Mistake on the second to last line of the second proof: you wrote that the expression is Not an element of the irrationals, which is the same as saying it IS rational, which is not a contradiction. You probably didn't want the tilda.
When you were looking for a number aside from 10, I said to myself "21!", and then you said it too. Great minds think alike! Or, mediocre minds something think like great minds and it helps them boost their confidence for two minutes of the day.
After 6=p^2/q^2 We can also transpose q^2 to the left and can say that 6 is a factor of P^2 hence a factor of p too. As 6 is a factor, p=6c for some integer c Substituting that in the original equation (6c)^2= 6q^2 After cancellations, we get that 6 is a factor of q^2 and hence a factor of 2 as well. But this contradicts with the fact that.p and q are co-primes
"6 is a factor of P^2 hence a factor of p too" I think this needs more justification as it relies to some extent on 6 not being a square (a special case of sqrt(6) not being rational) Compare "9 is a factor of p^2 hence 9 is a factor of p", which fails for say p = 6.
This is because when P^2/q^2 =6 which is P^2 = 6q^2 Meaning that P^2 has two factors 6 and q And hence p has two factors 6 and q because a theorem says that if p divides a^2 p also divides a
@@abhinavdiddigam2330 "p divides a^2 p also divides a" is true when p isn't a multiple of a square, so 12 dives 6^2 without 12 dividing 6. So true in this case for 6 but would need to be noted, especially if sqrt(6) WAS rational and an integer!
At 5:10, considering √6 to be rational. Any rational number can be written in the form of p/q where p and q are integers and they are co-primes and q not equal to 0 This is a basic introductory line in proving irrationality So, according to your question 12 can divide 6^2 but not 6. But twelve is not a prime number
I always had a question in it What if instead of √6 we use like 8 8= p/q 64= p²/q² 64q²= p² So 64 is a factor of p² So 64 is a factor of p Hence P=64m [ m is an integer] P²= 4096m² 64q²= 4096m² q²= 64m² So 64 is a factor of q² Hence 64 is a factor of q This contradicts our point that p q are co prime hence 8 is an irrational number But we all know 8 is a rational number
5:25 we can do that because if p and q share factors, then we can cancel them such that they are in the lowest form. but that is the same as if they share no factors. so that is why we can assume that statement.
*General Theorem* - Every non-negative integer is either a perfect square (whose roots are +/- an integer) or a square of an [+/-] irrational number. i.e. for any co-prime integers p and q, if (p/q)^2 = n = some non-negative integer then q = 1 (the rational number has to be an integer). √6 (and more generally any linear combination of square-roots of non-perfect square integers) can be proved to be irrational using this *General Theorem* . No point in proving any specific case such as √6 or √3 +/- √2.
For (b) I think we can make the argument airtight by writing sqrt(6) = ((sqrt(3)+sqrt(2))^2 - 5)/2. Under our supposition, the RHS is rational, but by (a), LHS is irrational, which gives an explicit contradiction :)
I agree. The way he presented it requires that the addition of irrational and rational numbers is an irrational number, which possibly needs another proof.
You had a misprinted sir, the square of the sum is not belong to rational. And you wrote irrational instead of rational. Before last line. By the way, you are the best.
As far as I‘m concerned everything is correct. He just stated, that the term before was a contradiction to his assumption. Therefore his assumption is false which is exactly the point of a proof by contradiction.
@@IchKenDich SAZGAR is quite correct. Mr Woo should have written Q at 17:51, not Q~ The full expression is '(sqrt 3 + sqrt 2)^2 is NOT an element of Q'. That is the contradiction. If (sqrt 3 + sqrt 2)^2 was not an element of Q~, then it would be an element of Q, so there would be no contradiction.
Might be clearer to show that since Q is closed under the four arithmetical operations, sqrt(6) = ((sqrt(3) + sqrt(2))^2 - 5)/2 is rational, contradicting the first part.
You assume, without prior knowledge, that the rationals are closed under arithmetic for ONLY rationals, i.e. that a rational * irrational for some operation * is always rational, which we don't know and aren't given. He assumes this too, which I don't really like. It's pretty easy to show that rational + irrational = rational for all rational/irrational
In my opinion, instead of doing all those standard proofs about sqrt(2) (or in this case sqrt(6)) proofs it is much easer seeing that the n-th root of an integer is either an integer or irrational. This requires no calculations and is more insightful.
