It's kind of obvious... For example, three-digit numbers are between 100 and 1000.
If v(n) is defined as , then
.
To Prove:
For all n in the Natural Numbers, v(n) = k if and only if 10^(k-1) <= n < 10^k
v(n) is the number of digits of n with respect to base 10....so for example, v(d) = 1 for all digits d, and v(10) = 2.
I can't seem to figure out how to work this proof. Any help would be appreciated!
Why in the world would you want to use that?
Perhaps the simplest way to prove "v(n)= k if and only if " is by induction on k.
If k= 1, the statement becomes "v(n)= 1 if and only if .
Given that the statement for k= K is true, can you prove it for k= K+1?