Response to question asked at 1:53 Let a,b integer co-prime non 0 such that a/b + b/a is in Z If a/b is in Z and isn't +/-1, b/a isn't in Z, thus a/b + b/a isn't in Z. Same for b/a. So a/b and b/a are either +/-1 or not in Z, if they are not in Z by multiplying both side with b: a + b^2/a = kb kb - a = b^2/a, left side in Z but right side isn't because a and b are co-prime Therefore there isn't any other solution other than a/b = +/-1
Alternatively, if you pick your favorite integer b, r=(-b+-√b²-4)/2. The only square of a rational number that's 4 less than the square of an integer is 0, which is what gives +-1.
For the question asked at 1:53: Just stick with r and solve the quadratic equation r + 1/r = k r² -kr + 1 = 0 Δ = k² - 4 We get a rational r if and only if Δ is a perfect square (Δ = d²). But note that b² is also a perfect square. We then get k² - d² = 4. The only squares which difference is only 4 are with k = ±2 and d = 0. This can be proven considering the following cases: For k=0, k=1 then and k=2 we can prove by hand. (we get respectively Δ=-4, Δ=-3 and Δ=0) For k>=3 then: If d = k-1 then k²-d² = k² - (k-1)² = 2k - 1 > 4 which is true for k>=3. If d > k-1 then k²-d² k² -(k-1)² > 4. This shows that k²-d² can never be 4 if k>=3 For k
Thank you, it was bugging me that the Fibonacci closed form is basically infinite pairs of irrational numbers adding to be integers, but this proves it.
2:20 It is *impossible* Proof by contradiction: Say there was a solution a,b such that a and b are relatively prime. That would mean: a/b+b/a €Z aa/ab+bb/ab €Z (a^2+b^2)/ab €Z => for some number n : a^2+b^2=nab Looking at the equation in terms of mod a : a^2+b^2 Ξ 0 mod a b^2 Ξ 0 mod a =>
Hmmm. I did it this way: I set r + 1/r = k where k is an integer. Solving for r yields r = (k +- sqrt(k^2 - 2^2)) / 2 and yes, I wrote the "- 4" in the sqrt as 2^2 for a reason. Now, the only way to have the material in sqrt() be rational is if the difference of these squares is a square. So k can = 2 (thus, r = 1) or k can = -2 (thus, r = -1) Because there are no Pythagorean triples with 2^2 as an element, the only r's for which this works is 1 and -1. Or, to put it another way, there are no squares higher than 2, that are distant from another square by 4.
Mathematics seems to be the continum of WTF moment~ haha we always look for better way non-stop to come up with a better solution~ Having a pause is quite natural for us to proceed the proof~ Very much to learn from him!! Thumbs up!
Love the WTF moment. I myself had to go "hold up, what did he just do?" when you went straight from the fractional representation to the negative exponent representation without writing down the intermediate step, so I paused the video to double check it was right, then I unpause the video just in time to see you stare at it wondering what the hell you just did, exactly like I had just done :)
You really don't need strong induction for this problem, and it is a bad example to show it in my opinion. You only prove the base case for two values and then for the induction step you only use that P(k-1) and P(k-2) are true in order to prove that P(k) is false. Of course it depends on nomenclature and how you want to classify things, but this is more of a variation of the classic induction than strong induction; I've been taught this is called 2 step induction and the regular induction is a 1 step induction. A more motivating example for strong induction would typically involve proving a formula for a sequence. A better example for strong induction: image.prntscr.com/image/hYIHka_zSs6VtzQlUpBFFg.png (I'm not the author. Granted, it's not the easiest example but it's the first that I was able to find by looking back at problems I've solved) Here you actually need the fact that P(1),P(2),...,P(k-1) are true in order to prove that P(k).
For your question about values of r, couldn’t you just solve the quadratic equation r^2 -kr + 1=0 for some integer k? The discriminant is k^2-4 which needs to be a perfect square for r to be rational, and the only way for k^2-4 to be a perfect square (by quick inspection of the growth of perfect squares) is for k^2=4 which means k=+/-2 which results in the trivial r=+/-1 as stated.
THANK YOU LOTS for your wonderful and amazing math videos!! You're one of my ALL-TIME FAVORITE TH-cam personalities and hosts and I LOVE, LOVE, LOVE your channel!!!! Keep up the grand work!!!! :) :)
Show that a/b + b/a can't be an integer, where a and b are non zero integers except when (a,b) = (a,a) or (a, -a) or (-a, a) or (-a, -a). I.e. when a = b, or a = -b. Suppose by contradiction that a/b + b/a is an integer k, and further we can assume gcd(a,b) = 1, as otherwise we can cancel the gcd(a,b) from the top and bottom of a/b and from b/a and their sum will still be k. Now a/b + b/a = k, with gcd(a,b) = 1. So, a^2 + b^2 = abk So, b^2 = abk - a^2= a(bk - a) Case: b^2 divides a, and hence b divides a. Hence, b = 1 or -1, and a = 1 or -1, as the gcd(a,b) = 1. Case: b^2 divides bk - a So, b divides bk - a, as b^2 divides bk - a So b divides a, because b divides bk Hence, b = 1 or -1, and a = 1 or -1, as gcd(a,b) =1 (as before). So, in either case, b = 1 or -1 and a = 1 or -1. Hence, the primative solutions occur when gcd(a,b) = 1, and are: (1,1) (1,-1) (-1,1) (-1,-1) Note: the primative solution results in k = -2 or 2. So the general solutions (when a = b) are: (a, a) (a, -a) (-a, a) (-a, -a) Note: the general solution results in k = -2 or 2 (same as the primitive solutions).
Nice video! And I appreciate the way you had to reconsider your proof -- that's how the real world works after all. Some fun extensions - The result isn't just true for the naturals, it is true for all r^n + 1/r^n where n is integer. - In fact the result is true under all rings, not just r+1/r an integer. - It's easy to show that r must be irrational except for the trivial r=1 using the quadratic equation and knowledge of the distribution of squares.
It is fairly easy to prove, that only rational r possible is 1 or -1, because r+1/r=n always results in quadratic of type: r^2-nr+1=0. where n is integer. How do we know if quadratic has rational roots? Well it is easy, Discriminant of the quadratic equation has to be perfect square of a rational number, because otherwise you will have irrational square root. So discriminant of quadratic above is n^2-4 and it has to be perfect square. Because n is integer basically what it results in is 2 integer numbers that are perfect squares and differ by 4. And the only 2 squares that satisfy this are 0 and 4 (it can't be some non integer rational square because those will be irreducible but n has to be integer and for squares greater than 4, gaps between squares are greater than 4 too). 4 is square of 2 and -2 substiting 2 and -2 as integer will give solutions 1 and -1 for there is no 3rd and 4th root though because discriminant is 0 so each of the equations has only 1 root (because +0=-0)
Sir can u do a video on Integration of [ 1/x^p] from (0 to infinity) You have done this particular integral with limits (0,1) some years ago I got the answer of (1,infinty) But am finding difficulties to solve this in (0,infinity) Do we just spilt (0,infinity) it into two parts (0,1) and (1,infinity) ?