Here’s how I would approach part b without part a. If x = sqrt(2)+sqrt(3) then x^2-5=2sqrt(6) so (x^2-5)^2=24 which gives us x^4-10x^2+1=0 The only possible rational roots of this polynomial are plus or minus 1. Thus, sqrt(2)+sqrt(3) is an irrational root.
@@reubenmanzo2054 I am assuming that you are referring to the rational root statement? From the rational root theorem, the possible rational roots are found by taking the divisors of the constant term and dividing by the divisors of the leading coefficient. Since both of these quantities are 1, then the only divisors are 1 and -1 and so the possible combinations are either 1/1, 1/(-1), (-1)/1, or (-1)/(-1). Thus, the only possible rational roots are 1, -1. Clearly, sqrt(2)+sqrt(3) is not equal to either of these. But, by construction, sqrt(2)+sqrt(3) is a root of the quartic polynomial and thus, must be an irrational root.
There's a much easier & faster method of proof to show that the square root of ANY non-square number is irrational. The square root of a positive integer n is the solution of the polynomial x^2 - n = 0. By the Rational Root theorem, the only rational candidates are factors of n, and since none of them work, the solution must be irrational! QED :)
@@AchtungBaby77 This proof is less formal than the one shown in the video. If you want it to be just as formal, you'll have to prove the theorem. However that's way more complicated and unnecessary.
What is ring?? What do you mean algebraic integer?? Remember to not use too much mathematical jargon. Not many people understand it, even students from math department
@@arolimarcellinus8541 This solution relies on theorems from abstract algebra (usually a 2nd or 3rd year undergrad course). A "ring" is a set with addition and multiplication, satisfying certain axioms. An "algebraic integer" is a root of a monic polynomial (i.e., leading coefficient is 1, all other coefficients are integers). e.g., 1, sqrt(3), (1+sqrt(5))/2 are algebraic integers, whereas 1/2, (1+sqrt(7))/2 are not. The first theorem I used is the fact that the set of algebraic integers is a ring. This isn't easy to prove but it's a standard fact from abstract algebra. This theorem implies that the sum of two algebraic integers is also an algebraic integer. The second theorem I used is the rational root theorem (much easier to prove), which says that a rational algebraic integer must be an integer. Hope this helps
Where is the *proof* that 5 + 2*sqrt(6) is irrational? We've proved that sqrt(6) is irrational, but nothing you've shown has proven that a linear combination of rationals with a single irrational is irrational. That is true - and not difficult to prove - but you can't just say "because it has an irrational tucked away inside" the expression. Rigorous proofs don't rely on "I say so".
Let A = √3+√2 and B = √3-√2. Therefore AB=3-2=1, so A=1/B and B=1/A. So, either both are rational or both are irrational. A+B=2√3 and A-B=2√2. Since √2 (and√3) are irrational, therefore both have to be irrational (since both are rational would be true only if √2 and √3 were rational). There A = √3+√2 is irrational (since both A and B are irrational). Hence, proved.
With the exception of perfect square numbers (4, 9, 16, 25 etc), the square roots of all positive integers are irrational and it can be proven by the same proof by contradiction as you described.
Hey Teacher, I was thinkin, when we square all the equation, by assuming we are working on Z+. I'll have: 6 = p²/q², because I'm on the positive integers, I can rewirte this equation as: q² = p²/6. Can't I work with the fact that p²/6 is equal to a perfect square number?
I guess you could say that both p² and q² are perfect square, therefore all of their prime factors must have a even power (i.e. 36=2²×3², 144=2⁴×3²), and that p must have at least 2² and 3² because it being divided by 6 SHOULD give you another integer, but if you think about division when you already factored the number, dividing by 6 means that q has 2 to the power of an odd number (I.e. 36/6=(2²*3²)/2*3=2¹*3¹) and it is impossible for a perfect square. Don't know if it's an accurate or valid proof, but thought it was at least a step in the right direction
I don't think I fully understand why at (a) p and q need to be coprime. If they indeed were to share factors, they would result in forming an integer, which surely is rational as well?
