I made a completely different approach: I substituted b with (1-a) and calculated a with a^2 + b^2 = 2. Then I calculated b. Your way indeed looks nicer, but I have a pair of values for each a and b afterwards. :-)
You found generalised solution, which is not bad, the thing about this question is that you may notice that powers are all power of 2 and therefore you may use square sum formula to simplify solution, that would be extra point. I mean you not just randomly squaring things as you may think from video.
Easy 2 equations and 2 unknowns! Finds out the numerator for b is 1 +/- sqrt(3). Quickly realizes that b^8 is more distribution than I would like to do in my life time.
That is how I did it, except you've missed a factor of 2. (This comes from quadratic formula.) a=(1+sqrt(3))/2, b=(1-sqrt(3))/2 (or vice versa.) Then can use (1+x)^8 = 1 + 8x + 28x^2 + 56 x^3 + 70 x^4 + 56 x^5 + 28 x^6 + 8 x^7 + x^8. (Coefficients are 8 choose n, so not too hard to figure out, or you can just write Pascal's triangle sufficiently far.) Then we plug in x=sqrt(3). Except! We're just about to add this to x=-sqrt(3), so all the odd terms will cancel, and we can ignore them. So we end up with a^8+b^8 = (1/2)^8 . 2 . (1+(28)(3)+(70)(9)+(28)(27)+(1)(81)) = 1552 . (1/2)^7 = 97/8. Sorry, I'm not quite as hand-holding as PreMath, because I explained in one paragraph instead of 8 minutes.
You can also use the first 2 equations to make a quadratic equation and solve for one of the variables by the quadratic formula. The other variable is solved by substituting the found variable into a+b=1. Of course, A^8+b^8 becomes a lot of work without a calculator.
But note that by symmetry, the quadratic formula gives the solutions for a and b at the same time. One value takes the plus sqrt and the other the minus sqrt.
@@neilmasson3609 Thats how I did it a = b = root(7)/2 and since we are raising to an even power, the minus sign does not matter. Got wrong answer though.
I did solve this the mechanical way, finding a and b = (1+√3)/2 and (1-√3)/2 (or vice versa). then squaring each of those values three times and summing them. I don't think in this particular case that was more work or less elegant, and it does give some other minor insights.
Very nice method .Thank you. I did it by getting values for a and b. I ended up with (1/2)^8 multiplied by (1+root3)^8 +(1-root3)^8. It worked out at 97\8 .
I like this kind of problems and how the answer is very general! A sequence u_n of the form a^n+b^n is a recurrent linear sequence of order two, meaning there are x and y such that u_{n+2}= xu_{n+1} + yu_n, just like the Fibonacci sequence. Since you know u_0 = 2 u_1 = 1 u_2 = 2, you have 3 initial conditions which allows you to find x and y. This is the best thing about maths, now that I said it was possible to do so, I can be satisfied and not even bother to compute u_8 😎
I subtitued a by 1 - b in a second equation to find b and when i find b i did the same in the equation a^2 + b^2= 2 to find a and after i can calculate a^8 + b^8. But your way is nicer than mine because to calculate the 8 power it take a lot of time without calculator! Great job.
I tried being fancy and using a^2 + b^2 = 2 is the equation for a circle with radius sqrt (2) then a+b = 1, is a straight line that intersects 2 points on it, either being a valid solution, but Im dumb and cant do math
I believe there is a simpler solution from first equation u make b=1-a then replace b in second gettin (a)^2 + (1-a)^2=2. rearrange to (2a^2)-(2a)-1 = 0 and solve the quadratic and u have 1.366 and -0.366 which is actually where the curve y=1-x and y^2=2-x^2 meet values are (2+✓12)/4 and (2-✓12)/4 for a and b
a + b = 1, ab = - 1/2, find a, b; then we'll know a^8 + b^8. The result will not be as clean as in this video but the requirement doesn't say how "clean" the result has to be.
What is zero to the power of 1? If it is zero it opens for two obvious and alternative solutions but zero to the power of 1 is above my understanding so please explain. If zero to the power of 1 is 1 (or anything else ≠ zero) it is ruled out. Or is it not even defined?
Thats a brilliant solution! However! You even take you time to explain how power multiplication works, why, why didn't you explain in the begining, that we should get to solution by noticing that all powers are power of 2, so thats why we should imidiatelly think about simplification by square sum formula? Otherwise that can just be solved for a and b.
