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
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.
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.
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.
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!!!! :) :)
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
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.
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 :)
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)
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!
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
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 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
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).
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.
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.
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
We can raise r+1/r to the power of n so we get r^n +1/r^n and a midterm. Since we are prooving r^n +1/r^n to be an integer, we must see if the midterm is an integer. We notice that the coefficiants are reapeating we can group them so we put the coefficiant times r*1/r, which cancels out, and in the printeses we have r^a+1/r^a, where a is a power smaller then n. We can then again repeat this procces till we get the r +1/r in the printeses. Whenever n is an even integer we get a coefficiant which cannot be grouped but acoording to Newtons Binomial the product after it will have the same powers so they will cancel out for r and 1/r. Since we prooven that the midterm is an integer, we can say that r^n +1/r^n is an integer.
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.
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 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
the general solution to r^2 - kr + 1 = 0 is (k +- sqrt(k^2 - 4))/2, which means that sqrt(k^2 - 4) needs to be rational for r to be rational, so k^2 - 4 = a^2 or (k - a)(k + a) = 4. since both k and a are integers, the must both be 2 or both be -2 (or 1 4 but any of those pairs will lead to a non-integer solution). k + a = k - a = 2 then k = 2, leading to r = 1 k + a = k - a = -2 then k = -2, leading to r = -1
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.
The secondary question imply that n^2-4 be a perfect square for an integer n greater than 2 (because of the solution of x+1/x =n , the radical part is n^2-4 by the quadratic formula ), hence the answer is no.
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
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*.
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)
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.
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.
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.
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) ?
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.
Offhand, the proof only depends on the fact that the integers form a subgroup of the real numbers under addition, right? So the same theorem applies to r+1/r with respect to any subgroup (e.g., the rational numbers, or the rational multiples of pi).
It is actually sufficient to say : suppose that the statement holds for n and n+1 and let s prove that it s true for n+2 . But we should in this case verifie it fo n=1 and n=2
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.
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?
It is good to know some real numbers r with the condition r + 1/r be an integer, of course the only 2 rational numbers holding it are r=1 or -1 (trivial case), also do, in fact we can prove all the real numbers of the form r=(a+sqrt(a*a-4))/2 or r=(a-sqrt(a*a-4))/2, with abs.value(a)>=2. Motive we hey the trivial caess
What am I missing? The last part of "Check:" (from 11:11) uses a circular argument as 'proof'. In the Inductive Hypothesis, truth is assumed. That ASSUMPTION is then used to demonstrate the truth of the final 'k-1' element of the "Check:". How can the assumption be used as as proof? Where am I going wrong? 🍌😷😮
We were given that the statement is true for n=1 and proved that it is true for n=2. In the inductive hypothesis we assume that it will be true for all n >= k, but really what we need (as seen in the "check" equation) is to assume that the two previous elements are integers (k-1 and k elements). And we know we can find two such elements because both n=1=k-1 and n=2=k are true. From there we can prove it for n=3, and then using n=2 and n=3 we can prove it for n=4 and so on. I hope that made sense :)
In my career I have caught a few professionals 'fudging' their results to match the theoretical predictions. I always chastised them for their un-scientific approach and told them that they should go back and change their assumptions to make the results fit the prediction. This is the true scientific way! 😃
@@ShacLowGP Thanks. I'm still seeing circularity, though. BPRP referred back to the ASSUMPTION as proof of the conclusion. If the previous calculations using k-1=1 and k=2 'proved' the assumption, it would no longer be an assumption; he still referred to it as an assumption in the last "Check:" term. Is this a 'feature' of "Strong" Induction? 🍌😷😀
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.
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 🤣
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.
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.
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.
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.
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!
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.
I like when he glitches😂😂
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
The glitch in the matrix: 8:59, and finally caught exception 9:17
hahaha!
8:59 blackpenredpen.exe stops working
8:59 me when i go to speak to my crush...
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.
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..
9:00 bprp.exe has stopped
No one:
My cat when it sees a bird outside the window: 8:59
8:58 I never thought I'd laugh this hard at a bprp video
12:40 the quarantine is turning bprp into pierre de fermat
The proof is too large to fit in the board
@@ansper1905 😂😂
love this
8:59 I though my neighbor turned off their wifi
8:59 me throughout the entire duration of my discrete math final exam
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! : ))))
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!
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
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.
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?
This is a question from the 2014/15 British maths olympiad 1 (albeit half of it)
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?
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.
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)
If we can assume true for k-1 we can assume true for k+1 and promblem
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!
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
9:17 HAHAHAHAHA Your curse makes me LMAO
lol
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.
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_*
:)
9:12 when there is the answer given in a test and you thougt you had ist correct before you checked the solution...
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
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).
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.
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.
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 !!
We can raise r+1/r to the power of n so we get r^n +1/r^n and a midterm. Since we are prooving r^n +1/r^n to be an integer, we must see if the midterm is an integer. We notice that the coefficiants are reapeating we can group them so we put the coefficiant times r*1/r, which cancels out, and in the printeses we have r^a+1/r^a, where a is a power smaller then n. We can then again repeat this procces till we get the r +1/r in the printeses. Whenever n is an even integer we get a coefficiant which cannot be grouped but acoording to Newtons Binomial the product after it will have the same powers so they will cancel out for r and 1/r. Since we prooven that the midterm is an integer, we can say that r^n +1/r^n is an integer.