Yes, but if they share factors, we can find p' and q' by cancelling the common factors of p and q, and p'/q'=p/q, and continue the proof with p' and q'.
I think the last statement needs more justification, just because 5+2*sqrt(6) contains an irrational number, I don't think it's trivial to say that makes it irrational, I think there should have been a couple more lines
Firstly thank you very much I have a qustion : in the end of the proof you said if (3½+2½)²is irrationnel then (3½+2½)is also irrationnel So whow can we be sur that means x² irrationel=>x irrationel I mean aren't we also in need to proove this as well. 😊😊😊
Here's another proof. Set a = sqrt(2)+sqrt(3), then a² = 5 + 2.sqrt(6), so (a²-5)² = 24, hence a is a root of x^4 - 10x² + 1, so if a was rational, it would be equal to +-1, which is contradictory
I am in 10th. And I feel more comfortable to apply our method to prove √6 is irrational.. 😂 I would solve this in 2min... But I got confused by seeing all these 😂😂
Because we know that the sum of a rational and an irrational is irrational, but we can't say anything about the sum of two irrational numbers. For example, 1-sqrt(2) and sqrt(2) are both irrational, but their sum is rational.
Just looking at 6=p^2/q^2 I realize that p^2 must be 6 times more than q^2, and therefore p must have a factor of 6^0.5 which isn't an integer and therefore isn't possible. Really interesting problem.
With this method, I can proove root 9 is irrational, if you take square root 9 irrational, then assume root 9 rational, then root 9 equal to p/q, then square both sides, do same method above, we can prrove root 9 irrational, please can you explain where did we do wrong.
Isn't the first proof over as soon as you find that 6*q^2 = p^2 because q and p are coprime therefore q can't be a factor of p. Further, this means that sqrt(x) which isn't a whole number is always irrational.
You skipped over the proofs that a Rational + Irrational = Irrational and Rational * Irrational = Irrational. That might sound trivial, but both an Irrational + Irrational and Irrational * Irrational can be either Irrational or Rational. Let's all be rational.
How would you go about proving that the square root of a non-composite number (like sqrt(7)) is irrational? Since you can't make the same argument that p^2 is even I imagine you'd have to prove that p and q share a factor of 7, if that's even possible.
I believe it is easier to realise that if p and q share no factors, p^2 and q^2 also have no factors in common (since prime decomposition is unique - this is easily proved), but each factor occurs an even number of times (since squaring doubles the occurrence of each factor). Using the same approach we just note. So we can show sqrt(7) is irrational as 7 is a factor of p, and 7 is a factor of q. Where as if the number we root is a square, then it can be the factor that is the repeated factor by which the squares differ, so we need to show 7 is not a square. This proof applies to other roots since a^(1/d) where d is integer just means each factor occurs some multiple of d times, in p^d and q^d. But this seems too simple a proof, so maybe I'm missing something.
I always had this question and you might find me stupid but let me ask this Literally any number is that no./1 right expressed as a fraction so by that defination p/q so then no number is irrational
im confused as to why this proved that sqrt 6 is irrational? doesnt all this prove is that sqrt 6 can be reduced by a factor of 2? why does assuming that p and q have no factors in common prove that something is rational and that if they can still be reduced then they are irrational? i think this is my main hang up. im so lost please help me fill in the gap.
Your question is actually a big point If you continue that and you keep simplifying that you will never get to a factor q/p=k/m and you fo it again it will again give you same results even/even this will continue for ever
When you need to study, but get stuck in all of he's interesting video's. Then on the exam you get an A. It's like wtf i didn't even open the math book.
Can you please provide an explanation for why pi is irrational even if it can be expressed as a ratio where pi=c/d, c and d being the circumference and diameter of a circle?
Not many understands what is relatively prime in the beginning. Even some uni students cannot understand what is that term. No factor is more straightforward.
