At least one of the numbers is less than 1.
A*b = a + b - 1
a*b - a = b - 1 (subtract a from both sides)
a(b-1) = b-1 (factor out a)
a = (b-1)/(b-1) (divide by b-1)
a = 1
The problem is symmetrical, so both numbers must be 1.
1*1 = 1 + 1 - 1
It is interesting that (b-1)/(b-1) becomes an indeterminate form when b=1, yet the answer still holds.
A*b = a + b - 1
a*b - a = b - 1 (subtract a from both sides)
a(b-1) = b-1 (factor out a)
a = (b-1)/(b-1) (divide by b-1)
a = 1
The problem is symmetrical, so both numbers must be 1.
1*1 = 1 + 1 - 1
It is interesting that (b-1)/(b-1) becomes an indeterminate form when b=1, yet the answer still holds.
