Thank you so much! I'm just a dad who hasn't done any math in 20 years and spent an hour last night helping his daughter with really basic trigonometry. But I can do that stuff in my head, so I thought I would indulge in some more challenging problems. I don't care how young or old someone is, this kind of content is _always_ a good use of one's time.
I've been watching your videos for the past two months. Before watching the solution, I always try to solve the questions on my own. This is the first question I've successfully solved so far! I feel really good :)
@@arnavraheja10Just plug x^2-kx+1 into the quadratic equation like normal. It will give you the correct values of x for k=-1,0,1. Note that for k=-2, the solution to the quadratic is x=0, which is NOT a valid solution since we'd end up dividing by 0. One interesting thing we notice is that for k=0, x=+-i. This means i has the interesting property that -i = 1/i. We don't actually need square roots to find our way to the imaginary numbers! Simply by defining i (and -i) to be numbers such that the multiplicitave inverse of i is equal to the additive inverse, we have found a way to construct the complex numbers. We can derive i * i = -1 with some simple algebraic manipulation. i * i (-1 * -i) * i -1* (1/i) * i -1 * (i/i) -1 * 1 -1
Great video! -k is the sum of the roots of the quadratic equation and x is the conjugate of 1/x. There is no conjugate of pair of integers that sums to 0, 1 or -1. The rest will satisfy the condition.
Before watching video. ±1 obviously. Then the graph behavior will be constantly approaching y = x (because 1/x becomes increasingly negligeable as x goes to infinity). Since all integer values of y are on the y=x line, and the graph is approaching but never touches the line (at least in the x>0 direction) there will be no solutions outside of the area near zero where (x + 1/x) - x ≥ 1. Just kind of a sketch for how to start. .... edit just saw that the teaser says there are infintely many more.... and I thought there wouldn't be any more... uh oh. Edit: O wait. Close to zero the graph goes to plus infinity so it will cross all integer y values past a certain point on the way up! (I'm still just looking at the positive side) The solutions will probably be algebraic irrationals. Edit.... I like the solution in the video very much
Nice - and deceptively simple! At first, I thought I saw the question as real x for which x - 1/x is an integer and immediately thought of the golden ratio as a solution. For that case, there are no integer restrictions on k.
One can also see the limits on the integer k and the number of solutions, by rearranging the equation to 1/x = k - x : clearly no intersections of the straight line y= k-x and the hyperbola y=1/x for -2< k
x² + 1 = kx x² - kx + 1 = 0 x = [k ± sqrt(k² - 4)]/2 Then begin subtituting values for k where k²-4>=0 k = 2 --> x = 1 k = 3 --> x = (3±sqrt5)/2 k = 4 --> x = (2±sqrt3) Etc And k = -2 --> x = -1 k = -3 --> x = (-3±sqrt5)/2 k = -4 --> x = (-2±sqrt3) etc However if we're looking for integer or rational values of x we only have 1 and -1 to give k = 2 and -2 respectively
Yaaaaay, the first problem I correctly solve from your channel 🎉🎉. Just started diving tip of my nose 😂 in Math olympiads and number theory, what a vaaast ocean 🌊🪸
@@jamesharmon4994 You can get "all" of them by using the formula in the video, which is pretty obvious. That aside, my reply above was to the other poster, not to you, but YT deletes @tags in replies. God knows why.
