There is no solution in the REAL domain, but there are two in the COMPLEX domain (four if you consider the m-n symmetry). If you ask for solutions *limited* to the Real domain, you have to state it in the definition of the problem.
You mean m=n.[-1±√3.i]/2? Anyway, this would suggest that for either m or n, you could choose any you want for one of the variables, but the other variable would have a specific value, depending on the value of the one you chose. In this case, if I pick something random, like 6 for n, then : m = 6.[-1±√3.i]/2 m = 3.[-1±√3.i] m = -3+3.sqrt(3).i For other values: n = 1; m = -0.5 ± (√3/2)i n = 2; m = 1 ± (√3)i n = 3; m = -1.5 ± [(3√3)/2]i n = 4; m = -2 ± (2√3)i n = 5; m = -2.5 ± [(5√3)/2]i And so on. Do these values work? Well: If I put (1/2) + (1/[-1 ± (√3)i]) = 1/[(1/2) + [-1 ± (√3)i]) into symbolabs, it says "True". So, long story short, m and n are "dependent variables" to each other. You can choose any value for m, but the value of n depends on what you choose for m. Likewise, you can choose any value for n, but m depends on the value of n.
Thanks for indicating this. I was so worried about trying to find a fixed number of solutions that I overlooked the fact that there was only one equation, but two variables. Because of this, I had a labor of a time trying to prove that m = 1 magnitude units with 120 degrees heading and n = 1 magnitude units with 240 degrees heading was a solution along with n and m being allowed to switch values with each other.
My solution: 1/m+1/n=1/(m+n) LHS=(m+n)/mn=1/(m+n) (m+n)²=mn where m+n≠0,mn≠0 Let m+n=A, then m,n are the roots of x²-Ax+A²=0 Solving this quadratic x= (1±√3i)A/2 So m,n=k(1±√3i) for any non-zero real k
1/m + 1/n = 1/(m+n) n/(mn) + m/(mn) = 1/(m+n) (m+n)/(mn) = 1/(m+n) (m+n)² = mn m² + 2mn + n² = mn m² + mn + n² = 0 (m+n)² + m² + n² = 0 Since m and n can't be 0, there is no real solution. However, since we have one equation and two variables, there are infinite complex solutions.
Another solution: from m^2+mn+n^2 one calculates the delta of the equation, assuming that the variable is m or n.. we obtain n^2-4. 1.n^2= -3n^2 or -3m^2 if the variable is n.. in both cases the delta is minor or equal to zero and therefore there aren't solution ( neither m nor n can be zero)
After simplifying the equation we get.. m^2+mn+n^2=0. Now multiplying both sides of the equation by (m - n) we get, (m - n)(m^2+mn+n^2)= 0 × (m - n) =>m^3 - n^3 = 0 =>m^3 = n^3
Adding 1/m and 1/n gets us (m+n)/nm. So (m+n)/nm=1/(m+n). Cross multiplying gets us (m+n)^2=nm -> m^2+2nm+n^2=nm -> m^2+nm+n^2=0. Substitute values into the qudaratic formula such that (a,b,c)=(1,n,n^2). We get [-n±sqrt(n^2-4(1)(n^2))]/2. Since ∆
@@gagadaddy8713 Maybe?.. I don't know, because when I plugged in (a,b,c)=(1,n.n^2) when we had the equation m^2+mn+n^2=0 into the quadratic formula I got two solutions for m, hence two solutions for (m,n). Is there any way one can prove there are infinite complex solutions to this equation?
Here’s what I think is a (slightly) faster solution starting at (m+n)^2=mn. Taking the square root of both sides, we get m+n=sqrt(mn). Over the nonnegative real numbers, this is impossible by AM-GM unless m=n and m+n=0, so m=n=0, making the initial equation undefined. Because mn=(m+n)^2>=0, the only other case is that both m and n are negative, in which case -m and -n should be positive and satisfy the equation, which I’ve already shown is impossible. However, the approach in the video is better if you want to find the complex solutions.