This kind of question was in the entrance exam of one highschool in vietnam i taked that exam last yesterday,i am looking forward to knowing my result and wondering if i can be the student of that highschool :(((
Ikr rt dude and i remember this one question which was asked in a cambridge interview and the exact same question was there in our ninth half yearly lmao
I think this is really well done. Unfortunately, I also don't understand why the ratio of numbers can't have any factors in common. If they had common factors, like 10/15, wouldn't the ratio just reduce to an equivalent value (2/3), thus giving us, in effect, the same ratio?
Like you implied, each rational number can be reduced to a "simplest form". If we start from that form it makes it easier to find the contradiction in the assumptions of the proof.
lol i'm 9 years old...............................................yet i do calculus............................................please belive me.........
You make math feel so much more accessible to me, thank you for your amazing videos!
I love all the Algebra 1 connections in your proof. It is a great way to show the beginning algebra tools are quite valuable
The second proof is easy.
sqrt(2)+sqrt(3)=1.41 + 1.73 = 3.14 = pi
And everyone knows pi is irrational.
sqrt2 + sqrt3 does not equal pi. false equivalence.
@@djhc4 That's a joke...
@@mr.schloopka1124 yikes, didnt realise.
wow, this is the best comment I ever seen!
hello everyone knows that pi=circumference/diamter, hence clearly rational
I loved my college professor who taught Proofs and Analysis, but this is so good! I finally understand this proof!
Mistake on the second to last line of the second proof: you wrote that the expression is Not an element of the irrationals, which is the same as saying it IS rational, which is not a contradiction. You probably didn't want the tilda.
Ye I think that tilda is just a typo
@@zachary4849 yeah
Flowery discussion not advisable, too lengthy, bad.
Mr. Eddie is a package of blessings
When you were looking for a number aside from 10, I said to myself "21!", and then you said it too. Great minds think alike! Or, mediocre minds something think like great minds and it helps them boost their confidence for two minutes of the day.
21 factorial? that's a big one
or maybe u guys are both deep in meme culture
@@andrewsong7760 😏
"Small minds beg to differ"
Best maths teacher❤❤❤
Awesome, as always! By the way, Eddie, which model is that Tablet you use for the classes? It looks wonderful.
It's probably an iPad, since the app he's using (Notebility/Notability unsure of the spelling) is only available on iOS
its either an Ipad pro or an Ipad air 4, judging off the apple pencil gen 2 and the fact that it has the navigation bar thingy at the bottom
Notability looks awesome - is that just iPad split screen, and screen share using zoom or something?
funny thing, i was just proving this and then this got recommended !
Naturally
Very much intuitive
After 6=p^2/q^2
We can also transpose q^2 to the left and can say that 6 is a factor of P^2 hence a factor of p too. As 6 is a factor, p=6c for some integer c
Substituting that in the original equation
(6c)^2= 6q^2
After cancellations, we get that 6 is a factor of q^2 and hence a factor of 2 as well.
But this contradicts with the fact that.p and q are co-primes
"6 is a factor of P^2 hence a factor of p too" I think this needs more justification as it relies to some extent on 6 not being a square (a special case of sqrt(6) not being rational) Compare "9 is a factor of p^2 hence 9 is a factor of p", which fails for say p = 6.
This is because when P^2/q^2 =6 which is P^2 = 6q^2
Meaning that P^2 has two factors 6 and q
And hence p has two factors 6 and q because a theorem says that if p divides a^2 p also divides a
@@abhinavdiddigam2330 "p divides a^2 p also divides a" is true when p isn't a multiple of a square, so 12 dives 6^2 without 12 dividing 6. So true in this case for 6 but would need to be noted, especially if sqrt(6) WAS rational and an integer!
At 5:10, considering √6 to be rational. Any rational number can be written in the form of p/q where p and q are integers and they are co-primes and q not equal to 0
This is a basic introductory line in proving irrationality
So, according to your question 12 can divide 6^2 but not 6. But twelve is not a prime number
I always had a question in it
What if instead of √6 we use like 8
8= p/q
64= p²/q²
64q²= p²
So 64 is a factor of p²
So 64 is a factor of p
Hence
P=64m [ m is an integer]
P²= 4096m²
64q²= 4096m²
q²= 64m²
So 64 is a factor of q²
Hence 64 is a factor of q
This contradicts our point that p q are co prime hence 8 is an irrational number
But we all know 8 is a rational number
5:25 we can do that because if p and q share factors, then we can cancel them such that they are in the lowest form. but that is the same as if they share no factors. so that is why we can assume that statement.
