KCC's Quizzes AQQ272 about a division giving its double

1/2.

Indeed:  a/a  = 2a,  implies  a is not 0 (we can't divide by 0), leaving:  a  = 2a^2,  or 1=2a.

Under modular arithmetic  (residue modulo) there is a unique solution to  1 = 2x  (mod  y)   -- Gauss' notation
 that is:                                                                                                            x = (y+1)/2 

So, since there is an infinite number of odd numbers, there is an infinite number of solutions, one for each (mod  y), y being a odd integer.

y=3,  x = (3+1)/2  = 2.  Indeed 2x = 4 =  1 (mod 3)   since 4 residue modulo 3  get a rest of 1.
y=5,  x = (5+1)/2  = 3. Indeed  2x = 6  = 1 (mod 5), 

etc. They all satisfy                  2x = 1  (mod y)

Under modular arithmetic  (residue modulo) there is a unique solution to  1 = 2x  (mod  y)   -- Gauss' notation
 that is:                                                                                                            x = (y+1)/2 

So, since there is an infinite number of odd numbers, there is an infinite number of solutions, one for each (mod  y), y being a odd integer.

y=3,  x = (3+1)/2  = 2.  Indeed 2x = 4 =  1 (mod 3)   since 4 residue modulo 3  get a rest of 1.
y=5,  x = (5+1)/2  = 3. Indeed  2x = 6  = 1 (mod 5), 

etc. They all satisfy                  2x = 1  (mod y)

Wow, I learn something... I was not aware about the modular arithmetic method...

Additionnal note: when y becomes larger, y+1  is approximatively equals to y  and  x/y  tends toward the "continuous" solution, that is, 1/2.  A little bit like quantum mech. (modular)  tends toward classical mechanic.

I am really impressed (i.e. Schrondinger is part of our quizzes now..)