My favorite proof: If √2=a/b with gcd(a,b)=1, then 2=a^2/b^2 and gcd(a^2,b^2)=1, so a^2/b^2 is the reduced form of 2. This implies a^2=2 and b^2=1, which is obviously impossible as a is an integer. This proof is my favourite as it is not only really simple, but also generalizes really easily to all irrational roots.
Yamin New You use the fundamental theorem of arithmetic to deduce that gcd(a^2,b^2)≠1 would imply that a^2 and b^2 must share a common prime factor which a and b must then share as well. By contraposition, gcd(a,b)=1 implies gcd(a^2,b^2)=1. In case you feel like this makes this proof relatively complicated, I want to mention that the classic proof uses similar arguments.
I think the most elementary proof (and hence, quite reasonably, the best solution) is the one that simply uses the ''extremal principle'' or ''infinite descent'', it is somehow known by different names because of the great importance it actually has in the development of further arguments and well know elementary proof techniques. If x=sqrt 2 is rational, the, by definition, there are a, b integers (positive wlog) such that a=bx. Thus there exist at least one positive integer M such that Mx is an integer. Let N be the smallest such M. Then N(x-1) is a positive integer. Observe N(x-1)(x) is also an integer (easy verification). By minimality we deduce N is not greater than N(x-1), which is clearly false. The contradiction finishes the proof. Recall the importance of such elementary arguments, by using the fundamental theorem of arithmetic, which can be of course proved in many different ways, you are using a more bigger fact than the one you want to prove. This is not so strange, sometimes in mathematics you develop a theory in which the problem you wanted to solve, in terms of the new ''lenguage'' and tools discovered, is basically trivial. But we may want to avoid this process at this level ;)
for the last proof I think you should've started something like this: 2√2 + 3 = 2√2 + 3 2 + 2√2 + 1 = 2(√2+1) + 1 (√2 + 1)*(√2 + 1) = (2 + 1/(√2+1))*(√2+1) √2 + 1 = 2 + 1/(√2+1) instead of just pulling the equation out of nothing with the implication that it's true. At the very least you could've multiplied both sides by √2+1 to show that they are in fact equal.
The best proofs are those that generalise, in this case to show that the square root of any positive integer is either an integer or irrational, or, better still, that the nth root of any positive integer is either an integer or irrational. You can do this using unique factorisation, as hinted at by the start of proof 2.
Hey, I just realised that your logo at the start of this video looks like part of a proof of the Pythagorean theorem, marking out the area for c^2 in black. From this we can see that four right-angled triangles around one square fit into a larger square. Using a bit of algebra we can show: (a+b)^2 = c^2 + 4*(0.5*a*b) a^2 + 2ab + b^2 = c^2 + 2ab a^2 + b^2 = c^2
And it took me this long to realise it. Does it usually have an outline? If not, that's probably why it slipped under my radar for so long (somewhere around the three-year mark, likely introduced via the A4 paper video). The profile icon lacking an outline hides the triangles when displayed on a white page.
That's been one of my favorite PT proofs, since I read long ago that it was devised by James A. Garfield, 20th U.S. president, who was a teacher before he entered politics.
I like this: The square root of any integer is always either an integer or irrational (since a rational non-integer number will always square to another rational non-integer number) sqrt(2)>sqrt(1)=1 and sqrt(2)
here's my favourite proof: sqrt(2) = a/b 2 = (a/b)^2 2 = a^2/b^2 but since the prime-factorizations of square numbers have to have all primes to an even power, one of these must not be a square number, and we have a contradiction.
This actually proves that the square roots of all natural numbers are irrational except for the ones of square numbers, which are whole numbers. Because the quotient of two square numbers must be also a square number.
@@ObiWahn68 "the quotient of two square numbers must be also a square number." *only if it is a natural number other than that, yes, it's a simple proof, but it's very effective.