PLEASE TRADUCIR ,,,,, SI A Y B SON NUMEROS DELCONJUNTO Q SIN INCLUIR A LOS NEGATIVOS QUE NO PUEDEN SER INCLUIDOS PORQUE AL ELEVARLOS AL CUADRADO SE CONVIERTEN EN POSITIVOS Y LA SUMA SRA MAYORDE 2 ....... ,,,,SUS VALORES SON MENORES QUE UNO ,,,,,,,,,,,AL ELEVARLOS AL CUADRADO NUNCA SU SUMA LLEGARA A DOS HAGAN LA PRUEBA CON CUALQUIER DECIMAL Y LO COMPROBARAN ..........
Довольно занудное решение. Уж легче тогда из первого выразить одну переменную, например, a=1-b и подставить во второе ( 1-b)^2 + b^2 -2 =0 , решить квадратное уравнение , найти по отдельности значения a и b , ну а затем и значение выражения a^8 + b^8. простите, но ваше решение - просто дрочево )
Там корни (1±√3)/2, хотя некоторые прогнали это через бином 8 степени. Если вместо 1 поставить 3 или ⁴√17, или, вообще, некоторое число к, то тут трудность сохранится - можно легко получить k⁸/8-k⁶-k⁴+12k²+2. А вот бином будет ещё каким дрочевом!
@@diedoktor Thanks! Much appreciated. Another question: So (from your response) it's true that adding, subtracting, multiplying, or dividing an irrational number by a rational number gives an irrational number?
@@diedoktor Ah! Thank you very much indeed. The examples you give are all "contrived" to produce a rational result but yes, they all require a second irrational number to produce a multiple (or whatever) of an irrational so that the operation produces a rational "factor" of the irrational numbers. I promise I'll eventually stop asking you questions (lol), but...is it true or false that an arithmetic operation of two "disconnected" irrational numbers cannot produce a rational number? Say, sqrt(2)*sqrt(3)? All the examples you give can have rationals factored out, or use multiple instances of the same irrational number: (sqrt(2)*sqrt(2))/sqrt(2). I'm sorry to be a pest; I'm just particularly interested in this "type" of question which is never really explained to regular students, and which mathematicians generally take for granted. Thanks again for your generosity. EDIT: I think my sqrt(2) example is wrong....but I'm sure you know what I was getting at...I think what I was meaning was (sqrt(2) + sqrt(2))/sqrt(2).
@@diedoktor Thanks for that. And yes, it's plain that raising to a power can change things. I consider the first example that you gave to be "cheating," of course. Maybe the answer isn't truly known -- which I would find satisfying. One of the other math channels (related to Numberphile) has put up a relatively recent video where he remarks that pi^pi^pi^pi is totally unknown and which -- for all anyone knows -- could turn out to be an integer. I'd still like to know -- without the cheating -- whether arithmetic done between two (or more) unrelated irrational numbers (that doesn't include a repeat, which, in effect, means an exponentiation has occurred). I would have thought this was known, or easily shown, or something. Maybe I'm just not conveying my question clearly; but...it would be exciting to hear that nobody really knows! I also once watched a video which discussed "non-computable" irrational numbers (I forget the proper terminology). Here's a video about that: th-cam.com/video/5TkIe60y2GI/w-d-xo.html . Here's a Khan Academy video which addresses the question I pose, and which suggests that one needs "tricks" to get rational numbers from combining irrational numbers: th-cam.com/video/16-GZWi66CI/w-d-xo.html. Thanks again for your input.
@@diedoktor Let me try a simpler question: The square roots of all prime numbers are irrational. Is the product of any two of those (different) square roots not irrational? And can that be proved one way or the other? I would have to guess that the product of any two would be irrational. It would be VERY interesting if "some" were rational. It would be interesting to find that no proof exists regarding the rationality of the products.
@@diedoktor Well, that's clear enough (presupposing that it has in fact been proved -- and I don't doubt that it has). I wonder whether it could be further generalized -- but maybe that''s an unreasonable request. Hmm, PLEASE forgive me if you think I'm trying to be obnoxious -- I promise you I'm not. But now a new question occurs to me: 0.333...is a rational number: 1/3. a) What can be said about the rationality of, say, 0.3334, followed by endless 3's? And what could be said about the product of that number times 0.33334 followed by endless 3's? Feel free to drop this anytime you like. I have NO RIGHT to keep at you with these things -- although I can assure you that I have no desire to actually waste your time. EDIT: Just to be clear, this is NOT a response to your side point about transcendental numbers.