Something nice to think about regarding this problem, is that for f(x) = x + 1/x we can show that f(1/x) = 1/x + 1/(1/x) = 1/x + x = x + 1/x = f(x) So if f(x)=k , then f(1/x)=k We saw in the video which values of x solve the equation, let's call them x1 and x2. The property above shows that x2 = 1/x1. A nice follow up exercise is to show this directly by calculating the reciprocal of one of the solutions. Following the same spirit, though less surprising, we notice that f(-x) = -x + 1/(-x) = - (x + 1/x) = -f(x) Meaning, if f(x) = k , then f(-x) = -k We can therefore show that for each one of the solutions of f(x)=k from the video, there is a fitting solution of f(x)=-k from the video which is it's additive inverse, by putting a negative sign next to it and reaching the other's form. Lastly, a point to consider is that the function f(x) has a local minimum at x=1 and maximum at x=-1. It also approaches infinity when x approaches 0 from above or to infinity. Similarly, it approaches minus infinity when x approaches 0 from bellow or to minus infinity. Since f(x) is continuous at any x=/0, we can conclude that f(x)=r has a solution for any r>=2 or r
@@jaspereijkenboom4716 Of course! It is actually an introductory level material, something that you could see in a university level calculus class. So if you finished high school or are of such a math level, it is the next step in math studies. Whether you go for math or another STEM degree, you will probably cover it on your first semester. Of course, this is just an application of calculus and not the material itself. Not to discourage, I just don't want to cause any misunderstanding
@@joelklein3501 hahaha, thanks for the answer! Actually, I almost finished a computer science bachelor and will do a applied mathematics bridge year + master degree after this. I took a calculus course of course and finished an introductory course to analysis. My question was actually to find a book or just some exercises that specifically ask for looking for identities? of some function, like in this case f(x) = f(1/x) for example. This was not a point of attention in my calculus course
@@jaspereijkenboom4716 Oh, alright! In my head I thought you meant stuff like intermediate value theorem (even though I didn't mentioned it by name explicitely). I never learned it explicitely from any textbook, but a subject that might interest you is Functional Equations. It's similar to Differential Equations, but instead of having an equation of a function and it's derivatives, it's more of an identity that the function itself satifsy. For example, find all the functions that satisfy f(xy) = f(x)*f(y) Or f(x) = f(1/x) A math youtube channel that I like to watch that solves such questions occassionally is Michael Penn. Besides that, my knowledge about it is more of accumulated tricks that I learned throughout my studies, and not anything that covers this in particular
Well, congrats. You just earned a subscribe for your presentation style. I'd say that's a pretty good compliment since the only other math youtuber I watch is 3blue1brown. Good job and keep it up.
Wow!! Key idea (the mechanics of quadratics aside) is placing constraints on what otherwise looks far too huge a space to deal with! Such a profound result, from such a direct and easy technique - if you recognize the tool needed: quadratic equation solution method! Super thanks. This is what math is really about!
Very nice. A variant of this exercise: show that if x+1/x is an integer, then x is not rational when x is different from 1 and -1. It is interesting because it can be solved by divisibility issues. Thank you very much. Greetings from Spain.
He didn't spend a long time finding it. He spent a long time explaining what it meant, but he found it very quickly. I'm not sure what you mean about using the range of x+1/x. Firstly, the range isn't obvious and requires a bit of work to figure it - longer than his method took. Secondly, what are you going to do with the range once you've found it?
""Wisdom is a shelter as money is a shelter, but the advantage of knowledge is this: Wisdom preserves those who have it." Ecclesiastes 7:12 (NIV). Thanks for another great video!
When I tried to do it, the first thing I tried was to use a function to figure out what the set of solutions would look like. I represented the function f(x) = x + 1/x, and plotted the graph. With that, it is easy to see how there is a minimum in x=1, where f(1)=2. Also, due to the inverse symetry of the function, you can analyze the positive side of the graph and afterwards extrapolate that to the negative side in order to figure out the other half of the solutions. To see what the solutions of the equations look like, all you have to do is see where the graph cuts the horizontal lines y=2, y=3, y=4, etc... The points where those horizontal lines cut the graph are your solutions, and you can clearly see that, besides the minimum point in (1,2), all other solutions come in pairs that complement each other, with one being greater than 1 and the other being smaller than 1. Due to the inverse symetry of the function, you can also expect the same type of distribution of the solutions in the negative side of the graph. Knowing that the solutions exist and how they are distributed, you can solve the equation in the same manner you did on this video. I guess I took a bit of an unnecessary step in the process to ensure the existence of the solutions and the distribution of them, but still, it gives another perspective of the results that can be interesting. :) Nice video, with very intuitive explanations. Keep up the good work. Cheers!