There is much simpler. One realises that this equation is equivalent to ab = a+b = 1 with a=m/(n+m) and b = n/(n+m). The equivalence is obtained by multiplying the original equation by nm/(n+m). This obviously has no solution since a and b are smaller than 1 but of product equal to 1.
A different approach would be to think what we are actually doing by looking how this equation is written We add two fractions and get a fraction with a bigger denominator, so somehow we got a smaller number which contradicts with how addition works. Unless one denominator is negative (if they both are then we can just multiply by -1)? Let's check 1/m+1/(-n)=1/(m-n) 1/m-1/n=1/(m-n) (The variables are positive, I'm putting a minus in front to make it a negative, also it doesn't matter which variable is which) Now, if nm then 1/m=1/(m-n)+1/n 1/m+1/(n-m)=1/n we again add fractions with smaller denominators and get a fraction with a bigger one so a smaller number That's why there are no real solutions, those fractions are impossible to be added up that way
Here is the check of my answer; suppose m is the complex conjugate of n, m = (-1/2+(sqrt3)*i/2),n= (-1/2-(sqrt3)*i/2) m+n= -1 m*n=1 (i'll leave you to work the multiplication out.) go to the third line of my answer., 1/(m+n)=(m+n)/m*n LHS= 1/-1 = -1 RHS= -1/1 = -1 LHS= RHS = -1 No matter what rules of complex numbers you use the equation will always balance. m and n are also the reciprocals or each other since m*n=1.
This is one of the so-called "freshman's dreams". This clearly states that you cannot distribute powers, and you cannot apply the distributive property of negative or fractional numbers. The reason why you cannot distribute negative and fractional powers as their Taylor series (derivative polynomial chain) has an infinite number of terms. Also, there is a second freshman's dream of (x+y)^2=x^2+y^2. In this equation, 0 is the only solution, and if it were true for all numbers, the Mathsverse will collapse, as the Pythagorean Theorem will cause a paradox, as if a^2+b^2=c^2, a+b=c, and that means the hypotenuse of a right triangle must equal the sum of its 2 legs, which is impossible as the triangle will just be a straight line.
I used the discriminant to obtain that n, m must be complex or zero, but we already know from the original equation that it's not defined if m or n equals zero.
You can't multiply infinity by zero naively. More carefully looking with limits will give you a contradiction. For example, letting m,n be equal and go to zero will give the contradiction 1 = 4.
How have you obtained numerical values for m and n after the substitution m = -a if a is also an unknown? By the way, (m, n) = (-2, 1) is not a solution to 1/m + 1/n = 1/(m + n), as I'm going to prove you now: Left-hand side: 1/(-2) + 1/(1) = -1/2 + 1 = 1/2 Right-hand side: 1/(-2 + 1) = 1/(-1) = -1 Is 1/2 = -1? Obviously not. So (m, n) = (-2, 1) is not a solution, as I've stated earlier.
Very interesting. There"s no answer in R（real number field）. On the other hand, it has solutions in C (complex number field) and F3 (finite field Z/3Z) In F3, the solutions are (m,n)=(1,1),(-1,-1). In this video, R is the range. So you are right.
@@Nikioko Well, in case D(m)=R,D(n)=C (D denotes domain) there are infinite solutions. And no solutions in case D(m)=D(n)=R. Likewise, no solutions in case D(m)=D(n)=F5, but some solutions in case D(m)=F5,D(n)=F25.