Once you prove r belongs to the integers you can prove r can only be one or minus one(r^n + (1/r^n) = r^2n + 1/r^n. As this results in an integer, r^n divides r^2n + 1. Since r^n divides r^2n, r^n must divide 1, meaning r = 1 or r = -1)
For your question (r+1/r &Z ,r is rationnal) The answer is 1 and -1 no more Pf : let a and b two intergers a/b + b/a & Z (a² + b² )/ab & Z So we have to get denominator equal to 1 So ab | a² and ab | b² b | a and a | b So |b| = |a| a/b =b/a =1 or -1 Thiis is my proof plz let me know if i had any mistake
it's correct but the step from: ab | a²+b² to ab | a² and ab | b² is way too fast for most teachers as they want to be sure you're not thinking a | b+c => a | b and a | c (which is false in general).
Does this actually strengthen the proof? As long as you have the base case doesn't everything following make stuff before trivial? Also the question said for n is the natural numbers excluding 0.
Wait, if a statement P is true for all elements of N, isn't it then also true for all elements of N*, since N* is a subset of N? So yeah, if r+1/r is given to be an integer, then we know r≠0, so r^0+1/r^0=1+1/1=2 which is an integer, obviously. Therefore the video's argument can be made on these two for N, which implies its truth for N*.
You can even start at n=0. [Note that r can be any real number but 0.] Base case, r⁰ + 1/r⁰ = 1 + 1 = 2. Check. For k > 0, expand: (r + 1/r)ᵏ by the binomial expansion formula. By our premise, r + 1/r ε ℤ. and thus, (r + 1/r)ᵏ ε ℤ. Now the symmetry of the binomial coefficients: (k) = ( k ) ( j ) (k-j) allows grouping the above expansion into a sum of pairs of terms of the form C(rᵐ + 1/rᵐ) where 0 ≤ m ≤ k, and m≡k (mod 2); C is a binomial coefficient, and therefore an integer. When k is even, there will be a lone "middle" term ( k ) (½k ) which, of course, is also an integer. So (r + 1/r)ᵏ is a sum of integers, which is an integer. Thus, with the proposition being true for all j ε [0, k-1], it follows that it is true for j = k. Therefore, it is true for all non-negative integers. Fred
In fact, it is impossible for r+1/r to be in Z given a rational r (not equal to 1). This is proven in Imayormaynotknowcalculus's solution to a problem shown at artofproblemsolving.com/community/c6h2056217 .
1:53 Let: *r + 1/r = k* k and r are whole numbers, and r is not equal to 0, so we can multiply both sides by r *r^2 + 1 = kr* *r^2 - kr + 1 = 0* Using the quadratic formula, we get: *r = k/2 [+/-] sqrt[k^2 - 4]/2* If we want r to be a whole number, a first condition we need is that k^2 - 4 should be a proper square, otherwise r will be irrational *k^2 - 4 = n^2* n is a whole number *k^2 - n^2 = 4* *(k + n)(k - n) = 4* Since k and n are whole numbers, k + n and k - n should be too, so both should divide 4. If we decompose 4 into the product of 2 whole numbers, we get *4 = 2×2, (-2)×(-2), 4×1, (-4)×(-1), 1×4, (-1)×(-4)* We equal k + n and k - n to any of the factors. We get 6 systems of equations: 1.- k + n = 2 k - n = 2 k = 2, n = 0 2.- k + n = -2 k - n = -2 k = -2, n = 0 This are the only ones that have whole answers, so *k = 2 , -2* If we substitute this on r = k/2 [+/-] sqrt[k^2 - 4]/2, we get that *_r = 1, -1_* :)
1:53 the expression: a/b + b/a can be expanded into (a^2 + b^2) / ab. Assuming a and b are coprime, the top part of the fraction isn’t divisible by either a nor b. Therefore, the expression cannot be an integer.
Is just the base case of n=1 still enough? the inductive step that shows it's true for n=p relies on it being true for n=p-2. But for n=2, while it's true, it's not written in the proof. So you need two base cases, am I wrong?
Here's my solution for the a/b + b/a, excuse my formatting as I'm on mobile. Demonstrate: for nonzero, distinct integers a and b, a/b + b/a is not an integer. First we switch up the expression. a/b + b/a = (a^2 + b^2)/ab = (a+b)^2/ab - 2, which is an integer iff (a+b)^2/ab = k is as well. We assume that k is an integer. Case 1) gcd(a, b) = 1 gcd(a, a+b) = 1 => a !| (a+b) [read: a does not divide a+b] => a !| (a+b)^2 => ab !| (a+b)^2 So k is not an integer. Case 2) gcd(a, b) = d > 1 Let a' = a/d, b' = b/d, note that gcd(a', b') = 1 k = (a'd + b'd)^2/a'd*b'd = d^2 * (a'+b')^2/(d^2*a'b') = (a'+b')^2/a'b', which is covered in Case 1. QED Edit so it turns out that b = -a is a solution ig. I'd love to hear advice on how to incorporate that particular case into my demonstration, as I'm still rather new in all of this :P I think it'd go like: The only exception in case 1 is a=1, b=-1 WLOG, where a | a+b and thus no contradiction occurs. And by extension, the particular case within Case 2 where a' = 1, b' = -1 will hold as well. Thus, the only solution for the general case is a = -b. QED for real this time.
There also are interest properties of this with tchebychev polynomials: there is a sequence of polynomials (Pn) of Z[X] such that r^n + 1/r^n = Pn(r + 1/r) for all n
La démonstration est fausse . La propriété est vraie pour n= 1 ,oui. Alors On a supposé qu’elle est vraie pour « n » ,et uniquement pour « n » ,et pour qu’elle soit valable ,il faut démontrer qu’elle soit valable pour « n+1 » ,l’importance ici c’est le passage de « n » à « n+1 » ,c’est à dire on démontre que l’incrémentation est valable ,et c’est ça l’objet de la démonstration ,or notre ami ,pour démontrer que c’est valable pour « n+1 » ,il a supposé qu’elle est valable pour « n-1 » en disant : puisque c’est valable pour « n » ,donc elle est valable pour « n-1 » !! Mais qui nous autorise que le passage de « n » à « n-1 » est permis? Du moment que l’objet de la démonstration en cours c’est « démontrer le passage de « n » à « n+1 » » qui est équivaut au passage de « n-1 » à « n » qui n’est pas encore validé .