I'm a brazilian who likes math and I'm learning english. Your channel was a geat discovery, the greatest in much time, becaue I really enjoy see the questions you bring up here and the form that you solve them, so I can join the useful with the pleasant, thank you very much for the videos 😁(and please, forgive me any writing problem, I'm learning yet 😅)
x + 1/x = k , where k is an integer x² - kx + 1 = 0 ... apply quadratic formula ... x = [k ± √(k² - 4)]/2 Since we want x to be real, (k² - 4) must be greater than or equal to 0 ==> |k| ≥ 2 All real solutions: x = [k ± √(k² - 4)]/2 where k is an integer and |k| ≥ 2
6:17 i think this is inaccurate to say since k will have to be an integer no matter what since we've defined it so. Using k values between -2 and 2 will still grant an integer for x+1/x but does not satisfy the condition that x has to be real. Love the videos man!
The formula at the end also works with -1 and 1. It will involve a complex number, but that doesn't matter because the imaginary component will cancel out as well.
It would be nice to see the first few approximations to see how fast they go intuitively and to see if there might be a recurrence relation between them.
I think you spent too much time on the |k|>=2 part. I think that was the easiest part of the video, but you explained it about 4 times. Thanks for your video and overall approach. Good explanation.
We know how to draw this function. WE know it has an asymptota y=x . So the vertical distance between those two functions is getting smaller and smaller as x approaches infiniry ...
Besides 1 and -1 all solutions will have radicals as squares whose distances is 4 apart seems only to exist with 2^2 and 0^2 and checking with radicals takes a bit of work.
I’m confused, this seems like magic. When you assumed x + 1/x is an integer k, that is only an assumption at that point, it’s not proven yet. So when you solve for k which makes it real, why does that necessarily make it an integer?
I believe the given solution is only a subset of ALL real values of x that satisfy the requirement. For any integer k such that | k | >= 2, the following real values of x also satisfy the requirement: k + sqrt(k^2 - 1) k - sqrt(k^2 - 1) - (k + sqrt(k^2 - 1)) - (k - sqrt(k^2 - 1)) Just my two cents.
Solution: Let's define the target value as t, so we have x + 1/x = t two obvious solutions are x = - 1 and x = 1 as t would be -2 and 2 respectively Now let's solve for x: x + 1/x = t |*x x² + 1 = tx |-tx x² - tx + 1 = 0 x = (t ± √(t² - 4))/2 so any t² ≥ 4 creates real values. which means, that t ≤ -2 OR t ≥ 2 So any x fulfilling the equation x = (t ± √(t² - 4))/2 with t being an integer ≤ -2 OR ≥ 2 will mean, that that integer is the result of x + 1/x
I just participated in an math quiz or smthg, and it was the same thing but it asked x² +1/x is an integer and the x value lies between (sqrt2 and sqrt3). Is it the same process but with a 3rd degree function?
Childishly easy, takes 5 seconds solving ! the given condition is equivalent to x divides x²+1 (and x is not zero), hence x divides 1, hence x = plus ou minus 1. QED.
@@MichaelGraham1980 right you are, I didn't even read the question properly, for some reason having convinced myself you were asking for integer only solutions. 😊
@PrimeNewtons I watched the video till end that's why I used -7 to check. Moreover I am not mathematician, but I like maths so videos like yours look interesting to ma
the quadratic then becomes x² - kx -1 = 0 The determinant becomes k² +4 >= 0 which means that the restriction on the k disappears, so k can be any integer. so x can be found with x = (k ± √(k² +4) ) / 2. Where k is an integer. for k = 0 we get the trivial solution of x = 1 or x = -1. All the other solutions contain some kind of square root.
x + 1/x = (x^2 + 1)/x If it is to be an integer, then x must divide x^2 + 1; thus x^2 + 1 = nx for some integer n x^2 - nx + 1 = 0 x = (n +- sqrt(n^2 - 4))/2 Thus, every number x satisfying the equation x + 1/x is given by (n +- sqrt(n^2 - 4))/2, for all integers n. For example, try the first few natural numbers: n = 1 --> x is not real n = 2 --> x = (2/2) = 1 n = 3 -> x = (-3 +- sqrt(5))/2 And so on.
