# Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x

Jump to navigation Jump to search

## Theorem

Let $b \in \R$ be a real number.

$\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor \iff b \in \Z_{> 1}$

where $\left\lfloor{x}\right\rfloor$ denotes the floor of $x$.

## Proof 1

### Necessary Condition

Let:

$\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor$

Let $x = b$.

Then:

$\left\lfloor{\log_b b}\right\rfloor = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$
 $\displaystyle \left\lfloor{\log_b b}\right\rfloor$ $=$ $\displaystyle \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$ $\displaystyle \leadsto \ \$ $\displaystyle \left\lfloor{1}\right\rfloor$ $=$ $\displaystyle \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$ Logarithm of Base $\displaystyle \leadsto \ \$ $\displaystyle 1$ $=$ $\displaystyle \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$ Real Number is Integer iff equals Floor

Aiming for a contradiction, suppose $b \notin \Z$.

 $\displaystyle \left\lfloor{b}\right\rfloor$ $<$ $\displaystyle b$ Floor of Non-Integer $\displaystyle \leadsto \ \$ $\displaystyle \log_b \left\lfloor{b}\right\rfloor$ $<$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$ $<$ $\displaystyle \left\lfloor{\log_b b}\right\rfloor$ $\displaystyle \leadsto \ \$ $\displaystyle \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$ $\ne$ $\displaystyle \left\lfloor{\log_b b}\right\rfloor$ which contradicts $\left\lfloor{\log_b b}\right\rfloor = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$

Thus by Proof by Contradiction:

$b \in \Z$

But for $\log_b$ to be defined, $b > 0$ and $b \ne 1$.

Hence:

$b \in \Z_{> 1}$

$\Box$

### Sufficient Condition

Let $b \in \Z_{> 1}$.

Let $\left\lfloor{\log_b x}\right\rfloor = n$.

Then:

 $\displaystyle n$ $=$ $\, \displaystyle \left\lfloor{\log_b x}\right\rfloor \,$ $\displaystyle$ $\displaystyle \iff \ \$ $\displaystyle n$ $\le$ $\, \displaystyle \log_b x \,$ $\, \displaystyle <\,$ $\displaystyle n + 1$ Integer equals Floor iff Number between Integer and One More $\displaystyle \iff \ \$ $\displaystyle b^n$ $\le$ $\, \displaystyle x \,$ $\, \displaystyle <\,$ $\displaystyle b^{n + 1}$ Power Function is Strictly Increasing over Positive Reals $\displaystyle \iff \ \$ $\displaystyle b^n$ $\le$ $\, \displaystyle \left \lfloor {x}\right \rfloor \,$ $\, \displaystyle <\,$ $\displaystyle b^{n + 1}$ Number not less than Integer iff Floor not less than Integer $\displaystyle \iff \ \$ $\displaystyle n$ $\le$ $\, \displaystyle \log_b \left\lfloor{x}\right\rfloor \,$ $\, \displaystyle <\,$ $\displaystyle n + 1$ Power Function is Strictly Increasing over Positive Reals $\displaystyle \iff \ \$ $\displaystyle n$ $=$ $\, \displaystyle \left \lfloor {\log_b \left\lfloor{x}\right\rfloor}\right \rfloor \,$ $\displaystyle$ Integer equals Floor iff Number between Integer and One More $\displaystyle \iff \ \$ $\displaystyle \left\lfloor{\log_b x}\right\rfloor$ $=$ $\, \displaystyle \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor \,$ $\displaystyle$ Definition of $n$

$\blacksquare$

## Proof 2

We have that:

Logarithm is Strictly Increasing

and:

Natural Logarithm Function is Continuous

Suppose that $\log_b x \in \Z$: let $\log_b x = n$, say.

Then:

$x = b^n$

It follows that $x \in \Z \iff b \in \Z$.

$\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor \iff b \in \Z_{> 1}$

$\blacksquare$