@@ivornworrell This particular exercice not not be used exactly in practice, however, the objects involved (this is a class of sequences similar to the Fibonacci sequence) are very useful and commonly used. Maybe it will become clearer with Fibonacci sequence: There are many ways you can define it but one of them is that F_n is equal 1/sqrt(5)(a^n - b^n) for some reals a and b and F_1=F_2=1. If you can resolve the problem in the video then you can understand how and why what I just said characterizes the Fibonacci sequence. In practice, being able to switch from a characterization of the Fibonacci sequence to another is extremely important to study its properties. Then you must know that the Fibonacci sequence has many many real applications, so this exercice has many indirect applications to real world as well
@@ivornworrell Sorry I don't really know if and when it is seen in university actually, but I'm sure it could be the case, as it does not involve a lot of theory
I think your method is as short as any, and therefore would have found favour with the examiners. My own approach, common with other commentators here was to start at the bottom end [ by this analogy you started at the top by expanding a^8 + b^8 ]. As others point out the values of a and b can be found as a = P + Q and b = P - Q where P =1/2 and Q = sqrr(3)/2. This allows quick evaluation of a*b; which might have been used in your method with some advantage. With a bit of manipulative skill we could have evaluated a^8 + b^8 by the use of the Pascal triangle. Since a = P+Q and b = P- Q the final polynomial expressing the addition of the 8 -power sum is reduced to factors having only even powers of P and Q. That said I think your method is shorter while the latter is closer to brute force.
Same to you my dear friend. Thank you so much. Please accept my sincerest gratitude as well for the incredible job you are doing in educating our kids. Keep it up
I solved it other way. Not this beautiful.I think ‘a’ and ‘b’ must be written like ‘a=1+x’ and ‘b= -x’ and this way I solved the second equation and then I put the results in the third equation and the result became 12,125 what is 12 1/8. The same result.
![CAMPปลิ้น | EP.75[2/2] คุยแบบนี้แล้วจะจบกี่โมงพ่ะย่ะค่ะ!!!](http://i.ytimg.com/vi/N6E8njWH7Zg/mqdefault.jpg)
very well explained bro, thanks for sharing this problem
Always welcome my dear friend 242
Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
I made a completely different approach: I substituted b with (1-a) and calculated a with a^2 + b^2 = 2. Then I calculated b. Your way indeed looks nicer, but I have a pair of values for each a and b afterwards. :-)
You found generalised solution, which is not bad, the thing about this question is that you may notice that powers are all power of 2 and therefore you may use square sum formula to simplify solution, that would be extra point. I mean you not just randomly squaring things as you may think from video.
Same.
It is an interesting method shown.
But a and b are so easy to find. Then square multiply square reaches the 8th power very quickly.
I agree with this solution, I don't like when someone solves equation using something we don't know how he found out
It’s kind of cool that the solution doesn’t involving solving for either a or b.
Easy 2 equations and 2 unknowns! Finds out the numerator for b is 1 +/- sqrt(3). Quickly realizes that b^8 is more distribution than I would like to do in my life time.
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
That is how I did it, except you've missed a factor of 2. (This comes from quadratic formula.) a=(1+sqrt(3))/2, b=(1-sqrt(3))/2 (or vice versa.) Then can use (1+x)^8 = 1 + 8x + 28x^2 + 56 x^3 + 70 x^4 + 56 x^5 + 28 x^6 + 8 x^7 + x^8. (Coefficients are 8 choose n, so not too hard to figure out, or you can just write Pascal's triangle sufficiently far.) Then we plug in x=sqrt(3). Except! We're just about to add this to x=-sqrt(3), so all the odd terms will cancel, and we can ignore them. So we end up with
a^8+b^8 = (1/2)^8 . 2 . (1+(28)(3)+(70)(9)+(28)(27)+(1)(81)) = 1552 . (1/2)^7 = 97/8.
Sorry, I'm not quite as hand-holding as PreMath, because I explained in one paragraph instead of 8 minutes.
@@michaelwoodhams7866 They didnt miss the factor of 2, it says 'the numerator for b is...'
You can also use the first 2 equations to make a quadratic equation and solve for one of the variables by the quadratic formula. The other variable is solved by substituting the found variable into a+b=1. Of course, A^8+b^8 becomes a lot of work without a calculator.