@@Channel-gc3em What? The first paragraph is repeating what I said: There is no real solution for both m and being real. At least one of the two (as they are interchangeable, it doesn't matter which one) must be complex, and then you get infinite solutions. But I don't know what you are talking about in the second part of your answer.
m=0 n=0 1/n=error 1/m=error error+error=error (because you can't add words so it's an error) 1/m+n=error SIMPLE! ...or they could both be infinity 1/infinity=0 infinity+infinity=infinity
That's not how it works... You can't add two expressions where the denominator is zero. Plus, 1/infinity is not 0, it is undefined. The limit for y=1/x shows that when x approaches infinity, y approaches 0.
should I've said no REAL solution exists as theres a non real solution by the complex plane Unless it specify the m,n are real numbers but the video didnt show that Theres also some other answers that will be ruled out too like infinity and negative infinity for both m and n As if we took 1/x and went higher and higher (lower for negative)for x we will approach 0 So we can assume 1/infinity =0 But as i said it wont be taken as a solution and will be out i just like to mention it
The Solution is wrong! if either m or n would be equal to 0, than in the intial equation there would be term 1/0 multiple times. There is in fact no real solution
@jash21222 No. This is a diophantine equation. From the equation produced m^2+mn+n^2=0, and for whatever value you plug in for m or n, the other variable has a resulting complex solution. Where did you get m and n are integers? It was not mentioned in the video I think.
@jash21222 Yes, but the problem never stated that we needed integer solutions. You don't have to stick with the definition every time. It's a diophantine equation but we're looking for any solution.
m²+mn+n²=0 Take modulo m --> n is some multiple of m n=um, u is any integer (1) Now take modulo n --> m=vn, v is any integer (2) Divide (1) by (2) n/m=(u/v)(m/n) --> (n/m)²=u/v n/m=±sqrt(u/v) n=±m×sqrt(u/v) (m,n)={(k,±ksqrt(u/v) | (k,u,v) is any triple integer} Am I correct logically?
@@gagadaddy8713 I have already said that. 0 is the the only solution we can derive but since plugging it in means dividing by it then it cannot be a solution and there are no real soulutions.
That's why it's hard, its a diophantine equation. And sometimes you can solve an equation if it contains multiple unknowns, or find m in terms of n in this case.
There is no solution in the REAL domain, but there are two in the COMPLEX domain (four if you consider the m-n symmetry). If you ask for solutions *limited* to the Real domain, you have to state it in the definition of the problem.
Actually, there are infinitely many complex solutions
What are the two solutions.
@@arshhussainnaqvi9276 n = 1/2 i (sqrt(3) m + i m) and n = -1/2 i (sqrt(3) m - i m).
@@jomalderez thanks!
@@jomalderez can u also tell how u got them pls
Well, I would argue that there are complex solutions with m=n.[-1+sqrt(3).i]/2 and vice versa
You mean m=n.[-1±√3.i]/2?
Anyway, this would suggest that for either m or n, you could choose any you want for one of the variables, but the other variable would have a specific value, depending on the value of the one you chose.
In this case, if I pick something random, like 6 for n, then :
m = 6.[-1±√3.i]/2
m = 3.[-1±√3.i]
m = -3+3.sqrt(3).i
For other values:
n = 1; m = -0.5 ± (√3/2)i
n = 2; m = 1 ± (√3)i
n = 3; m = -1.5 ± [(3√3)/2]i
n = 4; m = -2 ± (2√3)i
n = 5; m = -2.5 ± [(5√3)/2]i
And so on.
Do these values work?
Well:
If I put (1/2) + (1/[-1 ± (√3)i]) = 1/[(1/2) + [-1 ± (√3)i]) into symbolabs, it says "True".
So, long story short, m and n are "dependent variables" to each other.
You can choose any value for m, but the value of n depends on what you choose for m. Likewise, you can choose any value for n, but m depends on the value of n.
Thanks for indicating this. I was so worried about trying to find a fixed number of solutions that I overlooked the fact that there was only one equation, but two variables. Because of this, I had a labor of a time trying to prove that m = 1 magnitude units with 120 degrees heading and n = 1 magnitude units with 240 degrees heading was a solution along with n and m being allowed to switch values with each other.
Since we have one equation and two variables, there are infinite complex solutions.