Well... To be honest, I don't feel completely convinced about this proof you did. In my opinion, you have to assume too much things in the same statement... Maybe it's true whether case n=k is Z then it implies automatically that n=k-1 is also an Z. I just kind of confused! Maybe if you can prove separately that case n=k-1 is also an integer, like you did with case n=2 maybe I will feel better! haha BTW I'm the man who last year argued you that when you solve a differential equation and you have +-a constant... its just not another constant because you lose a solution! Well, you answered me that having initial conditions doesn't matter to lose the +-... But if you DON'T have it??? (Yes, I'm still confused about that but anyway, fortunately you are safe and that's the best!) Thanks for the video. :) #YAY
Thanks for your comments! I never intend to argue with anyone online. I am very happy to hear what you said at the end. Thank you very much and I hope you are doing well too! 😊
Show that the only integer solutions to a/b + b/a are a = b or a = -b, where a, b are integers, and where a ≠ 0 and b ≠ 0. Clearly, if a = b or a = -b, then a/b + b/a is an integer, which is 2 and -2 respectively. Let’s show that a = b or a = -b are the only conditions that result in a/b + b/a being an integer. Suppose, by way of contradiction, that a/b + b/a = k, where a, b, k are integers, and where a ≠ b and a ≠ -b. Let b = a + n, where n ≠ 0, as otherwise a = b, and where n ≠ -2a, as otherwise a = -b. Now as a and b are integers, then n is an integer too. Now, a/b + b/a = k and b = a + n ⇒ a/(a+n)+(a+n)/a = k ⇒ a^2 + (a + n)^2 = a(a + n)k ⇒ a^2+(a^2+2na+n^2)= ka^2+kna ⇒ 2a^2 + 2na + n^2 -ka^2 -kna = 0 ⇒ (2-k)a^2 + (2-k)na + n^2 = 0
⇒ a = [-(2-k)n ± √([(2-k)n)^2 - 4(2-k)n^2)]]/2(2-k), where k ≠ -2 … Equation (1) Note: When k = 2, then (2-k)a^2+(2-k)na + n^2=0 + 0 + n^2= 0. So, n = 0, which contradicts n ≠ 0 as the given condition. I.e. a = b. Now as -(2-k)n and 2(2-k) are integers in Equation (1), then it is required that √(((2-k)n)^2 - 4(2-k)n^2) be a prefect square, as otherwise a will be an irrational number. Consider ((2-k)n)^2 - 4(2-k)n^2, i.e. the discriminant of Equation (1). Now ((2-k)n)^2 - 4(2-k)n^2 = ((2-k)^2)n^2 - 4(2-k)n^2 =(2-k)n^2 * ((2-k)-4) =(2-k)n^2 * (-2-k) =(k-2)n^2 * (2+k) =(k-2)(k+2)n^2 =(k^2-4)n^2 Now for the discriminant to be a perfect square, it is required that (k^2-4)n^2 be a perfect square, which requires that k^2-4 be a perfect square, say m^2, so k^2 - m^2=4, where m is an integer. Now the only two perfect integer squares that differ by 4 are 0^2 & 2^2 and 0^2 & (-2)^2. I.e. k = 2 or-2.
It was discovered earlier that k = 2 is not admissible, as this leads to a = b. So, consider k = -2, and place it in Equation 1. Hence a = [-(2 - (--2))n ± √([(2 - (-2))n)^2 - 4(2 - (-2))n^2 )]]/2(2-k) ⇒ a = (-(2 - (-2))n ± √((4n)^2 - 4(4)n^2 ))/2(2-(-2)) ⇒ a = (-(2-(-2))n ±√(0))/2(2-(-2)) ⇒ a = -4n/8 ⇒ a = -n/2 ⇒ n = -2a, which contradicts n ≠ -2a, as the given condition. I.e. a = -b. The above approach shows that the only two solution are a = b or a = -b, as no other are admissible. Note: The above approach is via contradiction, but it could have been reformulated as a direct proof by just assuming a/b + b/a = k, where a, b, k are integers and deducing (as above) that the only solutions are a = b or a = -b.
2:09: because I don't want to type this out using the comments, I made a PDF of the proof by contradiction. Keep in mind, I am not sure how to write a proof accurately. However, I think this should be sufficient: drive.google.com/file/d/1enFL2OTiiKD4LMQv-WOh1OTO4A21gkJj/view?usp=sharing Let me know of any issues that may arise from how I proved this.
1:53 A proof that if r+1/r is an integer, then r cannot be a rational number not equal to 1, -1. Let r = n/m where n, m are co-prime, therefore m/n+n/m = (n² +m²)/mn = k, where k is an integer. Rearranging the equation we get n² + (mk)n + m² = 0, using the quadratic formula we get that n1, 2 = (-mk ± sqrt((mk)² - 4m²))/2 = (-mk ± m*sqrt(k² - 4))/2. n1, 2 are integers, so the whole equation must be an integer too. That means that sqrt(k² - 4) = t that is an integer. From that we get (k-t)(k+t) = 4, therefore k-t = 1, 2, 4 and k+t = 4, 2, 1. Checking for the solution, k±t cannot equal to 1 or 4, and if it equals 2, then k = 2, but if a solution k exists, than -k is also a solution, because if the "k²", so k = ±2. Therefore, n² ±2mn+ m² = (n±m)² = 0, so n = ±m, so the solution for r is: m, m, r=1, m, -m, r=-1. If a rational solution r exists, then r=±1. QED
Response to question asked at 1:53 It is possible only for a=b, if that is not the case either a>b or b>a. if we add a/b +b/a =a^2+b^2/ab to be integer means that there is a common factor ab that we can simplified lest say that a>b and we can say that a=kb were k is the product of the other prime factors in this case a^2/ab=k but in order to have an integer b^2 must also be divide by ab, divide by b is trivial but divide by a means that b=qa were q is the product of the other prime factors. Which implies that b=qkb which implies that qk=1 but a>b therefore k>1 and we have a contradiction. Same contradiction if we say that a
It seems that it is not possible for r to be rational unless r=1. Proof: let r+1/r =k where k is an integer, then solve the resulting quadratic equation. We can see that sqrt(k^2-4) can never be rational unless k=2.
I am grateful for the video but you are just too fast when you speak. This is Maths and you can't quickly assume we know and just move on as if we get. I would be more grateful if you would take it slow, step by step so that we get it.
It is a general fact that if a polynomial with integer coefficients has a rational root a/b in lowest terms, then b divides the coefficient of the leading term and a divides the coefficient of the constant term (look up Rational Root Theorem on Wiki). So if the leading term has coefficient 1, then rational roots must actually be integer roots. This is another way to see that if r+1/r is an integer and r=a/b in lowest terms then r must be an integer and this is only possible for r=1 or r=-1.
Wait, your video title is misleading. But at least I got to saw you glitch. I thought you were going to prove the Strong Induction itself. Anyway, great video.
