Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x/Proof 1

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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

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 $n$

$\blacksquare$


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