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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x} \iff b \in \Z_{> 1}$

where $\floor x$ denotes the floor of $x$.


Proof 1

Necessary Condition

Let:

$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x}$

Let $x = b$.

Then:

$\floor {\log_b b} = \floor {\log_b \floor b}$
\(\ds \floor {\log_b b}\) \(=\) \(\ds \floor {\log_b \floor b}\)
\(\ds \leadsto \ \ \) \(\ds \floor 1\) \(=\) \(\ds \floor {\log_b \floor b}\) Logarithm of Base
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \floor {\log_b \floor b}\) Real Number is Integer iff equals Floor


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

\(\ds \floor b\) \(<\) \(\ds b\) Floor of Non-Integer
\(\ds \leadsto \ \ \) \(\ds \log_b \floor b\) \(<\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \floor {\log_b \floor b}\) \(<\) \(\ds \floor {\log_b b}\)
\(\ds \leadsto \ \ \) \(\ds \floor {\log_b \floor b}\) \(\ne\) \(\ds \floor {\log_b b}\) which contradicts $\floor {\log_b b} = \floor {\log_b \floor b}$

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 $\floor {\log_b x} = n$.


Then:

\(\ds n\) \(=\) \(\, \ds \floor {\log_b x} \, \) \(\ds \)
\(\ds \leadstoandfrom \ \ \) \(\ds n\) \(\le\) \(\, \ds \log_b x \, \) \(\, \ds < \, \) \(\ds n + 1\) Integer equals Floor iff Number between Integer and One More
\(\ds \leadstoandfrom \ \ \) \(\ds b^n\) \(\le\) \(\, \ds x \, \) \(\, \ds < \, \) \(\ds b^{n + 1}\) Power Function is Strictly Increasing over Positive Reals
\(\ds \leadstoandfrom \ \ \) \(\ds b^n\) \(\le\) \(\, \ds \floor x \, \) \(\, \ds < \, \) \(\ds b^{n + 1}\) Number not less than Integer iff Floor not less than Integer
\(\ds \leadstoandfrom \ \ \) \(\ds n\) \(\le\) \(\, \ds \log_b \floor x \, \) \(\, \ds < \, \) \(\ds n + 1\) Power Function is Strictly Increasing over Positive Reals
\(\ds \leadstoandfrom \ \ \) \(\ds n\) \(=\) \(\, \ds \floor {\log_b \floor x} \, \) \(\ds \) Integer equals Floor iff Number between Integer and One More
\(\ds \leadstoandfrom \ \ \) \(\ds \floor {\log_b x}\) \(=\) \(\, \ds \floor {\log_b \floor x} \, \) \(\ds \) Definition of $n$

$\blacksquare$


Proof 2

We have that:

Logarithm is Strictly Increasing

and:

Real 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:

$x \in \Z \iff b \in \Z$

Thus by McEliece's Theorem (Integer Functions):

$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x} \iff b \in \Z_{>1}$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Conditions_for_Floor_of_Log_base_b_of_x_to_equal_Floor_of_Log_base_b_of_Floor_of_x&oldid=517614"