Stares silently for an inordinately long period of time at a whiteboard trying to find an error that may or may not exist saying, "what the fuck?" As a physics grad student I can confirm this is how actual math research is done 🤣
Let a and b be co-prime integers, if a/b + b/a is in Z then multiply by a and you get aa/b + b or multiply by b and you get a + bb/a, those numbers can only be integers if a and b are -1 or +1.
a/b + b/a = (a^2 + b^2)/(ab) which means that a|(a^2 + b^2) and b|(a^2 + b^2) which means a^2 = 0 (mod b) (because b^2 =0 (mod b) but this has no solutions if a and b are relatively prime which we can assume bc a fraction can always be reduced to least terms. Edit: Someone beat me to it :(
Proof without induction: let's define the function f(x) = r^x + r^-x we can conclude that f(1)*f(x) = f(x+1) + f(x-1) f(1) *f(1) = f(0) + f(2) -> f(1) is integer, f(0) is also an integer so f(2) is integer f(1)*f(2) = f(1) + f(3) -> f(1) and f(2) are integers so f(3) is integer f(1)*f(3) = f(2) + f(4) -> f(1) ,f(3) and f(2) are integers so f(4) is integer ..... with this method it is not hard to see that f(x) is always gonna be an integer.
r + 1/r = Z. By the quadratic formula, r = (Z +-sqrt(Z^2-4))/2. For both r and Z to be integers, both Z^2 and Z^2-4 must be perfect squares. So you’re looking for two perfect squares whose difference is 4; the only two such perfect squares are 0 and 4. Thus Z^2 must be 4 and Z^2-4 must be 0; thus Z must be 2 or -2. And plugging each of those values into the initial equation, r must be 1 or -1.
Another proof for 1:53: Assume that a and b are integers, coprime, and not equal to 1 or -1, and a/b + b/a in Z. a/b + b/a = (a^2+b^2)/(ab) This number is in Z, iff (a^2+2ab+b^2)/(ab) is in Z (adding 2ab/(ab) makes the number exactly 2 bigger) (a^2+2ab+b^2)/(ab) = (a+b)^2/(ab) But since a and b are integers, coprime, and not equal to 1 or-1, none of them divide (a+b)^2. Therefore a/b + b/a is not in Z
At 2:21, the only case that works is a=+/-b. You can simplify a/b+b/a to (a^2+b^2)/(ab), which has to be an integer. Therefore, ab must divide a^2+b^2. From this you get that both a divides b and b divides a, so a=+/-b. The only rational numbers that solve the equation are 1 and -1
By itself it would be incomplete but he already shows it to be true for n=1 or 2 (i.e. for integers 1 to k where k=2) beforehand - plugging in 2 shows it's true for n=3, then 3 would show it's true for n=4 and so on
If we solve the equation r^2-mr+1=0, the discriminant is sqrt(m^2-4). Then we set up the diophantine equation m^2-4=n^2 and conclude that m=(+/-)2 and r=(+/-)1 meaning that there are no nontrivial rational numbers satisfying the desired condition. How sad.
Sir, I have a request. Can you please make a video where you will show us a proof of the Lagrange's trigonometric identity for sin nx ? Proofs of The Lagrange's identity for cos nx can easily be found on net. But i cant find a proof for the sin nx identity....Pls help me Sir
Let r + 1/r = k in Z. Solving and wlg taking the root r = (k + sqrt(k^2 - 4))/2 we can show that r^n + 1/r^n for n in Z is a polynomial in k with integer coefficients. Further, the rational root test shows r can't be rational.
There is a rational solution other than +/-1 to the problem iff given c an integer where |c|>2 and d a rational number where d=/=0, does c and d exist that satisfy c^2-4=d^2.
Hi good day, i just wanted to ask, is it possible to find the area under the curve without knowing its function and only given is points. It seems impossible for me but i can't convince myself. I hope you could help me. Thank you.
I'm actually curious about what went through your head at 8:59 lol
I will let you know once this comment gets 100 likes. : )
@@blackpenredpen my other 100 fake accounts, here we go
Now I'm curious
@@blackpenredpen Your comment or InSane Samp's?
@@blackpenredpen Let us know!
I like when he glitches😂😂
9:17 we all know you are very kind, but this is pure gold.
I tried to restart the video because bprp didn't move anymore.
The glitch in the matrix: 8:59, and finally caught exception 9:17
hahaha!
8:59 me when i go to speak to my crush...
8:59 blackpenredpen.exe stops working
Response to question asked at 1:53
Let a,b integer co-prime non 0 such that a/b + b/a is in Z
If a/b is in Z and isn't +/-1, b/a isn't in Z, thus a/b + b/a isn't in Z. Same for b/a.
So a/b and b/a are either +/-1 or not in Z, if they are not in Z by multiplying both side with b:
a + b^2/a = kb
kb - a = b^2/a, left side in Z but right side isn't because a and b are co-prime
Therefore there isn't any other solution other than a/b = +/-1
r does not have to be a rational number, there may be no a,b are integer such that r = a/b
@@gbnam8 i was replying to 1:53 with a proof that indeed it's not possible except for the trivial case that r=a/b with a and b integers
You beat me to this!
Ahhh so nice!!!
Alternatively, if you pick your favorite integer b, r=(-b+-√b²-4)/2. The only square of a rational number that's 4 less than the square of an integer is 0, which is what gives +-1.
9:00 bprp.exe has stopped
No one:
My cat when it sees a bird outside the window: 8:59
For the question asked at 1:53:
Just stick with r and solve the quadratic equation
r + 1/r = k
r² -kr + 1 = 0
Δ = k² - 4
We get a rational r if and only if Δ is a perfect square (Δ = d²). But note that b² is also a perfect square. We then get k² - d² = 4.
The only squares which difference is only 4 are with k = ±2 and d = 0. This can be proven considering the following cases:
For k=0, k=1 then and k=2 we can prove by hand. (we get respectively Δ=-4, Δ=-3 and Δ=0)
For k>=3 then:
If d = k-1 then k²-d² = k² - (k-1)² = 2k - 1 > 4 which is true for k>=3.
If d > k-1 then k²-d² k² -(k-1)² > 4.
This shows that k²-d² can never be 4 if k>=3
For k
8:59 I though my neighbor turned off their wifi
12:40 the quarantine is turning bprp into pierre de fermat
The proof is too large to fit in the board
@@ansper1905 😂😂
love this
Thank you, it was bugging me that the Fibonacci closed form is basically infinite pairs of irrational numbers adding to be integers, but this proves it.
Yes! : ))))
2:20
It is *impossible*
Proof by contradiction:
Say there was a solution a,b such that a and b are relatively prime.
That would mean:
a/b+b/a €Z
aa/ab+bb/ab €Z
(a^2+b^2)/ab €Z
=> for some number n :
a^2+b^2=nab
Looking at the equation in terms of mod a :
a^2+b^2 Ξ 0 mod a
b^2 Ξ 0 mod a
=>
But a and b are not integers always
Eg. Fibonacci sequence
@@kamyak9468 there wasn't one mention of integers in this proof, it works
My favourite part about your proof is your creative method of drawing maths symbols. Well done!
You did a better job than I did. I forgot to mention they are relatively prime.
Hmmm. I did it this way: I set
r + 1/r = k where k is an integer. Solving for r yields
r = (k +- sqrt(k^2 - 2^2)) / 2 and yes, I wrote the "- 4" in the sqrt as 2^2 for a reason.
