A nice equation. You should know how surprising the answer is!
Vložit
- čas přidán 9. 05. 2023
- A fantastic equation which looks so hard but very easy to solve❗ I gave the students this problem as an exercise but no one solved correctly!
The answer is really surprising!
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
My solution:
1/m + 1/n = 1/(m+n)
Well, I would argue that there are complex solutions with m=n.[-1+sqrt(3).i]/2 and vice versa
After simplifying the equation we get.. m^2+mn+n^2=0.
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<0, n<0 because we can multiply both sides by -1 and get #1. Now only possible solutions are m<0, n>0, or m>0, n<0. Both numbers must have different signs (and m <> -n). Take the case where m>0, n<0, rewrite the equation as: 1/m - 1/|n| = 1/(m-|n|). 1) if m>|n|, 1/m < 1(m-|n|). rejected. 2) if m<|n|, multiply both sides by -1 and get #1.rejected. Therefore there is no real solution. The symmetry of the equation restricts its solution space.
No real solutions doesn't mean no solutions.
Set x = n/m to start. x = (-1+/- sqrt3i)/2
There is much simpler. One realises that this equation is equivalent to
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 ∆<0, there are no real roots, only complex solutions. Simplifying gets us the solution for m: m=-n/2±i*sqrt(3)n/2.
Teori hitungan ini dan praktisnya sudah terbukti dalam ilmu kelistrikan tentang resistor paralel dan seri.
Reminded me when the school Dean told off our math teacher for giving us a ‘No Solution’ exam question😂
4:02
M≠0
Add two numbers and get a smaller one? Not in the real domain
0,0 is also solution yielding infinity!!!
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.
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)
There are two solutions in COMPLEX domain
before anything, you should have stated upfront n and m are not 0 otherwise the fractions 1/n and 1/m are undefined