These videos are gold
*General Theorem* - Every non-negative integer is either a perfect square (whose roots are +/- an integer) or a square of an [+/-] irrational number. i.e. for any co-prime integers p and q, if (p/q)^2 = n = some non-negative integer then q = 1 (the rational number has to be an integer). √6 (and more generally any linear combination of square-roots of non-perfect square integers) can be proved to be irrational using this *General Theorem* . No point in proving any specific case such as √6 or √3 +/- √2.
For (b) I think we can make the argument airtight by writing sqrt(6) = ((sqrt(3)+sqrt(2))^2 - 5)/2. Under our supposition, the RHS is rational, but by (a), LHS is irrational, which gives an explicit contradiction :)
I agree. The way he presented it requires that the addition of irrational and rational numbers is an irrational number, which possibly needs another proof.
You had a misprinted sir, the square of the sum is not belong to rational. And you wrote irrational instead of rational. Before last line. By the way, you are the best.
17:53
As far as I‘m concerned everything is correct.
He just stated, that the term before was a contradiction to his assumption. Therefore his assumption is false which is exactly the point of a proof by contradiction.
@@IchKenDich SAZGAR is quite correct.
Mr Woo should have written Q at 17:51, not Q~
The full expression is '(sqrt 3 + sqrt 2)^2 is NOT an element of Q'. That is the contradiction.
If (sqrt 3 + sqrt 2)^2 was not an element of Q~, then it would be an element of Q, so there would be no contradiction.
Might be clearer to show that since Q is closed under the four arithmetical operations, sqrt(6) = ((sqrt(3) + sqrt(2))^2 - 5)/2 is rational, contradicting the first part.
You assume, without prior knowledge, that the rationals are closed under arithmetic for ONLY rationals, i.e. that a rational * irrational for some operation * is always rational, which we don't know and aren't given. He assumes this too, which I don't really like. It's pretty easy to show that rational + irrational = rational for all rational/irrational
I enjoy this proof a lot... thank you!
In my opinion, instead of doing all those standard proofs about sqrt(2) (or in this case sqrt(6)) proofs it is much easer seeing that the n-th root of an integer is either an integer or irrational. This requires no calculations and is more insightful.
In Unix context the tilde means home folder. And my home folder contains lots of irrational items.
Here’s how I would approach part b without part a. If x = sqrt(2)+sqrt(3) then x^2-5=2sqrt(6) so (x^2-5)^2=24 which gives us x^4-10x^2+1=0
The only possible rational roots of this polynomial are plus or minus 1. Thus, sqrt(2)+sqrt(3) is an irrational root.
And how did you establish that final statement?
@@reubenmanzo2054 I am assuming that you are referring to the rational root statement? From the rational root theorem, the possible rational roots are found by taking the divisors of the constant term and dividing by the divisors of the leading coefficient. Since both of these quantities are 1, then the only divisors are 1 and -1 and so the possible combinations are either
1/1, 1/(-1), (-1)/1, or (-1)/(-1).
Thus, the only possible rational roots are 1, -1. Clearly, sqrt(2)+sqrt(3) is not equal to either of these. But, by construction, sqrt(2)+sqrt(3) is a root of the quartic polynomial and thus, must be an irrational root.
In question (b), it's actually not necessary to use proof by contradiction - you can proof it without assuming $(\sqrt{2}+\sqrt{3})\in\Bbb{Q}$
Thank you for a great explanation of mathematical logic.
There's a much easier & faster method of proof to show that the square root of ANY non-square number is irrational.
The square root of a positive integer n is the solution of the polynomial x^2 - n = 0.
By the Rational Root theorem, the only rational candidates are factors of n, and since none of them work, the solution must be irrational! QED :)
What is the Rational Root theorem?
@@reubenmanzo2054 Google is your friend :)
@@AchtungBaby77 This proof is less formal than the one shown in the video. If you want it to be just as formal, you'll have to prove the theorem. However that's way more complicated and unnecessary.
My daughter loves your vids.