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.
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).
Loved the WTBIP moment, also this demystifies a bit the 1/phi +phi a tad :) Nice!
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
Someone get this brilliant man a bigger whiteboard!!!
the general solution to r^2 - kr + 1 = 0 is
(k +- sqrt(k^2 - 4))/2, which means that sqrt(k^2 - 4) needs to be rational for r to be rational,
so k^2 - 4 = a^2 or (k - a)(k + a) = 4.
since both k and a are integers, the must both be 2 or both be -2 (or 1 4 but any of those pairs will lead to a non-integer solution).
k + a = k - a = 2 then k = 2, leading to r = 1
k + a = k - a = -2 then k = -2, leading to r = -1
When did Induction get so buff?
But mathematicians love overpoweredly buffed methods
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.
The secondary question imply that n^2-4 be a perfect square for an integer n greater than 2 (because of the solution of x+1/x =n , the radical part is n^2-4 by the quadratic formula ), hence the answer is no.
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 ?
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.
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
I am disappointed you didn't draw a box at the end.
I ran out of space....
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*.
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)
Glitch in maths sloved by blackpenredpen
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.
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.
NO WAYYY! I was contemplating this question yesterday literally...
We could also make use of binomial theorem and prove it with weak induction
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.
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.
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.
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.
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?
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.
Offhand, the proof only depends on the fact that the integers form a subgroup of the real numbers under addition, right? So the same theorem applies to r+1/r with respect to any subgroup (e.g., the rational numbers, or the rational multiples of pi).
How do I create a strong induction hypothesis? What should I be looking out for?
9:36 shouldn't the 3rd term be 1/r^(1-k)?
An interesting choice of r (although not rational) is e^(i*pi/3).
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.
Please bring us some madness of differential geometry.
It is actually sufficient to say : suppose that the statement holds for n and n+1 and let s prove that it s true for n+2 . But we should in this case verifie it fo n=1 and n=2
XaXuser more like for n=1 and n=0, but n=0 is trivial, so we are good to go
that's regular induction, not strong induction
*my internet speed slows down for dt time*
My bprp video: 8:58
Me: 9:17
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 :(
(5:38) "Looks like Dragon Ball Z" :)
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
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?
It is good to know some real numbers r with the condition r + 1/r be an integer, of course the only 2 rational numbers holding it are r=1 or -1 (trivial case), also do, in fact we can prove all the real numbers of the form r=(a+sqrt(a*a-4))/2 or
r=(a-sqrt(a*a-4))/2,
with abs.value(a)>=2.
Motive we hey the trivial caess
Continuing my comentary if we take a=3 we get r=(3+sqrt(5))/2, it is funny to check that r+1/r=3, in general if a is any integer such that a>=2 or a
All the real numbers r described above are the only ones holding that r+1/r be an integer
you can also use Newton's law (a+b)^n for proof
What about r-1/r
What am I missing?
The last part of "Check:" (from 11:11) uses a circular argument as 'proof'.
In the Inductive Hypothesis, truth is assumed. That ASSUMPTION is then used to demonstrate the truth of the final 'k-1' element of the "Check:".
How can the assumption be used as as proof? Where am I going wrong?
🍌😷😮
We were given that the statement is true for n=1 and proved that it is true for n=2. In the inductive hypothesis we assume that it will be true for all n >= k, but really what we need (as seen in the "check" equation) is to assume that the two previous elements are integers (k-1 and k elements). And we know we can find two such elements because both n=1=k-1 and n=2=k are true. From there we can prove it for n=3, and then using n=2 and n=3 we can prove it for n=4 and so on. I hope that made sense :)
In my career I have caught a few professionals 'fudging' their results to match the theoretical predictions. I always chastised them for their un-scientific approach and told them that they should go back and change their assumptions to make the results fit the prediction. This is the true scientific way! 😃
@@ShacLowGP Thanks. I'm still seeing circularity, though. BPRP referred back to the ASSUMPTION as proof of the conclusion. If the previous calculations using k-1=1 and k=2 'proved' the assumption, it would no longer be an assumption; he still referred to it as an assumption in the last "Check:" term.
Is this a 'feature' of "Strong" Induction?
🍌😷😀
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
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
Since we did two base cases, couldn't we just say "Assume true for k, k+1, show true for k+2"?
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.
and when we stop assume and want know if its really the case ?
GOD BLESS YOU THANK YOU SO MUCH
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
Solve the integral of the fourth root of tan(x) dx on that blackboard 😈
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 🤣
Awesome beats at the end!
Appreciate it!
Video published today not uploaded.....
Was the intro Doraemon's song
1) 02:05 Challenge accepted :)
2) 08:58 Who else thought that this video lagged / freezed?
In a text book there is a complex numbers problem that goes like this: prove that if z satisfies z+1/z = 2 cosa then z^n + 1/z^n = 2cosna
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 do we need to prove for 2?
Giving a like for the Dragon Ball Z comment! I like these Induction videos.
9:20 Thought my internet had frozen
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.
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.