But note that by symmetry, the quadratic formula gives the solutions for a and b at the same time. One value takes the plus sqrt and the other the minus sqrt.
@@neilmasson3609 Thats how I did it a = b = root(7)/2 and since we are raising to an even power, the minus sign does not matter. Got wrong answer though.
I solved it this way and then did the binomial expansion. It's not too bad because all the odd terms in root 3 cancel out.
I had no idea how to solve this one, but you explained it well. Thank you for this!
I did solve this the mechanical way, finding a and b = (1+√3)/2 and (1-√3)/2 (or vice versa). then squaring each of those values three times and summing them. I don't think in this particular case that was more work or less elegant, and it does give some other minor insights.
@ENOW ANOLD AFFUEMBEY but YOU'RE not
I decided in the same way
Hello, how did you find a and b??
@@saulloyd9235 quadratic formula
@@saulloyd9235 Substitute b as (a-1) in the second equation and solve the quadratic.
My thoughts
1. Give it a go
2. persevere
3. And a way may present itself
Very nice method .Thank you. I did it by getting values for a and b. I ended up with (1/2)^8 multiplied by (1+root3)^8 +(1-root3)^8. It worked out at 97\8 .
Excellent John! Keep it up dear
I did it this way also, though I wish I could have come up with the PreMath solution; clever and elegant.
Yours is not a bad method (like mine) but I would prefer the tutor's.
Thank you sir. This problem is very intelligently solved. Thank you again in teacher's day.
Me wondering why step 2 wasn't used to just create the equation ab = -1/2 and the solve this along with a+b = 1 simultaneously.
nice solution sir thanks
NICE solution! great technique!
Thank you dear! Cheers!
Keep it up
Thank you very much Sir.
I like how you explain the small and "easy" details on the side without disturbing the overall sequence of the solution.
Good Job !
Thank you! From Brazil.
Smart and accurate.
Hi sir, what is the tool you use to write the text ?
I like this kind of problems and how the answer is very general! A sequence u_n of the form a^n+b^n is a recurrent linear sequence of order two, meaning there are x and y such that u_{n+2}= xu_{n+1} + yu_n, just like the Fibonacci sequence.
Since you know
u_0 = 2
u_1 = 1
u_2 = 2,
you have 3 initial conditions which allows you to find x and y.
This is the best thing about maths, now that I said it was possible to do so, I can be satisfied and not even bother to compute u_8 😎
Many thanks for you
I subtitued a by 1 - b in a second equation to find b and when i find b i did the same in the equation a^2 + b^2= 2 to find a and after i can calculate a^8 + b^8.
But your way is nicer than mine because to calculate the 8 power it take a lot of time without calculator!
Great job.
very nice problem. you makes math fun
Very nice explanation thank you sir
You're most welcome Sneka dear. Keep it up
Thanks 👍👍
Danke, das hat großen Spass gemacht … 😉
Why are you allowed to do step one? Isn't a^8 + b^8 different from a^8 + b^8 + 2(ab)^4? Like you have an extra term there, now.
I tried being fancy and using a^2 + b^2 = 2 is the equation for a circle with radius sqrt (2) then a+b = 1, is a straight line that intersects 2 points on it, either being a valid solution, but Im dumb and cant do math
Original way to look at it. If your goal was to get a quick estimate of the values of a and b, this would be a nice way to do it.
BEAUTIFUL SOLUTION, GREETINGS FROM BRAZIL
Very well!
I believe there is a simpler solution
from first equation u make b=1-a
then replace b in second gettin
(a)^2 + (1-a)^2=2. rearrange to (2a^2)-(2a)-1 = 0 and solve the quadratic and u have 1.366 and -0.366 which is actually where the curve y=1-x and y^2=2-x^2 meet
values are (2+✓12)/4 and (2-✓12)/4 for a and b
But obtaining a^8 + b^8 becomes complicated.
Brilliant
a + b = 1, ab = - 1/2, find a, b; then we'll know a^8 + b^8. The result will not be as clean as in this video but the requirement doesn't say how "clean" the result has to be.
Excellent problem, very well explained.
Excellent David! Keep it up dear
Nice sir
Thanks and welcome Phani dear
Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃
Is possible to find out a and b ?
good solution.. but have done it very slowly...