I love proof 4 best in the sense that I guess it is the most straightforward method that everyone can immediately get the point. Proof 1 is actually not that accessible; I once showed this version to a friend of mine who was an engineering student at Purdue. He was shocked and could not understand why would you ever want to prove something so obvious. To mathematicians, 'obvious' is a dangerous word, but from others' perspective, rigorous proof writing and little tricks like contradiction won't be popular outside mathematicians' own utopia. By saying utopia, I mean mathematicians have their own way of thinking, and we are proud of it, but others simply do not care.
Assume that Root 2 = a/b A must be even since A^2 = 2 x B^2. All even square numbers are divisible by 4, so B has to be even as well. It’s impossible to write in simplest form.
I like the following proof sketch: We can show using homomorphisms that for all field automorphisms of Q, the rationals, f(p)=p for all p in Q. Consider the field automorphism h from the field extensions Q(sqrt(2)) to Q(sqrt(2)) defined by h(sqrt(2)) = -sqrt(2). This is indeed an automorphism, but h(sqrt(2)) =/= sqrt(2). Thus sqrt(2) is not rational.
at 0:52, I don't understand how you conclude that because the left side of the equation, b^2, is even, that means b is even as well since any odd integer squared produces an even number. Pls explain! Thank you
The length of the outer square is a. The length of the red square is a-b. The length of the blue square is the length of the outer square minus two times the length of the red square. Therefore the length of the blue square is a-2*(a-b) = 2b-a
There is a smallest possible length, called the Planck distance. If you made a triangle with two sides with this Planck distance, the hypotenuse would not exist because it is not a multiple of this small distance.
Because that's where the contradiction comes from. We assume it is the lowest form but we can always find a lower one. There is no lowest form, which contradicts the assumption.
Correct. Any positive integer that is not a perfect square is irrational. Of course the square root of a fraction may be a fraction. For example, sqrt(9/4) = 3/2.
Well, let's get the first proof out of the way. Replace 2 with variable p. ./p = a/b (./p)b = a p(b^2) = a^2 Since one side is a power of p, the other must also be. Let's assume a/p = c. b^2 = p(c^2) Since one side is a power of p, the other must also be. We have proved that both a and b are multiples of p. This means they are not in simplest form, meaning that ./p cannot be rational. Let's substitute 4 in, and also change the variables. ./4 = 2/1 (./4)1 = 2 4(1^2) = 2^2 1^2 = 2(0.5^2) (I now know i was wrong....... let's figure out the long way a%4 = 0 then a^2 = (4n)^2 = 16n^2 = 4(4n^2) a%4 = 1 then a^2 = (4n + 1)^2 = 16n^2 + 2n + 1 = 4(4n^2) + 2n + 1 a%4 = 2 then a^2 = (4n + 2)^2 = 16n^2 + 4n + 4 = 4(4n^2 + n + 1)
If n is an integer such that s=sqrt(n) is rational, let a/b be its lowest form a^b=1 implies by Bezout that there exists two integers c and d such that ac+bd=1 Then 0=a-sb=(a-sb)(d-sc)=ad-s(ac-bd)+nbc=ad+nbc-s => s is an integer by construction. So if the square root of any integer is fractional, then it is irrational. 1
How come not being able to represent in lowest form makes it irrational? In that case i cannot represent 2 in lowest form since 2 = 0.000000---02/0.000000---01, i can keep on adding zeroes.
Nein nein nein, the lowest form basically means that a number could be expressed as two integers ,not fraction or decimal. For example, I can say that 2=2/1 but not 0.2/0.1 ,another one , we can express 1.5 as 3/2.Here 1.5 is not an integer but 2 is
"every rational number has a unique finite continued fraction" Actually every rational number has exactly two finite continued fractions. For instance 3/2 = 1+1/2 = 1+1/(1+1/1)
Yes, like Tipping Point Math said, there exist a lowest form for every rational. However, with the "rational" we see in these proofs, we see that, even if we write it in its presumed "lowest form," we can still reduce it. Thus, there is a form lower than the "lowest form" for this "rational," i.e. we can keep reducing the "rational" forever, without actually reaching an absolute lowest form. This can never be true for a rational. Reduced form may not have anything to do with irrationality directly, but the point is that, in our argument, we end up with something absurd, something that cannot be true. So the argument's only assumption - that sqrt(2) is rational - must be wrong.