Now, the only way to have the material in sqrt() be rational is if the difference of these squares is a square. So k can = 2 (thus, r = 1) or k can = -2 (thus, r = -1)
Because there are no Pythagorean triples with 2^2 as an element, the only r's for which this works is 1 and -1. Or, to put it another way, there are no squares higher than 2, that are distant from another square by 4.
8:58 I never thought I'd laugh this hard at a bprp video
This is a question from the 2014/15 British maths olympiad 1 (albeit half of it)
8:59 me throughout the entire duration of my discrete math final exam
Giving a like for the Dragon Ball Z comment! I like these Induction videos.
8:58 I watch things at 2x speed and that made me go "How am I at normal speed? Is this slowed?!"
Everything at 2x speed?
9:17 First cuss from bprp? Lol 😂
I thought his channel was family friendly
If you watch his older videos he often swears in those. Not to mention his marathon videos.
He cursed back a while ago to mean haters.
9:17 HAHAHAHAHA Your curse makes me LMAO
lol
Mathematics seems to be the continum of WTF moment~ haha we always look for better way non-stop to come up with a better solution~ Having a pause is quite natural for us to proceed the proof~ Very much to learn from him!! Thumbs up!
Love the WTF moment. I myself had to go "hold up, what did he just do?" when you went straight from the fractional representation to the negative exponent representation without writing down the intermediate step, so I paused the video to double check it was right, then I unpause the video just in time to see you stare at it wondering what the hell you just did, exactly like I had just done :)
bjr1822 haha well it happens all the time
It was a proof, so it was a WTS moment.
WTF= what’s the function?
You really don't need strong induction for this problem, and it is a bad example to show it in my opinion. You only prove the base case for two values and then for the induction step you only use that P(k-1) and P(k-2) are true in order to prove that P(k) is false. Of course it depends on nomenclature and how you want to classify things, but this is more of a variation of the classic induction than strong induction; I've been taught this is called 2 step induction and the regular induction is a 1 step induction. A more motivating example for strong induction would typically involve proving a formula for a sequence. A better example for strong induction: image.prntscr.com/image/hYIHka_zSs6VtzQlUpBFFg.png (I'm not the author. Granted, it's not the easiest example but it's the first that I was able to find by looking back at problems I've solved)
Here you actually need the fact that P(1),P(2),...,P(k-1) are true in order to prove that P(k).
9:12 when there is the answer given in a test and you thougt you had ist correct before you checked the solution...
For your question about values of r, couldn’t you just solve the quadratic equation r^2 -kr + 1=0 for some integer k? The discriminant is k^2-4 which needs to be a perfect square for r to be rational, and the only way for k^2-4 to be a perfect square (by quick inspection of the growth of perfect squares) is for k^2=4 which means k=+/-2 which results in the trivial r=+/-1 as stated.
Yes, I think it works! It didn't come to my mind when I was recording.
When did Induction get so buff?
But mathematicians love overpoweredly buffed methods
Next challenge
(1.3.5...(2n-1))/(2.4.6...(2n)) is less than or equal to 1/((3n + 1)^(1/2)), for any n integer ?
THANK YOU LOTS for your wonderful and amazing math videos!! You're one of my ALL-TIME FAVORITE TH-cam personalities and hosts and I LOVE, LOVE, LOVE your channel!!!! Keep up the grand work!!!! :) :)
Thank you very much. I am very happy to hear this!
Show that a/b + b/a can't be an integer, where a and b are non zero integers except when (a,b) = (a,a) or (a, -a) or (-a, a) or (-a, -a). I.e. when a = b, or a = -b.
Suppose by contradiction that
a/b + b/a is an integer k, and further we can assume gcd(a,b) = 1, as otherwise we can cancel the gcd(a,b) from the top and bottom of a/b and from b/a and their sum will still be k.
Now a/b + b/a = k, with gcd(a,b) = 1.
So, a^2 + b^2 = abk
So, b^2 = abk - a^2= a(bk - a)
Case: b^2 divides a, and hence b divides a.
Hence, b = 1 or -1, and a = 1 or -1,
as the gcd(a,b) = 1.
Case: b^2 divides bk - a
So, b divides bk - a, as b^2 divides bk - a
So b divides a, because b divides bk
Hence, b = 1 or -1, and a = 1 or -1,
as gcd(a,b) =1 (as before).
So, in either case, b = 1 or -1 and a = 1 or -1.
Hence, the primative solutions occur
when gcd(a,b) = 1, and are:
(1,1)
(1,-1)
(-1,1)
(-1,-1)
Note: the primative solution results in
k = -2 or 2.
So the general solutions (when a = b) are:
(a, a)
(a, -a)
(-a, a)
(-a, -a)
Note: the general solution results in
k = -2 or 2 (same as the primitive solutions).
NO WAYYY! I was contemplating this question yesterday literally...
Nice video! And I appreciate the way you had to reconsider your proof -- that's how the real world works after all.
Some fun extensions
- The result isn't just true for the naturals, it is true for all r^n + 1/r^n where n is integer.
- In fact the result is true under all rings, not just r+1/r an integer.
- It's easy to show that r must be irrational except for the trivial r=1 using the quadratic equation and knowledge of the distribution of squares.
It is fairly easy to prove, that only rational r possible is 1 or -1, because r+1/r=n always results in quadratic of type: r^2-nr+1=0. where n is integer.
How do we know if quadratic has rational roots? Well it is easy, Discriminant of the quadratic equation has to be perfect square of a rational number, because otherwise you will have irrational square root. So discriminant of quadratic above is n^2-4 and it has to be perfect square. Because n is integer basically what it results in is 2 integer numbers that are perfect squares and differ by 4. And the only 2 squares that satisfy this are 0 and 4 (it can't be some non integer rational square because those will be irreducible but n has to be integer and for squares greater than 4, gaps between squares are greater than 4 too). 4 is square of 2 and -2 substiting 2 and -2 as integer will give solutions 1 and -1 for there is no 3rd and 4th root though because discriminant is 0 so each of the equations has only 1 root (because +0=-0)
I thought strong induction was when you assumed the Induction Hypothesis was true for ALL values from 1 to k, not just k.
Zach L That is what he said
Loved the WTBIP moment, also this demystifies a bit the 1/phi +phi a tad :) Nice!
I am disappointed you didn't draw a box at the end.
I ran out of space....
Sir can u do a video on
Integration of [ 1/x^p] from (0 to infinity)
You have done this particular integral with limits (0,1) some years ago
I got the answer of (1,infinty)
But am finding difficulties to solve this in (0,infinity)
Do we just spilt (0,infinity) it into two parts (0,1) and (1,infinity) ?
Jack Summer Yes, but the problem is that the limits of the boundaries of the integral are infinite for all real values of p, so the integral diverges.