Nice solution and explanation. I have just a little remark. For every n ∈ Z value (under the aforementioned restriction) there are 2 x ∈ R solutions to the quadratic, which are reciprocals to each other. You could use this propriety when you tested the results (9:09), making the calculations easier.
x + 1/x = N x^2 - Nx + 1 = 0 x = [N +/- √(N^2 - 4)] / 2 N^2 - 4 = k^2 (k is odd if N is odd, k is even if N is even) N^2 - k^2 = 4 (N + k)(N - k) = 1 x 4 = 4 x 1 = 2 x 2 = -1 x -4 = -4 x -1 = -2 x -2 N = 2, x = 1 ✅ N = -2, x = -1 ✅
x doesn't need to be an integer only x+1/x = 'Ans' this Answer(Ans) needs to an integer and for that condition he solved the equation and got the values. He never claimed x needs to be an integer.
Thank you so much! I'm just a dad who hasn't done any math in 20 years and spent an hour last night helping his daughter with really basic trigonometry. But I can do that stuff in my head, so I thought I would indulge in some more challenging problems. I don't care how young or old someone is, this kind of content is _always_ a good use of one's time.
you're right !
I'll be 70 yo. in a few months and can't stop watching these videos!
🤣🤣
That's awesome. Please keep going. The mind is a powerhouse.
@ and maths is also an excellent method to hit a home run with Alzheimer's 🤣
I've been watching your videos for the past two months. Before watching the solution, I always try to solve the questions on my own. This is the first question I've successfully solved so far! I feel really good :)
I'm glad you were able to solve it!
Same here 🎉🎉🎉🎉 first problem, phew, I almost lost hope in solving one. But I do lack the knowledge so no giving up 🎉 💪💪💪🫡
We may as well find the complex solutions
If x is complex, you only need to remove the restriction for k
how are you going to solve for complex x without the k inequality restriction?
@@arnavraheja10Just plug x^2-kx+1 into the quadratic equation like normal. It will give you the correct values of x for k=-1,0,1. Note that for k=-2, the solution to the quadratic is x=0, which is NOT a valid solution since we'd end up dividing by 0.
One interesting thing we notice is that for k=0, x=+-i. This means i has the interesting property that -i = 1/i. We don't actually need square roots to find our way to the imaginary numbers! Simply by defining i (and -i) to be numbers such that the multiplicitave inverse of i is equal to the additive inverse, we have found a way to construct the complex numbers.
We can derive i * i = -1 with some simple algebraic manipulation.
i * i
(-1 * -i) * i
-1* (1/i) * i
-1 * (i/i)
-1 * 1
-1
@@arnavraheja10 let k = -1, 0, or 1
@@arnavraheja10 the only thing that k inequality does is guarantee that x is real, so if x is comples k is equal 1, 0 or -1
I had a feeling that it would have to be some weird irrational numbers.
Banger video!
it is
Great video! -k is the sum of the roots of the quadratic equation and x is the conjugate of 1/x. There is no conjugate of pair of integers that sums to 0, 1 or -1. The rest will satisfy the condition.
Thanks for that - I was wondering why the quadratic with k, which looked sensible,, excluded the two trivial solutions.
@@jamesweatherley7764 It actually doesn't; if k=+/-2 the discriminant is 0, so you are left with k/2 i.e. +/-1
Wonderful explanation sir. Thanks a lot. Your video improve quest to dive into the mathematics
Before watching video. ±1 obviously. Then the graph behavior will be constantly approaching y = x (because 1/x becomes increasingly negligeable as x goes to infinity). Since all integer values of y are on the y=x line, and the graph is approaching but never touches the line (at least in the x>0 direction) there will be no solutions outside of the area near zero where (x + 1/x) - x ≥ 1. Just kind of a sketch for how to start. .... edit just saw that the teaser says there are infintely many more.... and I thought there wouldn't be any more... uh oh. Edit: O wait. Close to zero the graph goes to plus infinity so it will cross all integer y values past a certain point on the way up! (I'm still just looking at the positive side) The solutions will probably be algebraic irrationals. Edit.... I like the solution in the video very much
Nice - and deceptively simple! At first, I thought I saw the question as real x for which x - 1/x is an integer and immediately thought of the golden ratio as a solution. For that case, there are no integer restrictions on k.