My solution:
1/m+1/n=1/(m+n)
LHS=(m+n)/mn=1/(m+n)
(m+n)²=mn where m+n≠0,mn≠0
Let m+n=A, then m,n are the roots of x²-Ax+A²=0
Solving this quadratic x=
(1±√3i)A/2
So m,n=k(1±√3i) for any non-zero real k
1/m + 1/n = 1/(m+n)
n/(mn) + m/(mn) = 1/(m+n)
(m+n)/(mn) = 1/(m+n)
(m+n)² = mn
m² + 2mn + n² = mn
m² + mn + n² = 0
(m+n)² + m² + n² = 0
Since m and n can't be 0, there is no real solution. However, since we have one equation and two variables, there are infinite complex solutions.
Another solution: from m^2+mn+n^2 one calculates the delta of the equation, assuming that the variable is m or n.. we obtain n^2-4. 1.n^2= -3n^2 or -3m^2 if the variable is n.. in both cases the delta is minor or equal to zero and therefore there aren't solution ( neither m nor n can be zero)
Большое спасибо!
That's what I did!
I assume that m 0 and n 0 and m -n. That restricts the solutions. 1) if m>0, n>0, 1/m > 1/(m+n). rejected. 2) same for m
After simplifying the equation we get.. m^2+mn+n^2=0.
Now multiplying both sides of the equation by (m - n) we get,
(m - n)(m^2+mn+n^2)= 0 × (m - n)
=>m^3 - n^3 = 0
=>m^3 = n^3
Adding 1/m and 1/n gets us (m+n)/nm. So (m+n)/nm=1/(m+n). Cross multiplying gets us (m+n)^2=nm -> m^2+2nm+n^2=nm -> m^2+nm+n^2=0. Substitute values into the qudaratic formula such that (a,b,c)=(1,n,n^2). We get [-n±sqrt(n^2-4(1)(n^2))]/2. Since ∆
There are two solutions in COMPLEX domain
m=1/2+√3i/2 & n=1/2-√3i/2 and
m=1/2-√3i/2 & n=1/2+√3i/2
No, this is a Diophantine equation, otherwise it is no meaning to have 2 variables in single equation.
@@gagadaddy8713 Maybe?.. I don't know, because when I plugged in (a,b,c)=(1,n.n^2) when we had the equation m^2+mn+n^2=0 into the quadratic formula I got two solutions for m, hence two solutions for (m,n). Is there any way one can prove there are infinite complex solutions to this equation?
@@gagadaddy8713 nvm there are infinitely many solutions in the complex plane. i realized my formula for m was based off of n
Here’s what I think is a (slightly) faster solution starting at (m+n)^2=mn. Taking the square root of both sides, we get m+n=sqrt(mn). Over the nonnegative real numbers, this is impossible by AM-GM unless m=n and m+n=0, so m=n=0, making the initial equation undefined. Because mn=(m+n)^2>=0, the only other case is that both m and n are negative, in which case -m and -n should be positive and satisfy the equation, which I’ve already shown is impossible. However, the approach in the video is better if you want to find the complex solutions.
Teori hitungan ini dan praktisnya sudah terbukti dalam ilmu kelistrikan tentang resistor paralel dan seri.
There is much simpler. One realises that this equation is equivalent to
ab = a+b = 1 with a=m/(n+m) and b = n/(n+m).
The equivalence is obtained by multiplying the original equation by nm/(n+m).
This obviously has no solution since a and b are smaller than 1 but of product equal to 1.
A different approach would be to think what we are actually doing by looking how this equation is written
We add two fractions and get a fraction with a bigger denominator, so somehow we got a smaller number which contradicts with how addition works.
Unless one denominator is negative (if they both are then we can just multiply by -1)?
Let's check
1/m+1/(-n)=1/(m-n)
1/m-1/n=1/(m-n)
(The variables are positive, I'm putting a minus in front to make it a negative, also it doesn't matter which variable is which)
Now, if nm then
1/m=1/(m-n)+1/n
1/m+1/(n-m)=1/n
we again add fractions with smaller denominators and get a fraction with a bigger one so a smaller number
That's why there are no real solutions, those fractions are impossible to be added up that way
No real solutions doesn't mean no solutions.
