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Homerigorous definition of the logarithm

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# rigorous definition of the logarithm

In this entry, we shall construct the logarithm as a Dedekind cut and then demonstrate some of its basic properties. All that is required in the way of background material are the properties of integer powers of real numbers.

###### Theorem 1.

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

###### Proof.

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

###### Theorem 2.

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

###### Proof.

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

###### Theorem 3.

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

###### Proof.

Since we assumed that $b>0$, we have that $x^{c}>y^{d}$ is equivalent to $x^{{bc}}>y^{{bd}}$. 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 $x>1$ and $y$ be real numbers. Then there exists an integer $n$ such that $x^{n}>y$.

###### Proof.

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

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

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

###### Theorem 5.

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

$\displaystyle L$ | $\displaystyle=\{r\in\mathbb{Q}\mid(\exists a,b\in\mathbb{Z})\quad b>0~{}\land~% {}r=a/b~{}\land~{}x^{a}\leq y^{b}\}$ | (1) | ||

$\displaystyle U$ | $\displaystyle=\{r\in\mathbb{Q}\mid(\exists a,b\in\mathbb{Z})\quad b>0~{}\land~% {}r=a/b~{}\land~{}x^{a}>y^{b}\}$ | (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 $x^{a}\leq y^{b}$ 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 $s\leq r$, 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 a real number. ∎

###### Definition 1.

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

## Mathematics Subject Classification

26A06*no label found*26A09

*no label found*26-00

*no label found*

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