A Mathematical Diversion

T: If c = a + b and x = a/c and y = b/c , then x + y = 1.

Proof:

1) if a = 0 and b = 0, then c = 0 but x = a/c and y = b/c give divide by zero errors
therefore x + y gives divide by zero errors.

2) if a = 0 and b = 1, then c = 1 and x = a/c = 0 and y = b/c = 1.
therefore x + y = 1

3) if a = 0 and b = 2, then c = 2 and x = a/c = 0 and y = b/c = 1
therefore x + y = 1

...

4) if a = 0 and b = infinity, then c = infinity and x = a/c = 0 and y = b / c = infinity / infinity = 1
therefore x + y = 1

5) if a = 1 and b = 0, then the answers are like 2 above.

6) if a = 1 and b = 1000, then c = 1001 and x = a/c = 1/1001 and y = b / c = 1000/1001
therefore x + y = (1 + 1001) / 1001 = 1001 / 1001 = 1

7) if a = 1 and b = infinity, then c = 1 + infinity and x = a/c = 1/infinity ~ 0 and y = b/c = infinity / (infinity + 1) ~ 1
therefore x + y = 1