Great video. Amazing work as always!
One can also see the limits on the integer k and the number of solutions, by rearranging the equation to 1/x = k - x : clearly no intersections of the straight line y= k-x and the hyperbola y=1/x for -2< k
x² + 1 = kx
x² - kx + 1 = 0
x = [k ± sqrt(k² - 4)]/2
Then begin subtituting values for k where k²-4>=0
k = 2 --> x = 1
k = 3 --> x = (3±sqrt5)/2
k = 4 --> x = (2±sqrt3)
Etc
And
k = -2 --> x = -1
k = -3 --> x = (-3±sqrt5)/2
k = -4 --> x = (-2±sqrt3)
etc
However if we're looking for integer or rational values of x we only have 1 and -1 to give k = 2 and -2 respectively
Yaaaaay, the first problem I correctly solve from your channel 🎉🎉. Just started diving tip of my nose 😂 in Math olympiads and number theory, what a vaaast ocean 🌊🪸
I never would have guessed there are infinitely many solutions!
The other 'funny' thing is that most of them are irrational, but none are transcendental.
@@dlevi67because transcendental numbers are number that isn’t root of any non-zero polynomial with integer coefficients
@ Yes, obviously.
@@jamesharmon4994 You can get "all" of them by using the formula in the video, which is pretty obvious.
That aside, my reply above was to the other poster, not to you, but YT deletes @tags in replies. God knows why.
@@jamesharmon4994 YT keeps deleting my replies for no reason. Just as it deleted the @tag in my reply to the other poster above.
Magnificent! The solution provides a review of the quadratic equation, the meaning of absolute value and its notation, set notation, etc. 🙂
Something nice to think about regarding this problem, is that for
f(x) = x + 1/x
we can show that
f(1/x) = 1/x + 1/(1/x) = 1/x + x = x + 1/x = f(x)
So if f(x)=k , then f(1/x)=k
We saw in the video which values of x solve the equation, let's call them x1 and x2. The property above shows that x2 = 1/x1.
A nice follow up exercise is to show this directly by calculating the reciprocal of one of the solutions.
Following the same spirit, though less surprising, we notice that
f(-x) = -x + 1/(-x) = - (x + 1/x) = -f(x)
Meaning, if f(x) = k , then f(-x) = -k
We can therefore show that for each one of the solutions of f(x)=k from the video, there is a fitting solution of f(x)=-k from the video which is it's additive inverse, by putting a negative sign next to it and reaching the other's form.
Lastly, a point to consider is that the function f(x) has a local minimum at x=1 and maximum at x=-1. It also approaches infinity when x approaches 0 from above or to infinity. Similarly, it approaches minus infinity when x approaches 0 from bellow or to minus infinity.
Since f(x) is continuous at any x=/0, we can conclude that f(x)=r has a solution for any r>=2 or r
Very interesting! Could I ask you, which (if any) university courses or math books would have exercises and go into topics like these?
@@jaspereijkenboom4716 Of course! It is actually an introductory level material, something that you could see in a university level calculus class. So if you finished high school or are of such a math level, it is the next step in math studies. Whether you go for math or another STEM degree, you will probably cover it on your first semester.
Of course, this is just an application of calculus and not the material itself. Not to discourage, I just don't want to cause any misunderstanding
@@joelklein3501 hahaha, thanks for the answer! Actually, I almost finished a computer science bachelor and will do a applied mathematics bridge year + master degree after this. I took a calculus course of course and finished an introductory course to analysis. My question was actually to find a book or just some exercises that specifically ask for looking for identities? of some function, like in this case f(x) = f(1/x) for example. This was not a point of attention in my calculus course
@@jaspereijkenboom4716 Oh, alright! In my head I thought you meant stuff like intermediate value theorem (even though I didn't mentioned it by name explicitely).
