>>11437763Let Pn be the statement that
sum [k = 0 to n] (a*10^-k) * b
=sum [k = 0 to n] (a*b*10^-k).
P1 is the statement that
(a + 0.1*a)*b = a*b + 0.1*b
By distributivity or whatever, this is true for all a, b. Thus P1 is true. Assume Pn is true, that is
sum [k = 0 to n] (a*10^-k) * b
=sum [k = 0 to n] (a*b*10^-k). Note that
sum [k = 0 to n+1] (a*10^-k) * b
=sum [k = 0 to n] (a*b*10^-k)+a*b*10^-(n+1)
=sum [k = 0 to n+1] (a*b*10^-k).
Thus Pn+1 is true, and therefore Pn is true for all n.
Thus Pn is true as n approaches infinity.
