Yet Another Definition of Euler's Number

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1. The Start

This note begin with the question: can we use the properties of $l n$ $l n$ to obtain the tangent-line inequality algebraically? (Note that the tangent-line inequality means the inequality $l n (𝑥) \leq 𝑥 - 1$ .) Here is what I tried.

First, let $𝐼 𝑛 = (𝑘 2 𝑛, 𝑘 2 𝑛 + 1]$ , where $𝑛 \in 𝑁$ , $𝑘 \in [1, 2)$ . Hence we know that this can cover $(1, + \infty)$ . In fact, you can check that the same can be made for $(0, 1)$ $(0, 1)$ , which is left to the reader. We next show that if the inequality holds on $𝐼 𝑛$ $𝐼 𝑛$ , then it holds on $𝐼 𝑛 + 1$ $𝐼 𝑛 + 1$ , which is basically an induction over intervals rather than integers. Here, our inductive hypothesis is $l n (𝑥) \leq 𝑥 - 1, \forall 𝑥 \in 𝐼 𝑛$ . Hence, we have $l n (2 𝑥) = l n (2) + l n (𝑥) \leq l n (2) + 𝑥 - 1, \forall 𝑥 \in 𝐼 𝑛$ . Since $𝐼 𝑛$ $𝐼 𝑛$ covers $(1, + \infty)$ , $𝑥 > 1 > l n (2)$ . Thus, $l n (2 𝑥) < 𝑥 + 𝑥 - 1 = 2 𝑥 - 1$ . Since $𝑥 \in 𝐼 𝑛$ , we obtain $2 𝑥 \in 𝐼 𝑛 + 1$ . Hence, $l n (𝑥) < 𝑥 - 1, \forall 𝑥 \in 𝐼 𝑛 + 1$ .

The problem is the proof of the inequality on $𝐼 0$ $𝐼 0$ , which is $(1, 4)$ $(1, 4)$ .

But you will soon find it as difficult as the orignal case, which I will explain later.

2. The Climax

I thought at first that the following theorem and proof would help.

Theorem 2.1. $l n (𝑘) < 𝑘, \forall 𝑘 > 0$ $l n (𝑘) < 𝑘, \forall 𝑘 > 0$

Proof. We first prove that $l n (𝑘) < 𝑘, \forall 𝑘 > 0$ . Since $𝑒 > 2$ , you can verify that $l n (𝑘) < l o g 2 (𝑘)$ . Thus, it suffices to show that $l o g 2 (𝑘) < 𝑘$ , which can be written as $𝑘 < 2 𝑘$ . We show it by induction.

$𝑘 \in (0, 1]$ Hence, $l o g 2 (𝑘) \leq 0 < 𝑘$ .
$l o g 2 (𝑘) < 𝑘$ ,also $𝑘 < 2 𝑘 \land 𝑘 > 1$ By $𝑘 > 1$ we have $𝑘 + 1 < 2 𝑘 < 2 \cdot 2 𝑘 = 2 𝑘 + 1$ .

Hence, $l n 𝑘 < 𝑘, \forall 𝑘 > 0$ .

□

The problem is very obvious. Here the inequality still holds for $𝑦 = l o g 2 (𝑥)$ . But in our case, the inequality is false for any other logarithm. The main problem is that this inequality depends on the property of $𝑒$ $𝑒$ , so that we cannot prove it unless we use the definition of $𝑒$ $𝑒$ , which makes the proof non-algebraic. Hence, our goal is impossible. Yet there is another question. Since this inequality is deeply dependent on the definition of $𝑒$ $𝑒$ , is it possible to define $𝑒$ $𝑒$ with this inequality? The answer is yes.

3. The End

This theorem and the following proof gives you exactly how to define $𝑒$ $𝑒$ with our inequality.

Theorem 3.1. There exists a unique smooth function $𝑓 : 𝑅 + \to 𝑅$ with the following property:
$𝑓 (𝑥 𝑦) = 𝑓 (𝑥) + 𝑓 (𝑦)$ ( $𝑥, 𝑦 > 0$ ).
$𝑓 (𝑥) \leq 𝑥 - 1$ ; equality holds if and only if $𝑥 = 1$ .
We define $𝑒$ $𝑒$ to be a positive real number such that $𝑓 (𝑒) = 1$ , and this definition is equivalent to the limit version.