Rearrange equation (1) to b:
b = 1 - a
Insert into equation (2):
a^2 + (1 - a)^2 = c
a^2 + a^2 - 2 * a + 1 = c
2 * a^2 - 2 * a + 1 - c = 0
a^2 - a + (1 - c)/2 = 0
Apply formula for quadratic equations:
a = 1/2 +- sqrt(1/4 - (1 - c)/2) = 1/2 +- 1/2 sqrt(2 * c - 1) = 1/2 * (1 +- sqrt(2 * c - 1))
b = 1 - a = 1/2 * (1 -+ sqrt(2 * c - 1))
Solution for a and b:
D = sqrt(2 * c - 1)
a1 = b2 = 1/2 * (1 + D)
a2 = b1 = 1/2 * (1 - D)
Calculation of a^8 + b^8:
a^8 = (1/2)^8 * (1 + 8*D + 28*D^2 + 56*D^3 + 70*D^4 + 56*D^5 + 28*D^6 + 8*D^7 + D^8)
b^8 = (1/2)^8 * (1 - 8*D + 28*D^2 - 56*D^3 + 70*D^4 - 56*D^5 + 28*D^6 - 8*D^7 + D^8)
a^8 + b^8 = (1/2)^8 * 2 * (1 + 28*D^2 + 70*D^4 + 28*D^6 + D^8)
= (1/2)^7 * (1 + 28*D^2 + 70*D^4 + 28*D^6 + D^8)
Applying for given numbers:
D = sqrt(2 * c - 1) = sqrt(3)
a^8 + b^8 = (1/2)^7 * (1 + 28 * 3 + 70 * 9 + 28 * 27 + 81)
= (1/2)^7 * (1 + 28 * 30 + 70 * 9 + 81)
= (1/2)^7 * (1 + 600 + 240 + 630 + 81)
= (1/2)^7 * 1552
= (1/2)^7 * (1600 - 48)
= (1/2)^7 * 16 * (100 - 3)
= (1/2)^7 * 2^4 * 97
= (1/2)^3 * 97
= 97/8
= 96/8 + 1/8
= 12 + 1/8
You are the best from Algeria
What is zero to the power of 1? If it is zero it opens for two obvious and alternative solutions but zero to the power of 1 is above my understanding so please explain. If zero to the power of 1 is 1 (or anything else ≠ zero) it is ruled out. Or is it not even defined?
Wish u happy Teachers day sir 💐💐💐💐
Thanks a lot Sneka dear. You are the best
The video is perfectly understandable at 1.75x speed.
This is so easy question for me😉
after doing linear algebra and complex analysis, this seems so trivial lol
2 + 2ab = 1
2ab = -1
ab = -0.5
Not sure how you're coming up with the first lines in each of the first 2 steps?
When I calculate it in my head, I have 48,5 because I make (1/64)(1+(28×3)+(70×9)+(28×27)+81)×2 simply.
Ok, for now the video is not readable.
this must be one of the easy question, as i managed to solve it
This was easy. Salam from azerbaijan
Very new! 🤗
I started with (a² + b²)^4, using the Newton's binomial formula, instead of (a^4 + b^4)² and got the same answer.
(a² + b²)^4 = a^8 + 4a^6 b^2 + 6a^4 b^4 + 4a^2 b^6 + b^8
Isolating (a^8 + b^8) we get
a^8 + b^8 = (a² + b²)^4 - 4a^6 b^2 - 6a^4 b^4 - 4a^2 b^6
Putting -4a²b² into evidence in the right hand side:
a^8 + b^8 = (a² + b²)^4 - 4(ab)²( a^4 + 6/4 a^2 b^2 + b^4)
Completing the square in the last parentheses:
a^8 + b^8 = (a² + b²)^4 - 4(ab)²( (a² + b²)² - (ab)²/2 )
Replacing a² + b² by 2 and ab by -1/2 (I skipped the step of calculating ab because I've done the same way as it's made in the video):
a^8 + b^8 = 2^4 - 4(-1/2)²( (2)² - (-1/2)²/2 )
a^8 + b^8 = 16 - 4(1/4)( 4 - (1/4)/2 )
a^8 + b^8 = 16 - 4 + 1/8 = 97/8
I hava a correct this solve thx so much
It's a very easy sum 👍
I think it's not a pre math , this is pro math
Is it just a coïncidence that solving a power 8 + b power 8 give us the first prime number separated by a gap of 8 which are 89 and 97
It is not obviously i have the secret of prime number and i will give it for humanity at the right moment
What is a = ? & b = ?