Here is the check of my answer;
suppose m is the complex conjugate of n,
m = (-1/2+(sqrt3)*i/2),n= (-1/2-(sqrt3)*i/2)
m+n= -1
m*n=1 (i'll leave you to work the multiplication out.)
go to the third line of my answer.,
1/(m+n)=(m+n)/m*n
LHS= 1/-1 = -1
RHS= -1/1 = -1
LHS= RHS = -1
No matter what rules of complex numbers you use the equation will always balance.
m and n are also the reciprocals or each other since m*n=1.
This is one of the so-called "freshman's dreams". This clearly states that you cannot distribute powers, and you cannot apply the distributive property of negative or fractional numbers. The reason why you cannot distribute negative and fractional powers as their Taylor series (derivative polynomial chain) has an infinite number of terms. Also, there is a second freshman's dream of (x+y)^2=x^2+y^2. In this equation, 0 is the only solution, and if it were true for all numbers, the Mathsverse will collapse, as the Pythagorean Theorem will cause a paradox, as if a^2+b^2=c^2, a+b=c, and that means the hypotenuse of a right triangle must equal the sum of its 2 legs, which is impossible as the triangle will just be a straight line.
I call a + b = c the sophomore's dream.
Set x = n/m to start. x = (-1+/- sqrt3i)/2
before anything, you should have stated upfront n and m are not 0 otherwise the fractions 1/n and 1/m are undefined
4:02 Set x=m/n to get x^2+x+1=0 which has no real solution as the discriminant is negative.
Reminded me when the school Dean told off our math teacher for giving us a ‘No Solution’ exam question😂
I used the discriminant to obtain that n, m must be complex or zero, but we already know from the original equation that it's not defined if m or n equals zero.
0,0 is also solution yielding infinity!!!
You can't multiply infinity by zero naively. More carefully looking with limits will give you a contradiction. For example, letting m,n be equal and go to zero will give the contradiction 1 = 4.
Add two numbers and get a smaller one? Not in the real domain
I came up with n = (m/2)(-1±isqrt(3))
But ♾️ = ♾️. Isn't that a solution?
Ha Ha haha 🤣
You can have different degrees of infinity,rule that out.
Division by zero is undefined; it is not infinity
لا ليس حلا في علم الرياضيات
Technically, infinities are not equal to each other.
Also dividing by 0 may as well give you a negative if you look at the graph of y=1/x
(m/n) = (-1 + i sqrt 3)/2 and the complex conjugate. Solutions are imaginary.
there are still complex solutions
That was very cool, Math Window!
The main condition is m=n and 1/m + 1/m = 1/n + 1/n = 1/m + 1/n = 1/0,5m = 1/0,5n.
nice explained, important example, not every equation leads to a solution
💖💖💖well teacher
he is not a good teacher.
It doesn' t specify that solution is in the real numbers.
This is just a "Freshman's dream" variation.
Let m=-a then after solving m equals to -2 and n equals to 1
How have you obtained numerical values for m and n after the substitution m = -a if a is also an unknown? By the way, (m, n) = (-2, 1) is not a solution to 1/m + 1/n = 1/(m + n), as I'm going to prove you now:
Left-hand side:
1/(-2) + 1/(1) = -1/2 + 1 = 1/2
Right-hand side:
1/(-2 + 1) = 1/(-1) = -1
Is 1/2 = -1?
Obviously not. So (m, n) = (-2, 1) is not a solution, as I've stated earlier.
@@diegocabrales thanks for correcting
Большое спасибо!
Very interesting.
There"s no answer in R（real number field）.
On the other hand, it has solutions in C (complex number field) and F3 (finite field Z/3Z)
In F3, the solutions are (m,n)=(1,1),(-1,-1).
In this video, R is the range. So you are right.