ALL these proofs did was prove root 2 is a stupid concept. The extension, to justify your claims is to make stupidity, by definition a legitimate number concept.
Look at that. I see how it's irrational. I don't get why in a satisfying way. Square root of 4 isn't irrational, but 2 is. Yea it makes sense with geometry and algebra sure but is 2 really so different than 4?
@@enantiodromia Well yes, but it also stands alone in being an even prime. Probably the simplest way to get a source of infinite randomness. I don't have the expertise to technically explain this down to the last detail but it seems to me that in a perfectly symmetrical Universe without CP violation equal parts matter and anti-matter and no time square root of 2 should be rational. Albeit to state that would be impossible in such a Universe. Maybe someone with the expertise could elaborate on that further but physics is not my field. I could be wrong but I find it fascinating that such a prime exists at all.
You did a bad job explaining proof 3 and 5 and over complicated proof 1. What's the purpose of making a video showing proofs to prole who already understand it? If there isn't and you're explaining it to people who don't, then you can't move that fast, by saying 'well if divide it at the length of b and make a smaller triangle dividing the hypotenuse at a-minus b then we can see these are congruent and if... Then... And if then.. then ... So see?! You can't just explain the proof using the mathematical logic, otherwise why even talk, just write it down. The point of making the video is supposed to be to explain the mathematical logic, not just assume that people understand it, because if they do they don't need you to make a video or speak, I'm going around in circles because that's what I think of your explanations. Pointless (not a pun) if you're not going to breakdown the logic express from your own angle (not a pun) that you feel would make it easier for people to relate to and grasp the concept. Otherwise this is less useful and just as dry as reading the proofs on Wikipedia
>her
>I'm not being irrational!
>me
>What's your unique finite continued fraction representation?
>???
>win
Pythagoras is gonna be super pissed when he sees this.
Geometry > algebra? I think you're being irrational.
LOL
ILLOGICAL STATEMENT DETECTED, SYSTEM 101 OVERLOAD
My favorite proof:
If √2=a/b with gcd(a,b)=1,
then 2=a^2/b^2 and gcd(a^2,b^2)=1,
so a^2/b^2 is the reduced form of 2.
This implies a^2=2 and b^2=1,
which is obviously impossible as a is an integer.
This proof is my favourite as it is not only really simple, but also generalizes really easily to all irrational roots.
How do you get the 1st implication where gcd(a^2,b^2)=1
Yamin New You use the fundamental theorem of arithmetic to deduce that gcd(a^2,b^2)≠1 would imply that a^2 and b^2 must share a common prime factor which a and b must then share as well. By contraposition, gcd(a,b)=1 implies gcd(a^2,b^2)=1.
In case you feel like this makes this proof relatively complicated, I want to mention that the classic proof uses similar arguments.
I think the most elementary proof (and hence, quite reasonably, the best solution) is the one that simply uses the ''extremal principle'' or ''infinite descent'', it is somehow known by different names because of the great importance it actually has in the development of further arguments and well know elementary proof techniques. If x=sqrt 2 is rational, the, by definition, there are a, b integers (positive wlog) such that a=bx. Thus there exist at least one positive integer M such that Mx is an integer. Let N be the smallest such M. Then N(x-1) is a positive integer. Observe N(x-1)(x) is also an integer (easy verification). By minimality we deduce N is not greater than N(x-1), which is clearly false. The contradiction finishes the proof.
Recall the importance of such elementary arguments, by using the fundamental theorem of arithmetic, which can be of course proved in many different ways, you are using a more bigger fact than the one you want to prove. This is not so strange, sometimes in mathematics you develop a theory in which the problem you wanted to solve, in terms of the new ''lenguage'' and tools discovered, is basically trivial. But we may want to avoid this process at this level ;)
Wow that's excellent
I would hope the error at 0:44 would be corrected. If we are replacing "a" with 2c, then a^2 would be 4c^2, not 2c^2.
