ) , youll never get a perfect square .
Hence 2 * b^2 = a^2 is impossible .
But does this also work for sqrt(3) or does this need another proof?
For even number only
Very satisfying
Could've said gcd(a,b) = 1
Amazing explanation! Great job.
Good
Job
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?
Thank uh....after a long time I m able to do this question....🙄😅
Great contradiction example!
ur welcome
I really love this channel but please don’t make it in Hindi cause I can’t understand
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.
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
Thanks for the video. Our discussion is kinda hard to understand.
Plz. Prove that a negative of an irrational no. Is an irrational no.?
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.
Doe this mean that irrational numbers can be written as a fraction if a and b ARE reducible?
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.
Ok, how do we express with mathematical symbols that a/b is an irreducible fraction??
Nice
At present, the root 2 value is computed to 10 trillion digits.
Maybe little digression inserted, odd x odd = odd
any one in 2021
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 :(
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)
thanks!
that is the question where i thought, where i stuck. thanks
@@TheCarpenterUnion you are really a savior even after nearly a decade! god bless you bro i was asking the same question
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?
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.
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?
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 TH-camr 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.
Yeah, I definitely like that proof much more, it leads to much less confusion. Thanks for the answer!
I still dont understand if a and b is reducible it mean that sqrt of 2 s irrational?
Excellent!!
first comment
Wouldn't it be easier and quicker to stop the proof at 3:00: 2b^2 = a^2?
Since 2b^2 is necessarily even, but a^2 is not necessarily even (e.g. 3x3), you have exceptions to the equation, wherein the numbers could be different by virtue of one being even and the other being odd, and thus the equation is already invalid... right?
You can stop around there, yes, but not quite for the reason you stated, as Gaben pointed out.
But you are on to something! If you make a slight change to your argument, it works!
The left hand side has an odd number of 2's in its prime factorization, but the right hand side has an even number of 2's in its prime factorization. Since prime factorizations are unique, it is impossible for the left- and right-hand sides to be equal, which is a contradiction.
MuffinsAPlenty : Excellent reasoning!
Why is it that the left hand side has an odd number of 2's in its prime factorization, but the right hand side has an even number of 2's in its prime factorization.
Hi, Tony Liu. For some reason, I didn't get notified about your response, and I stumbled upon this again, so sorry for being 8 months late on this!
Suppose you have a positive integer n, which has a prime factorization. You can get a prime factorization for n^2 by squaring the prime factorization for n. But squaring the prime factorization for n is the same thing as repeating each prime an additional time. So each prime appears _twice_ in the prime factorization of n^2 for each time it appears in the prime factorization of n. So no matter how many times 2 appears in the prime factorization of n, it appears twice as many times in the prime factorization of n^2. Hence, it appears an _even_ number of times in the prime factorization of n^2.
(For example, 30 = 2∙3∙5, so 30^2 = (2∙3∙5)^2 = 2^2∙3^2∙5^2; another example is 12 = 2^2∙3, so 12^2 = (2^2)^2∙3^2 = 2^4∙3^2)
But n could be any positive integer. In particular, the result must be true for both a and b. Since a and b are positive integers, 2 must appear an even number of times in the prime factorization of a^2 and in the prime factorization of b^2. Therefore, 2b^2 has 2 appear one more time than an even number of times, so it appears an odd number of times. But 2 can only appear an even number of times for a^2.
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.
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.
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)?
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.
√2 is not an integer.
It’s all mathematics... you passed that elevator. Thanks Khan
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.
Thumb up for the effort!
This is how these are done. We assume well-defindedness.
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?
whats the problem if a and b would have factors in common?
+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.
+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...
+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.
+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).
+DocWelcher now i understood ^^ Thank you!
But this doesn't prove it is irrational number. It just shows that assumption of irreducibility is wrong.
Since it show our assumption that it is rational is wrong then which one will be right? Obviously irrational.
@@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?
If both numbers are even are they reducible or not? First tell me that. I will help you with your problem.@@kornelijekovac9793
Thank ya :)
wow Khan sounds just like DJ Vlad!