The sum of two numbers is 1.
The sum of the reciprocals of the numbers is 1.
Find the sum of the squares of the numbers (without finding the numbers themselves.)
What do you notice?
Let the two numbers be x and y
Then, x+y =1, 1/x+1/y =1 => x+y=1, (x+ y)/(x*y) = 1
=> x+y=1 and xy=1/1=1
=> x^2+y^2=(x+y)^2-2xy=1^2-2*1=-1
So, the sum of squares of the numbers is -1
Now, if each of x and y were real, their sum of squares would be positive.
Since it is negative, it follows that at least one of x and y is not real.
If possible, let one of x any y be real and the other not real.
Then, their product would not be a real number. Contradiction, since xy is 1, a real number.
THUS WE NOTICE THAT, each of the numbers CANNOT be a real number.
*Of course, this finds active support in the solution of simultaneous equations, e.g., (x, y)=( (1+v3i)/2, 1-v3i)/2), ((1-v3i)/2, (1+v3i)/2)
Edited on March 22, 2022, 7:34 am
Edited on March 22, 2022, 7:36 am
Edited on March 22, 2022, 7:38 am
Puzzle Solution