I never learned it explicitely from any textbook, but a subject that might interest you is Functional Equations. It's similar to Differential Equations, but instead of having an equation of a function and it's derivatives, it's more of an identity that the function itself satifsy. For example, find all the functions that satisfy
f(xy) = f(x)*f(y)
Or
f(x) = f(1/x)
A math youtube channel that I like to watch that solves such questions occassionally is Michael Penn.
Besides that, my knowledge about it is more of accumulated tricks that I learned throughout my studies, and not anything that covers this in particular
@@joelklein3501 Ahhhh that's perfect! For some reason I just really like these functional equations. Will look into it. Thanks!
A bit besides the point, but what a clean blackboard! This is a lovely video. You explain mathematics in a very calming way.
Well, congrats. You just earned a subscribe for your presentation style. I'd say that's a pretty good compliment since the only other math youtuber I watch is 3blue1brown. Good job and keep it up.
Wow!! Key idea (the mechanics of quadratics aside) is placing constraints on what otherwise looks far too huge a space to deal with! Such a profound result, from such a direct and easy technique - if you recognize the tool needed: quadratic equation solution method! Super thanks. This is what math is really about!
Very nice. A variant of this exercise: show that if x+1/x is an integer, then x is not rational when x is different from 1 and -1. It is interesting because it can be solved by divisibility issues. Thank you very much. Greetings from Spain.
why spend so long to find abs(k) >= 2 when you can just use the range of the original function (x+1/x)
He didn't spend a long time finding it. He spent a long time explaining what it meant, but he found it very quickly. I'm not sure what you mean about using the range of x+1/x. Firstly, the range isn't obvious and requires a bit of work to figure it - longer than his method took. Secondly, what are you going to do with the range once you've found it?
Great Video and very nice explanation.
eq sign not aligned with a division sign gave me OCD, but I still love this video
""Wisdom is a shelter as money is a shelter, but the advantage of knowledge is this: Wisdom preserves those who have it." Ecclesiastes 7:12 (NIV). Thanks for another great video!
Cool. Totally surprised by the end result.
cool video! I would have loved to see the proof instead of a single test case. Maybe you can make another video with only the proof
Interesting video and very nice handwriting.
The math is nice. But that accent is the real beauty.
His smile means, killed the equation by them 😂
It's natural that there will be infinitely many options because this is just a rational but such a cool result!
When I tried to do it, the first thing I tried was to use a function to figure out what the set of solutions would look like. I represented the function f(x) = x + 1/x, and plotted the graph. With that, it is easy to see how there is a minimum in x=1, where f(1)=2. Also, due to the inverse symetry of the function, you can analyze the positive side of the graph and afterwards extrapolate that to the negative side in order to figure out the other half of the solutions.
To see what the solutions of the equations look like, all you have to do is see where the graph cuts the horizontal lines y=2, y=3, y=4, etc... The points where those horizontal lines cut the graph are your solutions, and you can clearly see that, besides the minimum point in (1,2), all other solutions come in pairs that complement each other, with one being greater than 1 and the other being smaller than 1. Due to the inverse symetry of the function, you can also expect the same type of distribution of the solutions in the negative side of the graph.
Knowing that the solutions exist and how they are distributed, you can solve the equation in the same manner you did on this video.
I guess I took a bit of an unnecessary step in the process to ensure the existence of the solutions and the distribution of them, but still, it gives another perspective of the results that can be interesting. :)
Nice video, with very intuitive explanations. Keep up the good work.
Cheers!
I'm a brazilian who likes math and I'm learning english. Your channel was a geat discovery, the greatest in much time, becaue I really enjoy see the questions you bring up here and the form that you solve them, so I can join the useful with the pleasant, thank you very much for the videos 😁(and please, forgive me any writing problem, I'm learning yet 😅)
Cool. I'm still trying to see at what point the restriction that k is an integer was used (not just said)...
it is required by the question. we are looking for integers.
x + 1/x = k , where k is an integer
x² - kx + 1 = 0
... apply quadratic formula ...
x = [k ± √(k² - 4)]/2
Since we want x to be real, (k² - 4) must be greater than or equal to 0 ==> |k| ≥ 2
All real solutions:
x = [k ± √(k² - 4)]/2
where k is an integer and |k| ≥ 2
6:17 i think this is inaccurate to say since k will have to be an integer no matter what since we've defined it so. Using k values between -2 and 2 will still grant an integer for x+1/x but does not satisfy the condition that x has to be real. Love the videos man!