There is no error. It says b^2 = 2 * c^2.
He already replaced a^2 by 4 * c^2 and divided by 2.
He divided both sides by 2
for the last proof I think you should've started something like this:
2√2 + 3 = 2√2 + 3
2 + 2√2 + 1 = 2(√2+1) + 1
(√2 + 1)*(√2 + 1) = (2 + 1/(√2+1))*(√2+1)
√2 + 1 = 2 + 1/(√2+1)
instead of just pulling the equation out of nothing with the implication that it's true.
At the very least you could've multiplied both sides by √2+1 to show that they are in fact equal.
2:00 Tipping Point Memes
How?
RS Extreme - dude. oh my god. Everywhere I go. this is getting insane, haha
The best proofs are those that generalise, in this case to show that the square root of any positive integer is either an integer or irrational, or, better still, that the nth root of any positive integer is either an integer or irrational. You can do this using unique factorisation, as hinted at by the start of proof 2.
Proof 1 is the best for me. Great video
Last one was the easiest to grasp and most elegent.
Hey, I just realised that your logo at the start of this video looks like part of a proof of the Pythagorean theorem, marking out the area for c^2 in black.
From this we can see that four right-angled triangles around one square fit into a larger square. Using a bit of algebra we can show:
(a+b)^2 = c^2 + 4*(0.5*a*b)
a^2 + 2ab + b^2 = c^2 + 2ab
a^2 + b^2 = c^2
That choice of logo was on purpose before the first video was released!
And it took me this long to realise it. Does it usually have an outline? If not, that's probably why it slipped under my radar for so long (somewhere around the three-year mark, likely introduced via the A4 paper video). The profile icon lacking an outline hides the triangles when displayed on a white page.
That's been one of my favorite PT proofs, since I read long ago that it was devised by James A. Garfield, 20th U.S. president, who was a teacher before he entered politics.
Radical, dude!
I like this:
The square root of any integer is always either an integer or irrational (since a rational non-integer number will always square to another rational non-integer number)
sqrt(2)>sqrt(1)=1 and sqrt(2)
That's hot
here's my favourite proof:
sqrt(2) = a/b
2 = (a/b)^2
2 = a^2/b^2
but since the prime-factorizations of square numbers have to have all primes to an even power, one of these must not be a square number, and we have a contradiction.
This actually proves that the square roots of all natural numbers are irrational except for the ones of square numbers, which are whole numbers. Because the quotient of two square numbers must be also a square number.
@@ObiWahn68 "the quotient of two square numbers must be also a square number."
*only if it is a natural number
other than that, yes, it's a simple proof, but it's very effective.
Number 5 is my favorite.
I really liked proof 4.
Franz Schubert Same
Geometric proofs = best proofs.
Another vote for proof 4
Mind blowing bro...why don't they teach in high school like this...
All the proofs revolved around contradicting that a/b is the lowest form representation.
We learned proof 1 in the yellow analysis book, but this time I vote for proof 4.
Proof 1 is the classical proof that everyone learns. It's nice, but I prefer some of the others. What's your favorite?
I love proof 4 best in the sense that I guess it is the most straightforward method that everyone can immediately get the point. Proof 1 is actually not that accessible; I once showed this version to a friend of mine who was an engineering student at Purdue. He was shocked and could not understand why would you ever want to prove something so obvious. To mathematicians, 'obvious' is a dangerous word, but from others' perspective, rigorous proof writing and little tricks like contradiction won't be popular outside mathematicians' own utopia. By saying utopia, I mean mathematicians have their own way of thinking, and we are proud of it, but others simply do not care.
The last one was the most interesting.
I don't understand the conclusion of proof 2. Can someone help?
Proof 2 tells us not only that we cannot write sqrt(2) as a/b, but also gives us a lower bound on the gap between those two values.
@@TippingPointMath You can make the error arbitrarily small.
Assume that Root 2 = a/b
A must be even since A^2 = 2 x B^2.