Hello, I sent you a proof that it's not possible through twitter, make sure to take a look!
twitter.com/SrSrHipolito/status/1257044040945733632 here's the post if anyone wants to take a look at it :)
Solve the integral of the fourth root of tan(x) dx on that blackboard 😈
If we can assume true for k-1 we can assume true for k+1 and promblem
Once you prove r belongs to the integers you can prove r can only be one or minus one(r^n + (1/r^n) = r^2n + 1/r^n. As this results in an integer, r^n divides r^2n + 1. Since r^n divides r^2n, r^n must divide 1, meaning r = 1 or r = -1)
For your question (r+1/r &Z ,r is rationnal)
The answer is 1 and -1 no more
Pf : let a and b two intergers
a/b + b/a & Z
(a² + b² )/ab & Z
So we have to get denominator equal to 1
So ab | a² and ab | b²
b | a and a | b
So |b| = |a|
a/b =b/a =1 or -1
Thiis is my proof plz let me know if i had any mistake
it's correct but the step from:
ab | a²+b²
to
ab | a² and ab | b²
is way too fast for most teachers as they want to be sure you're not thinking a | b+c => a | b and a | c (which is false in general).
You can strengthen the proof, because for n=0 its also truth (r is not 0, beacuse You have 1/r, and then r^0+1/r^0=2, and its truth too)
It’s actually true for all integers. But usually we break that into cases. So I only proved the case for pos integers
Does this actually strengthen the proof? As long as you have the base case doesn't everything following make stuff before trivial? Also the question said for n is the natural numbers excluding 0.
Wait, if a statement P is true for all elements of N, isn't it then also true for all elements of N*, since N* is a subset of N? So yeah, if r+1/r is given to be an integer, then we know r≠0, so r^0+1/r^0=1+1/1=2 which is an integer, obviously. Therefore the video's argument can be made on these two for N, which implies its truth for N*.
You can substitute r^n = u and solve the quadratic equation then state it has always a real solution.
Real solution, but what about integer solution for the peoblem?
You can even start at n=0. [Note that r can be any real number but 0.] Base case, r⁰ + 1/r⁰ = 1 + 1 = 2. Check.
For k > 0, expand:
(r + 1/r)ᵏ
by the binomial expansion formula. By our premise, r + 1/r ε ℤ. and thus, (r + 1/r)ᵏ ε ℤ.
Now the symmetry of the binomial coefficients:
(k) = ( k )
( j ) (k-j)
allows grouping the above expansion into a sum of pairs of terms of the form
C(rᵐ + 1/rᵐ)
where 0 ≤ m ≤ k, and m≡k (mod 2); C is a binomial coefficient, and therefore an integer.
When k is even, there will be a lone "middle" term
( k )
(½k )
which, of course, is also an integer.
So (r + 1/r)ᵏ is a sum of integers, which is an integer.
Thus, with the proposition being true for all j ε [0, k-1], it follows that it is true for j = k.
Therefore, it is true for all non-negative integers.
Fred
for the question if r can be rational you can also use the rational root theorem. The only possible rational roots of r^2 - nr + 1 are 1 or -1.
Wow. I don’t think there’s any simpler proof. Awesome.
with strong induction you might have to check more base cases. I think n=0 case is sound, so it works, but anyway n=2 case was already covered.
In fact, it is impossible for r+1/r to be in Z given a rational r (not equal to 1).
This is proven in Imayormaynotknowcalculus's solution to a problem shown at artofproblemsolving.com/community/c6h2056217 .
(5:38) "Looks like Dragon Ball Z" :)
1:53 Let:
*r + 1/r = k*
k and r are whole numbers, and r is not equal to 0, so we can multiply both sides by r
*r^2 + 1 = kr*
*r^2 - kr + 1 = 0*
Using the quadratic formula, we get:
*r = k/2 [+/-] sqrt[k^2 - 4]/2*
If we want r to be a whole number, a first condition we need is that k^2 - 4 should be a proper square, otherwise r will be irrational
*k^2 - 4 = n^2*
n is a whole number
*k^2 - n^2 = 4*
*(k + n)(k - n) = 4*
Since k and n are whole numbers, k + n and k - n should be too, so both should divide 4. If we decompose 4 into the product of 2 whole numbers, we get
*4 = 2×2, (-2)×(-2), 4×1, (-4)×(-1), 1×4, (-1)×(-4)*
We equal k + n and k - n to any of the factors. We get 6 systems of equations:
1.-
k + n = 2
k - n = 2
k = 2, n = 0
2.-
k + n = -2
k - n = -2
k = -2, n = 0
This are the only ones that have whole answers, so *k = 2 , -2*
If we substitute this on r = k/2 [+/-] sqrt[k^2 - 4]/2, we get that
*_r = 1, -1_*
:)
1:53 the expression: a/b + b/a can be expanded into (a^2 + b^2) / ab. Assuming a and b are coprime, the top part of the fraction isn’t divisible by either a nor b. Therefore, the expression cannot be an integer.
I got to a repeated infinite fraction when I tried solving m + 2 = (a + b)^2 / ab for b in terms of m and a and then I was like no way.
Is just the base case of n=1 still enough? the inductive step that shows it's true for n=p relies on it being true for n=p-2. But for n=2, while it's true, it's not written in the proof. So you need two base cases, am I wrong?
Here's my solution for the a/b + b/a, excuse my formatting as I'm on mobile.
Demonstrate: for nonzero, distinct integers a and b, a/b + b/a is not an integer.
First we switch up the expression. a/b + b/a = (a^2 + b^2)/ab = (a+b)^2/ab - 2, which is an integer iff (a+b)^2/ab = k is as well.
We assume that k is an integer.
Case 1) gcd(a, b) = 1
gcd(a, a+b) = 1
=> a !| (a+b) [read: a does not divide a+b]
=> a !| (a+b)^2
=> ab !| (a+b)^2
So k is not an integer.
Case 2) gcd(a, b) = d > 1
Let a' = a/d, b' = b/d, note that gcd(a', b') = 1
k = (a'd + b'd)^2/a'd*b'd = d^2 * (a'+b')^2/(d^2*a'b') = (a'+b')^2/a'b', which is covered in Case 1. QED
Edit so it turns out that b = -a is a solution ig. I'd love to hear advice on how to incorporate that particular case into my demonstration, as I'm still rather new in all of this :P
I think it'd go like:
The only exception in case 1 is a=1, b=-1 WLOG, where a | a+b and thus no contradiction occurs. And by extension, the particular case within Case 2 where a' = 1, b' = -1 will hold as well. Thus, the only solution for the general case is a = -b. QED for real this time.
There also are interest properties of this with tchebychev polynomials: there is a sequence of polynomials (Pn) of Z[X] such that r^n + 1/r^n = Pn(r + 1/r) for all n
Spooky coincidence. Those polynomials are the reason I clicked this video, they are on my whiteboard behind me !!
La démonstration est fausse .
La propriété est vraie pour n= 1 ,oui.