Algebraic integers form a ring. So sqrt(2)+sqrt(3) is an algebraic integer, but it is not an integer, so it is irrational. q.e.d.
What is ring?? What do you mean algebraic integer?? Remember to not use too much mathematical jargon. Not many people understand it, even students from math department
@@arolimarcellinus8541 This solution relies on theorems from abstract algebra (usually a 2nd or 3rd year undergrad course). A "ring" is a set with addition and multiplication, satisfying certain axioms. An "algebraic integer" is a root of a monic polynomial (i.e., leading coefficient is 1, all other coefficients are integers). e.g., 1, sqrt(3), (1+sqrt(5))/2 are algebraic integers, whereas 1/2, (1+sqrt(7))/2 are not.
The first theorem I used is the fact that the set of algebraic integers is a ring. This isn't easy to prove but it's a standard fact from abstract algebra. This theorem implies that the sum of two algebraic integers is also an algebraic integer. The second theorem I used is the rational root theorem (much easier to prove), which says that a rational algebraic integer must be an integer. Hope this helps
Love the proofs.
Where is the *proof* that 5 + 2*sqrt(6) is irrational? We've proved that sqrt(6) is irrational, but nothing you've shown has proven that a linear combination of rationals with a single irrational is irrational.
That is true - and not difficult to prove - but you can't just say "because it has an irrational tucked away inside" the expression. Rigorous proofs don't rely on "I say so".
Let A = √3+√2 and B = √3-√2. Therefore AB=3-2=1, so A=1/B and B=1/A. So, either both are rational or both are irrational. A+B=2√3 and A-B=2√2. Since √2 (and√3) are irrational, therefore both have to be irrational (since both are rational would be true only if √2 and √3 were rational). There A = √3+√2 is irrational (since both A and B are irrational). Hence, proved.
If you're going to restrict p and q to positive integers, wouldn't it save time to simply restrict them to naturals?
With the exception of perfect square numbers (4, 9, 16, 25 etc), the square roots of all positive integers are irrational and it can be proven by the same proof by contradiction as you described.
Hey Teacher, I was thinkin, when we square all the equation, by assuming we are working on Z+. I'll have:
6 = p²/q², because I'm on the positive integers, I can rewirte this equation as:
q² = p²/6. Can't I work with the fact that p²/6 is equal to a perfect square number?
after p^2/6 is a perfect square and therefore p is a multiple of 6 where do you go?
sub p=6k, 6 = 36k^2/q^2
1=6k^2/q^2
6k^2=q^2
and you just get brought back to the original equation
I guess you could say that both p² and q² are perfect square, therefore all of their prime factors must have a even power (i.e. 36=2²×3², 144=2⁴×3²), and that p must have at least 2² and 3² because it being divided by 6 SHOULD give you another integer, but if you think about division when you already factored the number, dividing by 6 means that q has 2 to the power of an odd number (I.e. 36/6=(2²*3²)/2*3=2¹*3¹) and it is impossible for a perfect square. Don't know if it's an accurate or valid proof, but thought it was at least a step in the right direction
What writing app are you using? :)
17:53 oops, said Q but wrote Q'
SAZGAR got there first it seems
I don't think I fully understand why at (a) p and q need to be coprime.
If they indeed were to share factors, they would result in forming an integer, which surely is rational as well?
Yes, but if they share factors, we can find p' and q' by cancelling the common factors of p and q, and p'/q'=p/q, and continue the proof with p' and q'.
I think the last statement needs more justification, just because 5+2*sqrt(6) contains an irrational number, I don't think it's trivial to say that makes it irrational, I think there should have been a couple more lines
trivial? Trivial means unimportant lol. Wrong word there I think.
man u r so helpful ......love ur explanation
Sir Eddie Woo could you teach us A level math under cambridge.
I couldn't promise you that a 4 unit (Ext.2) maths teacher from Australia would teach the A level syllabus/curriculum for no reason.
We proof them by logical is one way we can !
Is proof by contradiction the only method to prove that a number is irrational?
Firstly thank you very much
I have a qustion : in the end of the proof you said
if (3½+2½)²is irrationnel then (3½+2½)is also irrationnel
So whow can we be sur that means
x² irrationel=>x irrationel
I mean aren't we also in need to proove this as well.