All even square numbers are divisible by 4, so B has to be even as well. It’s impossible to write in simplest form.
He sounds like Mark Ruffalo. Thank you Dr. Banner
I like the following proof sketch:
We can show using homomorphisms that for all field automorphisms of Q, the rationals, f(p)=p for all p in Q.
Consider the field automorphism h from the field extensions Q(sqrt(2)) to Q(sqrt(2)) defined by h(sqrt(2)) = -sqrt(2).
This is indeed an automorphism, but h(sqrt(2)) =/= sqrt(2). Thus sqrt(2) is not rational.
at 0:52, I don't understand how you conclude that because the left side of the equation, b^2, is even, that means b is even as well since any odd integer squared produces an even number. Pls explain! Thank you
"any odd integer squared produces an even number"
Name a single odd integer whose square is even.
how do you get the length of the blue square to be 2b-a?
The length of the outer square is a.
The length of the red square is a-b.
The length of the blue square is the length of the outer square minus two times the length of the red square.
Therefore the length of the blue square is a-2*(a-b) = 2b-a
I have a feeling this is gonna blow up... I can sense it
Edit: if it doesn’t I will cry
Edit 2: why is every comment a really bad pun
There is a smallest possible length, called the Planck distance. If you made a triangle with two sides with this Planck distance, the hypotenuse would not exist because it is not a multiple of this small distance.
My favourite is the Matholodger proof.
the first, the square and the last were my favorites
why we use the hypothesis a/b is in its lowest form?
what we can say about 2/4?
is 2/4 is rational or irrational? ... i m confused.
Because that's where the contradiction comes from. We assume it is the lowest form but we can always find a lower one. There is no lowest form, which contradicts the assumption.
Does this mean that the square root of any number that isn’t a perfect square is irrational?
Correct. Any positive integer that is not a perfect square is irrational. Of course the square root of a fraction may be a fraction. For example, sqrt(9/4) = 3/2.
That's true but it doesn't follow from the content of the video
I dont know if Crit1kal is good at math.
Well, let's get the first proof out of the way.
Replace 2 with variable p.
./p = a/b
(./p)b = a
p(b^2) = a^2
Since one side is a power of p, the other must also be. Let's assume a/p = c.
b^2 = p(c^2)
Since one side is a power of p, the other must also be.
We have proved that both a and b are multiples of p.
This means they are not in simplest form, meaning that ./p cannot be rational.
Let's substitute 4 in, and also change the variables.
./4 = 2/1
(./4)1 = 2
4(1^2) = 2^2
1^2 = 2(0.5^2)
(I now know i was wrong....... let's figure out the long way
a%4 = 0 then a^2 = (4n)^2 = 16n^2 = 4(4n^2)
a%4 = 1 then a^2 = (4n + 1)^2 = 16n^2 + 2n + 1 = 4(4n^2) + 2n + 1
a%4 = 2 then a^2 = (4n + 2)^2 = 16n^2 + 4n + 4 = 4(4n^2 + n + 1)
If n is an integer such that s=sqrt(n) is rational, let a/b be its lowest form
a^b=1 implies by Bezout that there exists two integers c and d such that ac+bd=1
Then 0=a-sb=(a-sb)(d-sc)=ad-s(ac-bd)+nbc=ad+nbc-s => s is an integer by construction.
So if the square root of any integer is fractional, then it is irrational.
1
I think ac-bd should be ac+bd
good, but i think the backround music is annoying
How come not being able to represent in lowest form makes it irrational? In that case i cannot represent 2 in lowest form since 2 = 0.000000---02/0.000000---01, i can keep on adding zeroes.