Alors On a supposé qu’elle est vraie pour « n » ,et uniquement pour « n » ,et pour qu’elle soit valable ,il faut démontrer qu’elle soit valable pour « n+1 » ,l’importance ici c’est le passage de « n » à « n+1 » ,c’est à dire on démontre que l’incrémentation est valable ,et c’est ça l’objet de la démonstration ,or notre ami ,pour démontrer que c’est valable pour « n+1 » ,il a supposé qu’elle est valable pour « n-1 » en disant : puisque c’est valable pour « n » ,donc elle est valable pour « n-1 » !!
Mais qui nous autorise que le passage de « n » à « n-1 » est permis? Du moment que l’objet de la démonstration en cours c’est « démontrer le passage de « n » à « n+1 » » qui est équivaut au passage de « n-1 » à « n » qui n’est pas encore validé .
Well... To be honest, I don't feel completely convinced about this proof you did. In my opinion, you have to assume too much things in the same statement... Maybe it's true whether case n=k is Z then it implies automatically that n=k-1 is also an Z. I just kind of confused! Maybe if you can prove separately that case n=k-1 is also an integer, like you did with case n=2 maybe I will feel better! haha BTW I'm the man who last year argued you that when you solve a differential equation and you have +-a constant... its just not another constant because you lose a solution! Well, you answered me that having initial conditions doesn't matter to lose the +-... But if you DON'T have it??? (Yes, I'm still confused about that but anyway, fortunately you are safe and that's the best!) Thanks for the video. :) #YAY
Thanks for your comments! I never intend to argue with anyone online. I am very happy to hear what you said at the end. Thank you very much and I hope you are doing well too! 😊
@@blackpenredpen Sure! Just giving you more ideas for making new videos! ;)
@@blackpenredpen BTW I'm also well, THANKS!!
Strong inductionn theorem (where n is assumed true from 1 to some base case k) can actually be proven by contradiction.
Show that the only integer solutions to a/b + b/a are a = b or a = -b, where a, b are integers,
and where a ≠ 0 and b ≠ 0.
Clearly, if a = b or a = -b, then a/b + b/a is an integer, which is 2 and -2 respectively.
Let’s show that a = b or a = -b are the only conditions that result in a/b + b/a being an integer.
Suppose, by way of contradiction, that a/b + b/a = k, where a, b, k are integers, and where a ≠ b and a ≠ -b.
Let b = a + n, where n ≠ 0, as otherwise a = b, and where n ≠ -2a, as otherwise a = -b.
Now as a and b are integers, then n is an integer too.
Now, a/b + b/a = k and b = a + n
⇒ a/(a+n)+(a+n)/a = k
⇒ a^2 + (a + n)^2 = a(a + n)k
⇒ a^2+(a^2+2na+n^2)= ka^2+kna
⇒ 2a^2 + 2na + n^2 -ka^2 -kna = 0
⇒ (2-k)a^2 + (2-k)na + n^2 = 0
⇒ a = [-(2-k)n ± √([(2-k)n)^2 - 4(2-k)n^2)]]/2(2-k), where k ≠ -2 … Equation (1)
Note: When k = 2, then (2-k)a^2+(2-k)na + n^2=0 + 0 + n^2= 0.
So, n = 0, which contradicts n ≠ 0 as the given condition. I.e. a = b.
Now as -(2-k)n and 2(2-k) are integers in Equation (1), then it is required that √(((2-k)n)^2 - 4(2-k)n^2) be a prefect square, as otherwise a will be an irrational number.
Consider ((2-k)n)^2 - 4(2-k)n^2, i.e. the discriminant of Equation (1).
Now ((2-k)n)^2 - 4(2-k)n^2
= ((2-k)^2)n^2 - 4(2-k)n^2
=(2-k)n^2 * ((2-k)-4)
=(2-k)n^2 * (-2-k)
=(k-2)n^2 * (2+k)
=(k-2)(k+2)n^2
=(k^2-4)n^2
Now for the discriminant to be a perfect square, it is required that (k^2-4)n^2 be a perfect square, which requires that k^2-4 be a perfect square,
say m^2, so k^2 - m^2=4, where m is an integer.
Now the only two perfect integer squares that differ by 4 are 0^2 & 2^2 and 0^2 & (-2)^2. I.e. k = 2 or-2.
It was discovered earlier that k = 2 is not admissible, as this leads to a = b.
So, consider k = -2, and place it in Equation 1.
Hence a = [-(2 - (--2))n ± √([(2 - (-2))n)^2 - 4(2 - (-2))n^2 )]]/2(2-k)
⇒ a = (-(2 - (-2))n ± √((4n)^2 - 4(4)n^2 ))/2(2-(-2))
⇒ a = (-(2-(-2))n ±√(0))/2(2-(-2))
⇒ a = -4n/8
⇒ a = -n/2
⇒ n = -2a, which contradicts n ≠ -2a, as the given condition. I.e. a = -b.
The above approach shows that the only two solution are a = b or a = -b, as no other are admissible.
Note: The above approach is via contradiction, but it could have been reformulated as a direct proof by just assuming a/b + b/a = k, where a, b, k are integers and deducing (as above) that the only solutions are a = b or a = -b.
2:09: because I don't want to type this out using the comments, I made a PDF of the proof by contradiction. Keep in mind, I am not sure how to write a proof accurately. However, I think this should be sufficient:
drive.google.com/file/d/1enFL2OTiiKD4LMQv-WOh1OTO4A21gkJj/view?usp=sharing
Let me know of any issues that may arise from how I proved this.
I forgot a
e b. Also, obviously, a
e0. and b
e0..
It isn't possible. Let n be an integer. r^2-nr+1=0 quadratic formula (n+-sqrt(n^2-4))/2=r
If -1
1:53
A proof that if r+1/r is an integer, then r cannot be a rational number not equal to 1, -1.
Let r = n/m where n, m are co-prime, therefore m/n+n/m = (n² +m²)/mn = k, where k is an integer. Rearranging the equation we get n² + (mk)n + m² = 0, using the quadratic formula we get that
n1, 2 = (-mk ± sqrt((mk)² - 4m²))/2 = (-mk ± m*sqrt(k² - 4))/2.
n1, 2 are integers, so the whole equation must be an integer too. That means that sqrt(k² - 4) = t that is an integer.
From that we get (k-t)(k+t) = 4, therefore k-t = 1, 2, 4 and k+t = 4, 2, 1.
Checking for the solution, k±t cannot equal to 1 or 4, and if it equals 2, then k = 2, but if a solution k exists, than -k is also a solution, because if the "k²",
so k = ±2.
Therefore, n² ±2mn+ m² = (n±m)² = 0, so n = ±m, so the solution for r is: m, m, r=1, m, -m, r=-1.
If a rational solution r exists, then r=±1.