Proof. We first show that $𝑓$ $𝑓$ is unique. Since $𝑓 (1) = 𝑓 (1 \cdot 1) = 2 𝑓 (1)$ , we obtain $𝑓 (1) = 0$ . Apply $𝜕 𝑦$ $𝜕 𝑦$ to both sides of $𝑓 (𝑥 𝑦) = 𝑓 (𝑥) + 𝑓 (𝑦)$ , we have $𝑥 𝑓' (𝑥 𝑦) = 𝑓' (𝑦)$ . Let $𝑦 = 1$ , we obtain $𝑥 𝑓' (𝑥) = 𝑓' (1)$ . Hence, $𝑓' (𝑥) = 𝐶 𝑥$ , where $𝐶 = 𝑓' (1)$ . Since $𝑓 (𝑥) \leq 𝑥 - 1$ , we let $𝑔 (𝑥) ≔ 𝑓 (𝑥) - (𝑥 - 1)$ . Because the equality holds if and only if $𝑥 = 1$ , we know that $𝑔$ $𝑔$ has the maximum value at $1$ $1$ , Hence $𝑔' (1) = 0$ . It follows that $𝑔' (𝑥) = 𝑓' (𝑥) - 1$ . Substitute this into the previous expression to obtain $𝑓' (1) = 1$ . Hence, $𝑓' (𝑥) = 1 𝑥$ . Thus, we obtain

𝑓 (𝑥) = \int 𝑥 1 1 𝑥 d 𝑥 .

𝑓 (𝑥) = \int 𝑥 1 1 𝑥 d 𝑥 .

𝑓 (𝑥) = \int 𝑥 1 1 𝑥 d 𝑥 .

Hence, $𝑓$ $𝑓$ is unique.

We next show that our definition of $𝑒$ $𝑒$ is equivalent to the limit version. Let $𝐹 (𝑥) ≔ 𝑓 (𝑥 𝑦) - 𝑓 (𝑥) - 𝑓 (𝑦)$ , where $𝑦$ $𝑦$ is a constant. Since $𝐹' (𝑥) = 1 𝑥 - 1 𝑥 = 0$ , $𝐹 (𝑥)$ $𝐹 (𝑥)$ is a constant. Because $𝐹 (1) = 𝑓 (𝑦) - 𝑓 (1) - 𝑓 (𝑦) = 0$ , we have $𝐹 (𝑥) = 0$ . Thus, we obtain

𝑓 (𝑥 𝑦) = 𝑓 (𝑥) + 𝑓 (𝑦) .

𝑓 (𝑥 𝑦) = 𝑓 (𝑥) + 𝑓 (𝑦) .

𝑓 (𝑥 𝑦) = 𝑓 (𝑥) + 𝑓 (𝑦) .

Since $𝑓$ $𝑓$ is defined upon $𝑅 +$ $𝑅 +$ , $𝑓' (𝑥) = 1 𝑥 > 0$ . Hence $𝑓$ $𝑓$ is injective. Let $e x p$ $e x p$ be the inverse function of $𝑓$ $𝑓$ . By the inverse function theorem, $e x p' (𝑥) = e x p (𝑥)$ . By $𝑓 (𝑥 𝑦) = 𝑓 (𝑥) + 𝑓 (𝑦)$ , we have $e x p (𝑥 + 𝑦) = e x p (𝑥) e x p (𝑦)$ . Since $𝑓$ $𝑓$ is smooth, we know that $e x p$ $e x p$ is smooth. By continuity, we can denote $e x p$ $e x p$ as $𝑒 𝑥$ $𝑒 𝑥$ , where $𝑒 = e x p (1)$ (equivalent to $𝑓 (𝑒) = 1$ ). From $𝑓 (𝑥) \leq 𝑥 - 1$ , we know that $𝑒 𝑥 \geq 𝑥 + 1$ , which implies $𝑒 1 𝑛 \geq 1 𝑛 + 1$ , which is

𝑒 \geq (1 + 1 𝑛) 𝑛 .