Technically, it also works with some complex numbers : i + 1/i = i - i = 0
You're imaging things
@@Why553-k5b_1Photography :D
nice clear explanation. Good job!
The formula at the end also works with -1 and 1. It will involve a complex number, but that doesn't matter because the imaginary component will cancel out as well.
Where he from? great videos.
It would be nice to see the first few approximations to see how fast they go intuitively and to see if there might be a recurrence relation between them.
Wow,
such briliant 👏👏👏
I think you spent too much time on the |k|>=2 part. I think that was the easiest part of the video, but you explained it about 4 times.
Thanks for your video and overall approach. Good explanation.
We know how to draw this function. WE know it has an asymptota y=x . So the vertical distance between those two functions is getting smaller and smaller as x approaches infiniry ...
Besides 1 and -1 all solutions will have radicals as squares whose distances is 4 apart seems only to exist with 2^2 and 0^2 and checking with radicals takes a bit of work.
I’m confused, this seems like magic. When you assumed x + 1/x is an integer k, that is only an assumption at that point, it’s not proven yet. So when you solve for k which makes it real, why does that necessarily make it an integer?
Solved before viewing. I found a countably infinite number of solutions. Remarkable.
I believe the given solution is only a subset of ALL real values of x that satisfy the requirement.
For any integer k such that | k | >= 2, the following real values of x also satisfy the requirement:
k + sqrt(k^2 - 1)
k - sqrt(k^2 - 1)
- (k + sqrt(k^2 - 1))
- (k - sqrt(k^2 - 1))
Just my two cents.
Solve it reversely. Can we find an integer K such that the expression is an integer and x is also an integer?
That was really cool.
Solution:
Let's define the target value as t, so we have
x + 1/x = t
two obvious solutions are x = - 1 and x = 1 as t would be -2 and 2 respectively
Now let's solve for x:
x + 1/x = t |*x
x² + 1 = tx |-tx
x² - tx + 1 = 0
x = (t ± √(t² - 4))/2
so any t² ≥ 4 creates real values.
which means, that t ≤ -2 OR t ≥ 2
So any x fulfilling the equation x = (t ± √(t² - 4))/2 with t being an integer ≤ -2 OR ≥ 2 will mean, that that integer is the result of x + 1/x
Thank you for this!
I just participated in an math quiz or smthg, and it was the same thing but it asked x² +1/x is an integer and the x value lies between (sqrt2 and sqrt3). Is it the same process but with a 3rd degree function?
I was overthinking the last step
my idea was to set sin((x+1/x)pi)=0 and solve for it, but that's definitely not an easy way. No idea why that was my first thought tho
It's not a video, just feels like, it's goosebumps in me..😅
Childishly easy, takes 5 seconds solving ! the given condition is equivalent to x divides x²+1 (and x is not zero), hence x divides 1, hence x = plus ou minus 1. QED.
0/10 thank you for playing, better luck next time
@@MichaelGraham1980 right you are, I didn't even read the question properly, for some reason having convinced myself you were asking for integer only solutions. 😊
Which class problem is this in your country?
I thought the solution would be interesting after realizing 3=Φ²+1/Φ². I was *slightly* disappointed
Very well explained.
Tank you. Now I understand.
The only integer X for which this is true is x = -1 or 1. You can deduce this from the quadratic equation.
Great ❤. Thank u
I messed up, cause I was thinking that x had to be an integer as well.
Wow! Amazing! I love maths!!!
How do you come up with such interesting problem?
This came as a thought while making the previous video
@PrimeNewtons Thank you for answer.
I'm korean and I really enjoy your video.
So x =-7
- 7 -1/7 is intege?
No. Did you watch the video?
@PrimeNewtons I watched the video till end that's why I used -7 to check.