QED
Response to question asked at 1:53
It is possible only for a=b, if that is not the case either a>b or b>a. if we add a/b +b/a =a^2+b^2/ab to be integer means that there is a common factor ab that we can simplified lest say that a>b and we can say that a=kb were k is the product of the other prime factors in this case a^2/ab=k but in order to have an integer b^2 must also be divide by ab, divide by b is trivial but divide by a means that b=qa were q is the product of the other prime factors. Which implies that b=qkb which implies that qk=1 but a>b therefore k>1 and we have a contradiction. Same contradiction if we say that a
Do practice ! 🤣🤣🤣 Great
It seems that it is not possible for r to be rational unless r=1. Proof: let r+1/r =k where k is an integer, then solve the resulting quadratic equation. We can see that sqrt(k^2-4) can never be rational unless k=2.
I am grateful for the video but you are just too fast when you speak. This is Maths and you can't quickly assume we know and just move on as if we get. I would be more grateful if you would take it slow, step by step so that we get it.
It is a general fact that if a polynomial with integer coefficients has a rational root a/b in lowest terms, then b divides the coefficient of the leading term and a divides the coefficient of the constant term (look up Rational Root Theorem on Wiki). So if the leading term has coefficient 1, then rational roots must actually be integer roots. This is another way to see that if r+1/r is an integer and r=a/b in lowest terms then r must be an integer and this is only possible for r=1 or r=-1.
Wait, your video title is misleading. But at least I got to saw you glitch. I thought you were going to prove the Strong Induction itself. Anyway, great video.
If q+1/q=m for rational q, integer m, then q is a rational root of x^2-mx+1. But then q=+-1 by rational root theorem.
Stares silently for an inordinately long period of time at a whiteboard trying to find an error that may or may not exist saying, "what the fuck?"
As a physics grad student I can confirm this is how actual math research is done 🤣
Let a and b be co-prime integers, if a/b + b/a is in Z then multiply by a and you get aa/b + b or multiply by b and you get a + bb/a, those numbers can only be integers if a and b are -1 or +1.
I subscribed you a long time ago because of that magical black ball(currently it's black)
Since we did two base cases, couldn't we just say "Assume true for k, k+1, show true for k+2"?
1) 02:05 Challenge accepted :)
2) 08:58 Who else thought that this video lagged / freezed?
What about r-1/r
This is from Azerbaijan Junior Mathematics Olympiad
How do I create a strong induction hypothesis? What should I be looking out for?
It is not possible for r to be rational such that r + 1/r is an integer greater than 2. You can use the rational root theorem to prove it.
a/b + b/a = (a^2 + b^2)/(ab) which means that a|(a^2 + b^2) and b|(a^2 + b^2) which means a^2 = 0 (mod b) (because b^2 =0 (mod b) but this has no solutions if a and b are relatively prime which we can assume bc a fraction can always be reduced to least terms.
Edit: Someone beat me to it :(
Proof without induction:
let's define the function f(x) = r^x + r^-x
we can conclude that f(1)*f(x) = f(x+1) + f(x-1)
f(1) *f(1) = f(0) + f(2) -> f(1) is integer, f(0) is also an integer so f(2) is integer
f(1)*f(2) = f(1) + f(3) -> f(1) and f(2) are integers so f(3) is integer
f(1)*f(3) = f(2) + f(4) -> f(1) ,f(3) and f(2) are integers so f(4) is integer
.....
with this method it is not hard to see that f(x) is always gonna be an integer.
Tazheb Kuroyzo ...that's literally just induction, what you just did, just in a non-rigorous heuristic manner
We could also make use of binomial theorem and prove it with weak induction
An interesting choice of r (although not rational) is e^(i*pi/3).
ええーー
具体例なく証明しちゃうのあまり慣れてないから、なかなかスリルがありました
中文使用者：
這個方法稱為完整歸納法，是數學歸納法的一種，但是要假設n從1到k都成立，證明n=k+1時也成立，則n=任意正整數時成立
Glitch in maths sloved by blackpenredpen
9:36 shouldn't the 3rd term be 1/r^(1-k)?
why do you speak so fast! I can barely understand you!
Someone get this brilliant man a bigger whiteboard!!!
r + 1/r = Z.
By the quadratic formula, r = (Z +-sqrt(Z^2-4))/2.
For both r and Z to be integers, both Z^2 and Z^2-4 must be perfect squares. So you’re looking for two perfect squares whose difference is 4; the only two such perfect squares are 0 and 4. Thus Z^2 must be 4 and Z^2-4 must be 0; thus Z must be 2 or -2. And plugging each of those values into the initial equation, r must be 1 or -1.
Another proof for 1:53:
Assume that a and b are integers, coprime, and not equal to 1 or -1, and a/b + b/a in Z.
a/b + b/a = (a^2+b^2)/(ab)
This number is in Z, iff (a^2+2ab+b^2)/(ab) is in Z (adding 2ab/(ab) makes the number exactly 2 bigger)
(a^2+2ab+b^2)/(ab) = (a+b)^2/(ab)
But since a and b are integers, coprime, and not equal to 1 or-1, none of them divide (a+b)^2. Therefore a/b + b/a is not in Z
and when we stop assume and want know if its really the case ?
At 2:21, the only case that works is a=+/-b. You can simplify a/b+b/a to (a^2+b^2)/(ab), which has to be an integer. Therefore, ab must divide a^2+b^2. From this you get that both a divides b and b divides a, so a=+/-b. The only rational numbers that solve the equation are 1 and -1
This proof sounds a bit like cheating to me. It's like saying it's true because I assume it to be true for integers from 1 to k.
By itself it would be incomplete but he already shows it to be true for n=1 or 2 (i.e. for integers 1 to k where k=2) beforehand - plugging in 2 shows it's true for n=3, then 3 would show it's true for n=4 and so on
I found another secret place!
you’re a time traveler!
2 weeks ago?! 😆
😄😄
Hows this possible
If we solve the equation
r^2-mr+1=0,
the discriminant is sqrt(m^2-4). Then we set up the diophantine equation m^2-4=n^2 and conclude that m=(+/-)2 and r=(+/-)1 meaning that there are no nontrivial rational numbers satisfying the desired condition. How sad.
Why didnt you solve it using binomial expansion
If r+1/r =k then r=(k±sqrt(k²-4))/2 which is rational only for k=2 and r=1.
I forgot k=-2,r=-1
Sir, I have a request. Can you please make a video where you will show us a proof of the Lagrange's trigonometric identity for sin nx ? Proofs of The Lagrange's identity for cos nx can easily be found on net. But i cant find a proof for the sin nx identity....Pls help me Sir
Let r + 1/r = k in Z. Solving and wlg taking the root r = (k + sqrt(k^2 - 4))/2 we can show that r^n + 1/r^n for n in Z is a polynomial in k with integer coefficients. Further, the rational root test shows r can't be rational.
There is a rational solution other than +/-1 to the problem iff given c an integer where |c|>2 and d a rational number where d=/=0, does c and d exist that satisfy c^2-4=d^2.
Hi good day, i just wanted to ask, is it possible to find the area under the curve without knowing its function and only given is points. It seems impossible for me but i can't convince myself. I hope you could help me. Thank you.