Nein nein nein, the lowest form basically means that a number could be expressed as two integers ,not fraction or decimal. For example, I can say that 2=2/1 but not 0.2/0.1 ,another one , we can express 1.5 as 3/2.Here 1.5 is not an integer but 2 is
That's why we can't say that root 2 as p/q where p,q are integers and q is not equal to 0
Why is that √2+1 = 2 + 1/(√2+1)?
how can we know that 2b-a = (a-b) + (a-b), shouldn't it be (2b-a)^2 = (a-b)^2 + (a-b)^2
proof no 4 is awesome
What about any root that is not an integer it is irrational
Thanks
"every rational number has a unique finite continued fraction"
Actually every rational number has exactly two finite continued fractions. For instance 3/2 = 1+1/2 = 1+1/(1+1/1)
I hadn't seen the second one before so that was my favorite.
damnit i don't understand why showing your assumption of being in lowest form is wrong is proof of irrationality
Every fraction can be written in lowest form. Some of the proofs end up showing that the lowest form is violated, hence a contradiction.
Yes, like Tipping Point Math said, there exist a lowest form for every rational. However, with the "rational" we see in these proofs, we see that, even if we write it in its presumed "lowest form," we can still reduce it. Thus, there is a form lower than the "lowest form" for this "rational," i.e. we can keep reducing the "rational" forever, without actually reaching an absolute lowest form. This can never be true for a rational.
Reduced form may not have anything to do with irrationality directly, but the point is that, in our argument, we end up with something absurd, something that cannot be true. So the argument's only assumption - that sqrt(2) is rational - must be wrong.
Nice vid!!!!
I understand the first proof but not the last 4 proofs.
Beautiful video
There is a mistake at 0:50
a=2c >>> a2=4c2 and not 2c2
No it's not a mistake. There is a thing called dividing by two
A number squared isn’t always even.
Can the square root of an irrational number be rational?
No, because the square of a rational is rational.
You are aweeeeesome
ALL these proofs did was prove root 2 is a stupid concept. The extension, to justify your claims is to make stupidity, by definition a legitimate number concept.
2 be squared or not 2 b squared...
The simplest one,of course.
just here for the epic meme
Look at that. I see how it's irrational. I don't get why in a satisfying way. Square root of 4 isn't irrational, but 2 is. Yea it makes sense with geometry and algebra sure but is 2 really so different than 4?
Yes, 4 is a perfect square, 2 is not.
@@enantiodromia Well yes, but it also stands alone in being an even prime. Probably the simplest way to get a source of infinite randomness. I don't have the expertise to technically explain this down to the last detail but it seems to me that in a perfectly symmetrical Universe without CP violation equal parts matter and anti-matter and no time square root of 2 should be rational. Albeit to state that would be impossible in such a Universe. Maybe someone with the expertise could elaborate on that further but physics is not my field. I could be wrong but I find it fascinating that such a prime exists at all.
Senpai Notice Me
4 is awesome!
(x/v)^2= 2. Simple. ..
You made a mistake in the first proof:
a=2c
a^2 = (2c)^2 = 4c^2
You wrote a^2 = 2c^2
Martin Wiesner Because he devided both sides by 2
No, he didn't. He wrote b^2 = 2c^2, which is correct.
48 130 th sub!
You did a bad job explaining proof 3 and 5 and over complicated proof 1. What's the purpose of making a video showing proofs to prole who already understand it? If there isn't and you're explaining it to people who don't, then you can't move that fast, by saying 'well if divide it at the length of b and make a smaller triangle dividing the hypotenuse at a-minus b then we can see these are congruent and if... Then... And if then.. then ... So see?! You can't just explain the proof using the mathematical logic, otherwise why even talk, just write it down. The point of making the video is supposed to be to explain the mathematical logic, not just assume that people understand it, because if they do they don't need you to make a video or speak, I'm going around in circles because that's what I think of your explanations. Pointless (not a pun) if you're not going to breakdown the logic express from your own angle (not a pun) that you feel would make it easier for people to relate to and grasp the concept. Otherwise this is less useful and just as dry as reading the proofs on Wikipedia
To se u nas zove resavska skola
But I knew that -_- I thoght everyone knew that
0:50 that should be 4c^2
nah it was 2b² = 4c² -> b² = 2c²
0/10 Worst music ever.
For proof 1, if you plug in 2c for a, you'd get 4c^2, not 2c^2.
Notice that the left-hand-side was also divided by 2, so 4c^2 became 2c^2.