Moreover I am not mathematician, but I like maths so videos like yours look interesting to ma
You are confusing x with k.
0:00 interesting
So x^2 + 1 = nx -> x^2 - nx + 1 = 0 with n an
So x= (n +- sqrt(n^2 -4) ) / 2n
as x is real, we need n^2-4 >= 0 so n >= 2 or n
Very interesting question and solved in a simple way.
I ❤ this!
Absolutely wonderous.
What about for X-1/X
the quadratic then becomes x² - kx -1 = 0
The determinant becomes k² +4 >= 0 which means that the restriction on the k disappears, so k can be any integer.
so x can be found with x = (k ± √(k² +4) ) / 2. Where k is an integer.
for k = 0 we get the trivial solution of x = 1 or x = -1. All the other solutions contain some kind of square root.
x + 1/x = (x^2 + 1)/x
If it is to be an integer, then x must divide x^2 + 1; thus
x^2 + 1 = nx for some integer n
x^2 - nx + 1 = 0
x = (n +- sqrt(n^2 - 4))/2
Thus, every number x satisfying the equation x + 1/x is given by (n +- sqrt(n^2 - 4))/2, for all integers n.
For example, try the first few natural numbers:
n = 1 --> x is not real
n = 2 --> x = (2/2) = 1
n = 3 -> x = (-3 +- sqrt(5))/2
And so on.
For n=1: Imagine it is.
I misunderstood the question, I love the solution though!
If you change the problem to x - 1/x = int then phi is a solution and that makes me happy :)
Thx!
i found 1 and was proud of myself.
X=n/2±(√(n²/4-1)) where |n| is greater than or equal to 2
And n is an integer
Infinitely many >1, infinitely maNy < 1 but which?? Oh and all’ the negative ones too(
Ah quadratic equation in x, okok😂
Good
Is it just me more he looks like dr foreman from house md 😂😂
Magnifique.
Amazing result
Réponse inattendue !!!
Très amusant !
x = [a +/- (a^2-4)^(1/2)]/2 a is an integer so a^2 >= 4
If x=0 the solution is not a non-integer
√-1 is integer, proof by comment
456 voted X I think
Clever.
Nice solution and explanation.
I have just a little remark. For every n ∈ Z value (under the aforementioned restriction) there are 2
x ∈ R solutions to the quadratic, which are reciprocals to each other. You could use this propriety when you tested the results (9:09), making the calculations easier.
еще бы знать английский, но что-то прикольное говорит
I am glad you still watched the video, even when you didn't understand everything I said.
@@PrimeNewtons у вас очень интересный канал, с интересом смотрю
This guy r un...🗿🗿
±1.
Infinity 1/infinity =0 but + infinity= infinity
I know it doesn't matter but sqrt(45) is 3sqrt(5) !!!!!!!!!!!!
1
x + 1/x = N
x^2 - Nx + 1 = 0
x = [N +/- √(N^2 - 4)] / 2
N^2 - 4 = k^2 (k is odd if N is odd, k is even if N is even)
N^2 - k^2 = 4
(N + k)(N - k) = 1 x 4 = 4 x 1 = 2 x 2 = -1 x -4 = -4 x -1 = -2 x -2
N = 2, x = 1 ✅
N = -2, x = -1 ✅
Good😂
If x=7 then x + 1/× is not an integer. (7+1/7). What is my fault?
Here your x does not follow the property x= (k±√k²-4)/2 7:05
k is the integer where you fill in 7, which gives a value of x. Fill that x into x + 1/x and it will be an integer.
x doesn't need to be an integer only x+1/x = 'Ans' this Answer(Ans) needs to an integer and for that condition he solved the equation and got the values. He never claimed x needs to be an integer.
He solves for k=7 to get x. Then, using that x, he reverses the process to get back to k=7.
🙏
x^2+1=kx, k integer
x^2-kx+1=0
x=(k+-sqr(k^2-4))/2 ; |k|>=2.
Too much words and calculus for succes a easy problem.
No, it was an almost perfect explanation!
@juergenzhang9133 May be almost perfect but too long.
There was no calculus at all here
Not 0, real fraction between -1, 1