Is it true that x in base n is always larger than or equal to x in base n-1?

Question

I'm an elementary education major, but one of the problems in my elementary mathematics class asked to compare the values of numbers in different bases, without converting to base ten. One example was whether 32 in base 10 is larger than, equal to, or smaller than 32 in base 5. I realized that the larger base you go up, the more each place value can hold, so I used that to solve the problems, and those specific problems that I answered are correct, but it made me wonder if you can generalize this rule. If you have a number in base n, is it always larger than or equal to a number in base n-1? Take the case where the number is larger than or equal to n, and the place digits can hold higher numbers before overflowing in base n compared to base n-1, so x in base n will always be larger than x in base n-1. Take the case where x is smaller than n, and overflowing place values don't matter, so x in base n is equal to x in base n-1 (e.g. 6 in base 10 is equal to 6 in base 8). Or at least, this is true for non-negative x and positive n, I think.

Would this be a valid proof? Thank you!

Answer 1

A more rigorous proof would be the following. Suppose the digits of the numeral are $$a_ka_{k-1}\cdots a_2a_1a_0.$$ If we read this as a base-$n$ number, then it is equal to $$a_k\cdot n^k+a_{k-1}\cdot n^{k-1}+\cdots +a_2\cdot n^2+a_1\cdot n + a_0.$$ If we read it as a base-$(n-1)$ number, then it is equal to $$a_k\cdot (n-1)^k+a_{k-1}\cdot (n-1)^{k-1}+\cdots +a_2\cdot (n-1)^2+a_1\cdot (n-1) + a_0.$$ The first number is greater than or equal to the second because $n^\ell \geq (n-1)^\ell$ for every nonnegative integer $\ell$.

Answer 2

@lccu's answer proves this for positive integers of more than one digit. For a single digit, the base is irrelevant, as you noted.

But why stop at integers? Don't forget you can continue with digits after the equivalent of the decimal point. In this case you'll have negative powers of the base, which are now smaller for base $n+1$ than for base $n$.

Whatever the base, any digits after the point will represent a value in the interval $(0,1)$. So changing base can only change the value of the fractional part by $<1$.

On the other hand, any increase in the integer part must be $\geq 1$, and will therefore be bigger than any decrease in the fractional part.

So (skipping some simple steps) we end up with the following. Let $S$ be the string of digits (including the point if needed), let $A$ be its value in base $n$, and let $B$ be its value in base $n+1$. Then:

• If $S$ has two or more digits before the point, $A<B$.
• If $S$ has one digit before the point and no nonzero digits after, then $A=B$.
• If $S$ has one digit before the point and any nonzero digits after, then $A>B$. (The digit before the point is allowed to be $0$.)

All the above assumes that $S$ represents a positive number, and that the base is a positive number $>1$. If $S$ represents a negative number, then the inequalities are reversed.

What I've not worked out is what happens if you choose some funny base like $-\frac12$ (I'd suggest investigating base $-2$, aka negabinary, first.)