# Thread: Base-Ten Representaion of Integers proof

1. ## Base-Ten Representaion of Integers proof

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!

2. It's kind of obvious... For example, three-digit numbers are between 100 and 1000.

If v(n) is defined as $\displaystyle \lfloor\log_{10}n\rfloor+1$, then

$\displaystyle \lfloor\log n\rfloor+1=k\Leftrightarrow$
$\displaystyle \lfloor\log n\rfloor=k-1\Leftrightarrow$
$\displaystyle k-1\le\log n<k\Leftrightarrow$
$\displaystyle 10^{k-1}\le n<10^k$.

3. Hint - What is the smallest and larget 'k' digit number (in decimal representation)?

4. Can we prove this using the fact that n < 10^n ? Instead of the log10n ? The procedure seems to be correct, but I need to use the face that if n is in the Naturals, then n < 10^n

Thanks for all the help!

5. Why in the world would you want to use that?

Perhaps the simplest way to prove "v(n)= k if and only if $\displaystyle 10^{k-1}\le n< 10^k$" is by induction on k.

If k= 1, the statement becomes "v(n)= 1 if and only if $\displaystyle 1\le n< 10$.

Given that the statement for k= K is true, can you prove it for k= K+1?