Thats a brilliant solution!
However! You even take you time to explain how power multiplication works, why, why didn't you explain in the begining, that we should get to solution by noticing that all powers are power of 2, so thats why we should imidiatelly think about simplification by square sum formula? Otherwise that can just be solved for a and b.
Is it really a math olympiad question and if yes then which one
Hurray ! I found it
12,125
Can you just solve what is A and what is B? That is more interesting to know.
I solved it basically the same way.
a=0.75, b=0.25
(root 1)^8 + (root 1) ^8 = 2
Very nice 👍🇧🇩🇧🇩🇧🇩
Thank you so much Sabbir dear 😀 You are awesome
Keep it up dear
Ans is 97/8. I was able to do this sum within a minute or so
Excellent! Keep it up dear
12 and 1/8
Being a 10th grader very easy question for me
I had done a minor mistake and my answer turns out to be 79/8 😂
The answer is 97/8. Easy but technical olympiad question. ( I have solved it 3 min)
Excellent Rahman! You are awesome
Keep it up dear
97/8
PLEASE TRADUCIR ,,,,, SI A Y B SON NUMEROS DELCONJUNTO Q SIN INCLUIR A LOS NEGATIVOS QUE NO PUEDEN SER INCLUIDOS PORQUE AL ELEVARLOS AL CUADRADO SE CONVIERTEN EN POSITIVOS Y LA SUMA SRA MAYORDE 2 ....... ,,,,SUS VALORES SON MENORES QUE UNO ,,,,,,,,,,,AL ELEVARLOS AL CUADRADO NUNCA SU SUMA LLEGARA A DOS HAGAN LA PRUEBA CON CUALQUIER DECIMAL Y LO COMPROBARAN ..........
i am happy i solved it :) ( i'm 16 btw)
Hmm I got 18.76. I got a quadratic in "a" solved it and then found "b". Then raised each to 8th power. Must have made an error somewhere.
Yep. The quadratic route does work; it's how I solved it. Likely you had a math error somewhere. I also got 12.125 or 97/8.
a^8+b^8=2ans
Frankly, lack of algebra knowledge, unable to solve it haha
2^8+(-1)^8 = 257.
I am giving this answer just by calculating in my head. I shall see the answer.
Довольно занудное решение. Уж легче тогда из первого выразить одну переменную, например, a=1-b и подставить во второе ( 1-b)^2 + b^2 -2 =0 , решить квадратное уравнение , найти по отдельности значения a и b , ну а затем и значение выражения a^8 + b^8. простите, но ваше решение - просто дрочево )
Там корни (1±√3)/2, хотя некоторые прогнали это через бином 8 степени. Если вместо 1 поставить 3 или ⁴√17, или, вообще, некоторое число к, то тут трудность сохранится - можно легко получить k⁸/8-k⁶-k⁴+12k²+2. А вот бином будет ещё каким дрочевом!
it's simple.
I'm not expert but if a = 0 & b=1 it works too....
And a8+b8=2
naa, it doesnt work for the a^2+b^2 = 2
25.625
My question: Can we determine from this result whether or not a and/or b are rational?
@@diedoktor Thanks! Much appreciated. Another question: So (from your response) it's true that adding, subtracting, multiplying, or dividing an irrational number by a rational number gives an irrational number?
@@diedoktor Ah! Thank you very much indeed. The examples you give are all "contrived" to produce a rational result but yes, they all require a second irrational number to produce a multiple (or whatever) of an irrational so that the operation produces a rational "factor" of the irrational numbers. I promise I'll eventually stop asking you questions (lol), but...is it true or false that an arithmetic operation of two "disconnected" irrational numbers cannot produce a rational number? Say, sqrt(2)*sqrt(3)? All the examples you give can have rationals factored out, or use multiple instances of the same irrational number: (sqrt(2)*sqrt(2))/sqrt(2). I'm sorry to be a pest; I'm just particularly interested in this "type" of question which is never really explained to regular students, and which mathematicians generally take for granted. Thanks again for your generosity. EDIT: I think my sqrt(2) example is wrong....but I'm sure you know what I was getting at...I think what I was meaning was (sqrt(2) + sqrt(2))/sqrt(2).