😊😊😊
Here's another proof. Set a = sqrt(2)+sqrt(3), then a² = 5 + 2.sqrt(6), so (a²-5)² = 24, hence a is a root of x^4 - 10x² + 1, so if a was rational, it would be equal to +-1, which is contradictory
And how did you establish the other roots weren't rational? This is the 'begging the question' fallacy.
@@reubenmanzo2054 No fallacy here. More explanation for the last step : en.wikipedia.org/wiki/Rational_root_theorem
I am in 10th. And I feel more comfortable to apply our method to prove √6 is irrational.. 😂 I would solve this in 2min... But I got confused by seeing all these 😂😂
I don't understand? I mean, he use the same method to prove that root 6 is an irrational that we use
it's literally the same method what are you on about 😐
@@laks._. it's not..he's using a different method. Maybe you use the same method as him but we don't. And it's kinda confusing.
If you say that 5 + 2sqrt(6) is irrational because one of its terms is irrational why can't you say that about sqrt(2) + sqrt(3)?
Oh, I guess you can but that was all about using irrationality of sqrt(6)
Because we know that the sum of a rational and an irrational is irrational, but we can't say anything about the sum of two irrational numbers. For example, 1-sqrt(2) and sqrt(2) are both irrational, but their sum is rational.
Just looking at 6=p^2/q^2 I realize that p^2 must be 6 times more than q^2, and therefore p must have a factor of 6^0.5 which isn't an integer and therefore isn't possible. Really interesting problem.
pointing out that you missed the possibility of 6^0.5 times p resulting in an integer - p being an integer its possible only if 6^0.5 is rational
With this method, I can proove root 9 is irrational, if you take square root 9 irrational, then assume root 9 rational, then root 9 equal to p/q, then square both sides, do same method above, we can prrove root 9 irrational, please can you explain where did we do wrong.
Why assume that p and q share no factor?
Make a video on Goldbach Conjecture
Isn't the first proof over as soon as you find that 6*q^2 = p^2 because q and p are coprime therefore q can't be a factor of p.
Further, this means that sqrt(x) which isn't a whole number is always irrational.
You skipped over the proofs that a Rational + Irrational = Irrational and Rational * Irrational = Irrational. That might sound trivial, but both an Irrational + Irrational and Irrational * Irrational can be either Irrational or Rational. Let's all be rational.
can we state that if p^2 is even, q^2 should be even as their ratio is an even number ?
In India these are class 10th ncert questions
who?
@@damianflett6360what who ?
I wish it was so good if I had such a teacher in my India.
I mean are you really teaches high school students...
How would you go about proving that the square root of a non-composite number (like sqrt(7)) is irrational? Since you can't make the same argument that p^2 is even I imagine you'd have to prove that p and q share a factor of 7, if that's even possible.
I believe it is easier to realise that if p and q share no factors, p^2 and q^2 also have no factors in common (since prime decomposition is unique - this is easily proved), but each factor occurs an even number of times (since squaring doubles the occurrence of each factor). Using the same approach we just note. So we can show sqrt(7) is irrational as 7 is a factor of p, and 7 is a factor of q. Where as if the number we root is a square, then it can be the factor that is the repeated factor by which the squares differ, so we need to show 7 is not a square. This proof applies to other roots since a^(1/d) where d is integer just means each factor occurs some multiple of d times, in p^d and q^d. But this seems too simple a proof, so maybe I'm missing something.
What app/program is he using for this?
I always had this question and you might find me stupid but let me ask this
Literally any number is that no./1 right expressed as a fraction so by that defination p/q so then no number is irrational
2=4/2
but you can't say √6=2√6/2
Because 2 times square root of six is irrational so you can't do it
im confused as to why this proved that sqrt 6 is irrational? doesnt all this prove is that sqrt 6 can be reduced by a factor of 2? why does assuming that p and q have no factors in common prove that something is rational and that if they can still be reduced then they are irrational? i think this is my main hang up. im so lost please help me fill in the gap.
Your question is actually a big point
If you continue that and you keep simplifying that you will never get to a factor
q/p=k/m and you fo it again it will again give you same results even/even this will continue for ever
which country are you from?
