rigorous definition of the logarithm

In this entry, we shall construct the logarithm as a Dedekind cutand then demonstrate some of its basic properties. All that isrequired in the way of background material are the properties ofinteger powers of real numbers.

Theorem 1.

Suppose that $a\mathrm{,}b\mathrm{,}c\mathrm{,}d$ are positive integers such that$a\mathrm{/}b\mathrm{=}c\mathrm{/}d$ and that $x\mathrm{>}\mathrm{0}$ and $y\mathrm{>}\mathrm{0}$ are real numbers.Then $x^a\mathrm{\le }{y}^b$ if and only if $x^c\mathrm{\le }{y}^d$.

Proof.

Cross multiplying, the condition $a/b=c/d$ is equivalent to $ad=bc$.By elementary properties of powers, $x^a\le y^b$ if and only if$x^{ad}\le y^{bd}$. Likewise, $x^c\le x^d$ if and only if $x^{bc}\le y^{bd}$ which, since $bc=ad$, is equivalent to $x^{ad}\le y^{bd}$.Hence, $x^a\le y^b$ if and only if $x^c\le x^d$.∎

Theorem 2.

Suppose that $a\mathrm{,}b\mathrm{,}c\mathrm{,}d$ are positive integers such that$a\mathrm{/}b\mathrm{\le }c\mathrm{/}d$ and that $x\mathrm{>}\mathrm{1}$ and $y\mathrm{>}\mathrm{0}$ are real numbers.If $x^c\mathrm{\le }{y}^d$ then $x^a\mathrm{\le }{y}^b$.

Proof.

Since we assumed that $b>0$, we have that $x^c\le y^d$ is equivalentto $x^{bc}\le y^{bd}$. Likewise, since $d>0$, we have that $x^a\le y^b$ is equivalent to $x^{ad}\le y^{bd}$. Cross-multiplying, $a/b\le c/d$is equivalent to $ad\le bc$. Since $x>1$, we have $x^{ad}\le x^{bc}$.Combining the above statements, we conclude that $x^c\le y^d$ implies$x^a\le y^b$.∎

Theorem 3.

Suppose that $a\mathrm{,}b\mathrm{,}c\mathrm{,}d$ are positive integers such that$a\mathrm{/}b\mathrm{>}c\mathrm{/}d$ and that $x\mathrm{>}\mathrm{1}$ and $y\mathrm{>}\mathrm{0}$ are real numbers.If $x^a\mathrm{>}{y}^b$ then $x^c\mathrm{>}{y}^d$.

Proof.

Since we assumed that $b>0$, we have that $x^c>y^d$ is equivalentto $x^{bc}>y^{bd}$. Likewise, since $d>0$, we have that $x^a>y^b$ is equivalent to $x^{ad}>y^{bd}$. Cross-multiplying, $a/b>c/d$is equivalent to $ad>bc$. Since $x>1$, we have $x^{ad}>x^{bc}$.Combining the above statements, we conclude that $x^c>y^d$ implies$x^a>y^b$.∎

Theorem 4.

Let $x\mathrm{>}\mathrm{1}$ and $y$ be real numbers.Then there exists an integer $n$ such that $x^n\mathrm{>}y$.

Proof.

Write $x=1+h$. Then we have ${(1+h)}^n\ge 1+nh$ for allpositive integers $n$. This fact is easily proved by induction.When $n=1$, it reduces to the triviality $1+h\ge h$. If${(1+h)}^n\ge 1+nh$, then

$${(1+h)}^{n+1}=(1+h){(1+h)}^n\ge (1+h)(1+nh)=1+(n+1)h+n{h}^2\ge 1+(n+1)h.$$

By the Archimedean property, there exists an integer $n$ suchthat $1+nh>y$, so $x^n>y$.∎

Theorem 5.

Let $x\mathrm{>}\mathrm{1}$ and $y$ be real numbers.Then the pair of sets $\mathrm{(}L\mathrm{,}U\mathrm{)}$ where

$L$$=\{r\in \mathbb{Q}\mid (\exists a,b\in \mathbb{Z})\mathit{\hspace{1em}}b>0\wedge r=a/b\wedge x^a\le y^b\}$(1)

$U$$=\{r\in \mathbb{Q}\mid (\exists a,b\in \mathbb{Z})\mathit{\hspace{1em}}b>0\wedge r=a/b\wedge x^a>y^b\}$(2)

forms a Dedekind cut.

Proof.

Let $r$ be any rational number. Then we have $r=a/b$ for some integers$a$ and $b$ such that $b>0$. The possibilities $x^a\le y^b$ and$x^a>y^b$ are exhaustive so $r$ must belong to at least one of $U$ and$L$. By theorem 1, it cannot belong to both. By theorem 2, if $r\in L$and $s\le r$, then $s\in L$ as well. By theorem 3, if $r\in U$and $s>r$, then $s\in U$ as well. By theorem 4, neither $L$ nor $U$are empty. Hence, $(L,U)$ is a Dedekind cut and defines areal number.∎

Definition 1.

Suppose $x\mathrm{>}\mathrm{1}$ and $y\mathrm{>}\mathrm{0}$ are real numbers. Then,we define ${\log }_x\mathit{}y$ to be the real number defined by the cut $\mathrm{(}L\mathrm{,}U\mathrm{)}$ ofthe above theorem.