𝑒 \geq (1 + 1 𝑛) 𝑛 .

Also, $𝑒 - 1 𝑛 + 1 \geq 1 - 1 𝑛 + 1$ , which is

𝑒 - 1 𝑛 + 1 \geq 𝑛 𝑛 + 1 .

𝑒 - 1 𝑛 + 1 \geq 𝑛 𝑛 + 1 .

Thus, $𝑒 1 𝑛 + 1 \leq 1 + 1 𝑛$ . Hence, we obtain

(1 + 1 𝑛) 𝑛 \leq 𝑒 \leq (1 + 1 𝑛) 𝑛 + 1, \forall 𝑛 \in 𝑁

(1 + 1 𝑛) 𝑛 \leq 𝑒 \leq (1 + 1 𝑛) 𝑛 + 1, \forall 𝑛 \in 𝑁

(1 + 1 𝑛) 𝑛 \leq 𝑒 \leq (1 + 1 𝑛) 𝑛 + 1, \forall 𝑛 \in 𝑁

Since $𝑓$ $𝑓$ and $e x p$ $e x p$ are well-defined, $𝑒$ $𝑒$ exists uniquely. Hence, $l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛$ $l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛$ and $l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 + 1$ $l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 + 1$ converges. Because $l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 + 1 l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛$ $l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 + 1 l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛$ = $l i m 𝑛 \to \infty (1 + 1 𝑛) = 1$ implies $l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 = l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 + 1$ . By the squeeze theorem, we obtain

𝑒 = l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 .

𝑒 = l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 .

𝑒 = l i m 𝑛 \to \infty (1 + 1 𝑛) 𝑛 .

□

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#import "@preview/noteworthy:0.4.0": noteworthy
#import "/themes/site/notes.typ": *
#show: web-equations
#import "/typ/shared.typ": *

#show: tylenotes.with(
  title: "Yet Another Definition of Euler's Number",
  date: "2026-09-18",
  tags: ("real", "lean"),
  summary: "Provide a new way of defining e from the tangent-line inequality.",
)

= The Start

This note begin with the question:
can we use the properties of $ln$ to obtain the tangent-line inequality algebraically?
(Note that the tangent-line inequality means the inequality $ln(x) <= x - 1$.)
Here is what I tried.

First, let $I_n = (k 2^n,k 2^(n+1)]$, where $n in N$, $k in [1,2)$.
Hence we know that this can cover $(1,+infinity)$.
In fact, you can check that the same can be made for $(0,1)$, which is left to the reader.
We next show that if the inequality holds on $I_n$, then it holds on $I_(n+1)$,
which is basically an induction over intervals rather than integers.
Here, our inductive hypothesis is $ln(x) <= x - 1, forall x in I_n$.
Hence, we have $ln(2 x) = ln(2) + ln(x) <= ln(2) + x - 1, forall x in I_n$.
Since $I_n$ covers $(1,+infinity)$, $x > 1 > ln(2)$.
Thus, $ln(2 x) < x + x - 1 = 2 x - 1$. Since $x in I_n$, we obtain $2 x in I_(n+1)$.
Hence, $ln(x) < x - 1, forall x in I_(n+1)$.

The problem is the proof of the inequality on $I_0$, which is $(1,4)$.

But you will soon find it as difficult as the orignal case,
which I will explain later.

= The Climax

I thought at first that the following theorem and proof would help.

#theorem[
  $ ln(k) < k, forall k > 0 $
]

#proof[
  We first prove that $ln(k) < k, forall k > 0$.
  Since $e > 2$, you can verify that $ln(k) < log_2(k)$.
  Thus, it suffices to show that $log_2(k) < k$, which can be written as $k < 2^k$.
  We show it by induction.
  + $k in (0,1]$ Hence, $log_2(k) <= 0 < k$.
  + $log_2(k) < k$,also $k < 2^k and k > 1$ By $k>1$ we have $k+1 < 2 k < 2 dot 2^k = 2^(k+1)$.
  Hence, $ln k < k, forall k > 0$.
]

The problem is very obvious. Here the inequality still holds for $y=log_2(x)$.
But in our case, the inequality is false for any other logarithm.
The main problem is that this inequality depends on the property of $e$,
so that we cannot prove it unless we use the definition of $e$,
which makes the proof non-algebraic. Hence, our goal is impossible.
Yet there is another question. Since this inequality is deeply dependent on the definition of $e$,
is it possible to define $e$ with this inequality?
The answer is yes.