When you need to study, but get stuck in all of he's interesting video's. Then on the exam you get an A. It's like wtf i didn't even open the math book.
Please help me... Why do they need to be co prime... Rational numbers dont always exist as co prime
Can you please provide an explanation for why pi is irrational even if it can be expressed as a ratio where pi=c/d, c and d being the circumference and diameter of a circle?
For pi to be rational, c and d would both have to be whole numbers.
@@jelmerterburg3588 exactly.
Wait why do p and q have to share no factors?
Can't you right any whole number, which is rational, as just n/1 and then that's it's ratio?
Instead of saying no factor, say p and q are relatively prime.
Not many understands what is relatively prime in the beginning. Even some uni students cannot understand what is that term. No factor is more straightforward.
????? He said coprime lol
Can someone tells me what is the app he is using?
This kind of question was in the entrance exam of one highschool in vietnam i taked that exam last yesterday,i am looking forward to knowing my result and wondering if i can be the student of that highschool :(((
High-school already asked to prove math?? Woow....must be a very elite school
Suppose√2 + √3 is rational. (√2 + √3)(√2 - √3) = -1 = rational → 2 - √3 is rational. (√2 + √3)/(√2 - √3) = -(√2 + √3)² = -5 - 2√6 = rational + irrational = irrational → √2 - √3 = rational/irrational = irrational. Contradiction. So √2 + √3 is irrational.
Cant believe i'm watching this for entertaining purpose
This problem 2nd one came in my 10th board
2nd line is wrong
3:19
6q² = p² so 6 | p²
By gauss p² | 6 so p² = 6. But 6 is not a perfect square
3:14 & 6:28 & 12:56.
lmao we learn this in 10th grade here in India
we were literally doing the exact sum yesterday
Ikr rt dude and i remember this one question which was asked in a cambridge interview and the exact same question was there in our ninth half yearly lmao
Everybody does:)
I think this is really well done. Unfortunately, I also don't understand why the ratio of numbers can't have any factors in common. If they had common factors, like 10/15, wouldn't the ratio just reduce to an equivalent value (2/3), thus giving us, in effect, the same ratio?
Like you implied, each rational number can be reduced to a "simplest form". If we start from that form it makes it easier to find the contradiction in the assumptions of the proof.
BE RATIONAL, SQRT(3)+SQRT(2). How dare you not be, this is a disgrace to the Pythagorean cult. No number can be irrational....
U could have just proved that p and q are not coprime
This video is way too long for what it tries to show.
Here in India these questions are in chapter 1 of our mathematics class 10 and are basic one.
super
square root of 3 irrational please try to reply this comment and give the proof.
Pls be direct to the point
? Thank You.
√3 + √2 = √5 solved
does anybody watch blackpenredpen? if so then please reply this comment.
Talking unnecessarily too much.
Absolutely talks too much. Get to point more efficiently. Watch on 2x speed.
a= golden ratio (φ)= (1+√5)/2 , b= golden ratio conjugate (Φ)= ( √5 - 1 )/2 ,
My new formula's
We know,
beta [ m , n ] =[ Γ(m)Γ(n) ]/ [ Γ(m+n) ]
1) Beta [ b, 1] = [ Γ(b)Γ(1) ]/ [ Γ(1+b) ] = Γ(b)/Γ(a)
= [ 2 Γ(2b) ]/ [ Γ(a+b) ] = [ a Γ(2b+1) ]/ [ Γ(a+b)] = a
2) Beta [ a, 1] = [ Γ(a)Γ(1) ]/ [ Γ(1+a) ] = Γ(a)/ Γ(a^(2))
= [ -2 Γ(-2b) ]/ [ Γ(-a-b) ] = [ b Γ(1-2a) ]/ [ Γ(-a-b)]
= b
3) (a)! = a * ( (b)! )
4) (b)! = b * ( (a)! )
lol i'm 9 years old...............................................yet i do calculus............................................please belive me.........
Too slow
please come to usa....you should work for NASA....dont believe what you here its nice most places
Not a best mathematician😂
i love you
:)
>3
1st.) First! \0/
2nd