(And some solutions in F7,F13,F19,etc)
Since we have one equation and two variables, there are infinite complex solutions. Choose a real value for m, and you get a complex value for n.
@@Nikioko
Well, in case D(m)=R,D(n)=C (D denotes domain) there are infinite solutions.
And no solutions in case D(m)=D(n)=R.
Likewise, no solutions in case D(m)=D(n)=F5, but some solutions in case D(m)=F5,D(n)=F25.
no solutions → no solution
@@Channel-gc3em What? The first paragraph is repeating what I said: There is no real solution for both m and being real. At least one of the two (as they are interchangeable, it doesn't matter which one) must be complex, and then you get infinite solutions.
But I don't know what you are talking about in the second part of your answer.
m=0
n=0
1/n=error
1/m=error
error+error=error (because you can't add words so it's an error)
1/m+n=error
SIMPLE!
...or they could both be infinity
1/infinity=0
infinity+infinity=infinity
That's not how it works... You can't add two expressions where the denominator is zero. Plus, 1/infinity is not 0, it is undefined. The limit for y=1/x shows that when x approaches infinity, y approaches 0.
clever thinking tho
I could solve this problem in less than a minute 🤩😍
From 4:30 onward, it's waste of time and effort. It's already very clear m or n has to be 0, and obviously none of them can be 0.
should I've said no REAL solution exists as theres a non real solution by the complex plane
Unless it specify the m,n are real numbers but the video didnt show that
Theres also some other answers that will be ruled out too like
infinity and negative infinity for both m and n
As if we took
1/x and went higher and higher (lower for negative)for x we will approach 0
So we can assume 1/infinity =0
But as i said it wont be taken as a solution and will be out i just like to mention it
M≠0
N≠0
The Solution is wrong!
if either m or n would be equal to 0, than in the intial equation there would be term 1/0 multiple times. There is in fact no real solution
@jash21222 The video said "No solution" he's saying "No real solution".
@jash21222 No. This is a diophantine equation. From the equation produced m^2+mn+n^2=0, and for whatever value you plug in for m or n, the other variable has a resulting complex solution. Where did you get m and n are integers? It was not mentioned in the video I think.
@jash21222 Yes, but the problem never stated that we needed integer solutions. You don't have to stick with the definition every time. It's a diophantine equation but we're looking for any solution.
m/n =[(-1±i√3)/2]
Спасибо.
There is a solution but it is imaginary solution in algebra moonquake is correct one
Valde facile! m= --n/2 + i n√3/2, --n/2 ---i n√3/2( imaginarias)
Beautiful :)
m²+mn+n²=0
Take modulo m
--> n is some multiple of m
n=um, u is any integer (1)
Now take modulo n
--> m=vn, v is any integer (2)
Divide (1) by (2)
n/m=(u/v)(m/n) --> (n/m)²=u/v
n/m=±sqrt(u/v)
n=±m×sqrt(u/v)
(m,n)={(k,±ksqrt(u/v) | (k,u,v) is any triple integer}
Am I correct logically?
هذا الحل ليس صحيحا رياضيا مطلقا
لانه ممنوع التقسيم على صفر 😮😂
He has said that at the end of the video
We exclude 0 from our solutions, and since only 0 is our solution then there are no real solutions
@@_rd_5043 Zero can't be a solution as 1/m and 1/n is undefined.
@@gagadaddy8713 I have already said that.
0 is the the only solution we can derive but since plugging it in means dividing by it then it cannot be a solution and there are no real soulutions.
k/0=±∞
Not indeterminated,
±∞±∞=±∞
Ok ∞-∞ it's indeterminated
Bruh you cant “solve” an equation if it contains multiple unknowns, maybe if you had a system of equations. What is this video lol.
That's why it's hard, its a diophantine equation. And sometimes you can solve an equation if it contains multiple unknowns, or find m in terms of n in this case.
looooooooooooooooool
What is this. Kindergarden math
هذا الفيديو لا يستحق المشاهدة
Done so boringly, sorry.