= The End
This theorem and the following proof gives you exactly how to define $e$ with our inequality.

#theorem[
  There exists a unique smooth function $f: R_+ -> R$ with the following property:
  + $f(x y) = f(x) + f(y)$ ($x,y>0$).
  + $f(x) <= x - 1$ ; equality holds if and only if $x=1$.
  We define $e$ to be a positive real number such that $f(e)=1$,
  and this definition is equivalent to the limit version.
]

#proof[
  We first show that $f$ is unique.
  Since $f(1) = f(1 dot 1) = 2 f(1)$, we obtain $f(1) = 0$.
  Apply $partial y$ to both sides of $f(x y) = f(x) + f(y)$, we have $x f'(x y) = f'(y)$.
  Let $y = 1$, we obtain $x f'(x) = f'(1)$.
  Hence, $f'(x) = C / x$, where $C = f'(1)$.
  Since $f(x) <= x - 1$, we let $g(x) := f(x) - (x-1)$.
  Because the equality holds if and only if $x = 1$,
  we know that $g$ has the maximum value at $1$, Hence $g'(1) = 0$.
  It follows that $g'(x) = f'(x) - 1$.
  Substitute this into the previous expression to obtain
  $f'(1) = 1$. Hence, $f'(x) = 1/x$.
  Thus, we obtain $ f(x) = integral_1^x 1/x dif x. $
  Hence, $f$ is unique.

  We next show that our definition of $e$ is equivalent to the limit version.
  Let $F(x) := f(x y) - f(x) - f(y)$, where $y$ is a constant.
  Since $F'(x) = 1/x - 1/x = 0$, $F(x)$ is a constant.
  Because $F(1) = f(y) - f(1) - f(y) = 0$, we have $F(x) = 0$.
  Thus, we obtain $ f(x y) = f(x) + f(y). $
  Since $f$ is defined upon $R_+$, $f'(x) = 1/x > 0$. Hence $f$ is injective.
  Let $exp$ be the inverse function of $f$.
  By the inverse function theorem, $exp'(x) = exp(x)$.
  By $f(x y) = f(x) + f(y)$, we have $exp(x+y) = exp(x)exp(y)$.
  Since $f$ is smooth, we know that $exp$ is smooth.
  By continuity, we can denote $exp$ as $e^x$, where $e = exp(1)$(equivalent to $f(e) = 1$).
  From $f(x) <= x - 1$, we know that $e^x >= x + 1$,
  which implies $e^(1/n) >= 1/n + 1$, which is
  $ e >= (1 + 1/n)^n. $
  Also, $e^(-1/(n+1)) >= 1 - 1/(n+1)$, which is
  $ e^(-1/(n+1)) >= n/(n+1). $
  Thus, $e^(1/(n+1)) <= 1 + 1/n$.
  Hence, we obtain $ (1 + 1/n)^n <= e <= (1 + 1/n)^(n + 1), forall n in N $
  Since $f$ and $exp$ are well-defined, $e$ exists uniquely.
  Hence, $lim_(n->infinity)(1+1/n)^n$ and $lim_(n->infinity)(1+1/n)^(n+1)$ converges.
  Because $(lim_(n->infinity)(1+1/n)^(n+1)) / (lim_(n->infinity)(1+1/n)^n)$ = 
  $lim_(n->infinity)(1+1/n) = 1$ implies
  $lim_(n->infinity)(1+1/n)^n = lim_(n->infinity)(1+1/n)^(n+1)$.
  By the squeeze theorem, we obtain $ e=lim_(n->infinity)(1+1/n)^n. $

]