@@diedoktor Thanks for that. And yes, it's plain that raising to a power can change things. I consider the first example that you gave to be "cheating," of course. Maybe the answer isn't truly known -- which I would find satisfying. One of the other math channels (related to Numberphile) has put up a relatively recent video where he remarks that pi^pi^pi^pi is totally unknown and which -- for all anyone knows -- could turn out to be an integer. I'd still like to know -- without the cheating -- whether arithmetic done between two (or more) unrelated irrational numbers (that doesn't include a repeat, which, in effect, means an exponentiation has occurred). I would have thought this was known, or easily shown, or something. Maybe I'm just not conveying my question clearly; but...it would be exciting to hear that nobody really knows! I also once watched a video which discussed "non-computable" irrational numbers (I forget the proper terminology). Here's a video about that: th-cam.com/video/5TkIe60y2GI/w-d-xo.html . Here's a Khan Academy video which addresses the question I pose, and which suggests that one needs "tricks" to get rational numbers from combining irrational numbers: th-cam.com/video/16-GZWi66CI/w-d-xo.html. Thanks again for your input.
@@diedoktor Let me try a simpler question: The square roots of all prime numbers are irrational. Is the product of any two of those (different) square roots not irrational? And can that be proved one way or the other? I would have to guess that the product of any two would be irrational. It would be VERY interesting if "some" were rational. It would be interesting to find that no proof exists regarding the rationality of the products.
@@diedoktor Well, that's clear enough (presupposing that it has in fact been proved -- and I don't doubt that it has). I wonder whether it could be further generalized -- but maybe that''s an unreasonable request. Hmm, PLEASE forgive me if you think I'm trying to be obnoxious -- I promise you I'm not. But now a new question occurs to me: 0.333...is a rational number: 1/3. a) What can be said about the rationality of, say, 0.3334, followed by endless 3's? And what could be said about the product of that number times 0.33334 followed by endless 3's? Feel free to drop this anytime you like. I have NO RIGHT to keep at you with these things -- although I can assure you that I have no desire to actually waste your time. EDIT: Just to be clear, this is NOT a response to your side point about transcendental numbers.
a= 1,368
b=0,368
Tre Bian
But what are a and b? lol
Just brain-exercise, or real world applications?
Math has great applications in real world problems. AI is based on math and computer algorithms. Cheers
@@PreMath Can you please give an example of how to apply this particular math problem in real world situation? thx
@@ivornworrell This particular exercice not not be used exactly in practice, however, the objects involved (this is a class of sequences similar to the Fibonacci sequence) are very useful and commonly used. Maybe it will become clearer with Fibonacci sequence:
There are many ways you can define it but one of them is that F_n is equal 1/sqrt(5)(a^n - b^n) for some reals a and b and F_1=F_2=1. If you can resolve the problem in the video then you can understand how and why what I just said characterizes the Fibonacci sequence. In practice, being able to switch from a characterization of the Fibonacci sequence to another is extremely important to study its properties. Then you must know that the Fibonacci sequence has many many real applications, so this exercice has many indirect applications to real world as well
@@swenji9113 thx, is this first year university Math?
@@ivornworrell Sorry I don't really know if and when it is seen in university actually, but I'm sure it could be the case, as it does not involve a lot of theory
Somebody gave a dislike. Does that mean they disagree with the answer?
No worries John.
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Not the answer but the problem itself.
Found this one
If I didn’t solve it does that mean i have low iq ?
Absolutely no.
49/4 - 1/8. Please press the calculator!! No point torturing youself or yr students.
Don't drink soda.
is this really an olympiad question
8
I think your method is as short as any, and therefore would have found favour with the examiners. My own approach, common with other commentators here was to start at the bottom end [ by this analogy you started at the top by expanding a^8 + b^8 ].
As others point out the values of a and b can be found as a = P + Q and b = P - Q where P =1/2 and Q = sqrr(3)/2. This allows quick evaluation of a*b; which might have been used in your method with some advantage.
With a bit of manipulative skill we could have evaluated a^8 + b^8 by the use of the Pascal triangle. Since a = P+Q and b = P- Q the final polynomial expressing the addition of the 8 -power sum is reduced to factors having only even powers of P and Q. That said I think your method is shorter while the latter is closer to brute force.
a = ?
b = ?
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1
Common.
Answer should be 8.
2
I solved it other way. Not this beautiful.I think ‘a’ and ‘b’ must be written like ‘a=1+x’ and ‘b= -x’ and this way I solved the second equation and then I put the results in the third equation and the result became 12,125 what is 12 1/8. The same result.