The Real Numbers Part1 Natural Numbers
Introduction
Before start our tour of recognizing real numbers, I will introduce natural numbers first. This is a trivial thing for most of us, because we all use them in real life, and we know most of the formulars like: $a + b = b + a$, $ab = ba$ and so on. But only few of us know why this is true.
At first, we define some symbols:
DEFINITION 1.0.1: If $a$ and $b$ are numbers, and they are the same; then we have $a = b$ which means $a\ equals\ to\ b$
Accordingly , the following are true:
Obviously, $a = a$
If $a = b$; then $b = a$.
If we have another number $c$, and $a = b$ is given, $b = c$; then $a = c$.
DEFINITION 1.0.2: If $a$ and $b$ are numbers, and they are not the same; then we have $a \neq b$, which means $a \ not equals to b$.
So, if $a \neq b$; then $b \neq a$.
Peano Axioms
I will begin from some axioms as known as Peano Axioms, which means we assume these are true and there’s no need to prove it. Then we can deduce all theorems of natural numbers.
Some books are defined Natural Number without 0, but I will include it for convenience.
Here is what Peano Axioms define:
AXIOM 1.1.1: $0$ is a natural number.
AXIOM 1.1.2: If $a$ is a natural number, there is exactly one natural number called a successor of $a$, which is denoted by $a^{++}$.
If $a$ is complicated, it will be enclosed in parentheses for preventing from ambiguities, like $(a + 1)^{++}$.
Thus if $a$ and $b$ are natural numbers, if $a = b$ , then $a^{++} = b^{++}$.
AXIOM 1.1.3: For each natural number $a$ , $a^{++} \neq 0$.
AXIOM 1.1.4: If $a$ and $b$ are natural numbers and $a^{++} = b^{++}$ , then $a = b$.
AXIOM 1.1.5:
Induction.
Let $P(n)$ be any property pertaining to a natural number $n$:
- $P(0)$ is true,
- if when $P(k)$ is true, then so is $P(k^{++})$;
- then we can say that $P(n)$ holds for all natural number $n$.
DEFINITION 1.1.1: $0^{++} = 1$, $1^{++} = 2$, and so on.
From the Peano Axioms above, we can deduce many theorems of natural numbers:
THEOREM 1.1.1: Let $a$ and $b$ are
Natural Numbers, If $a \neq b$, then $a^{++} \neq b^{++}$.
Proof:
If $a^{++} = b^{++}$, then we know that $a = b$, (AXIOM 1.1.4)
so it’s a contradiction.
THEOREM 1.1.2: Let $a$ and $b$ are
Natural Numbers, If $a^{++} \neq b^{++}$ then $a \neq b$.
Proof:
If $a = b$, then $a^{++} = b^{++}$ , (AXIOM 1.1.2)
so it’s a contradiction.
Addition
Now, we can define addition for Natural Numbers:
DEFINITION 1.2.1: Let $a$ and $b$ are
Natural Numbers, the addition of $a$ and $b$ denoted by $a + b$ and
- $0 + a = a$
- $a^{++} + b = (a + b)^{++}$
THEOREM 1.2.1: If $a$ and $b$ are natural numbers, then $a + b$ has one and only one value.
Proof:
When $a = 0$, then
$0 + b = b$, (DEFINITION 1.2.1), and obviously, $a + b$ exists and has only one value $b$;
Inductive Hypothesis: suppose that $a + b$ has one and only one value holds for $a = k$;
when $a = k^{++}$, then
$k^{++} + b$
$= (k + b)^{++}$ (DEFINITION 1.2.1)
according to Inductive Hypothesis, we know that $(k + b)$ has one and only one value,
so is $(k + b)^{++}$. (DEFINITION 1.2.1)
This completes the proof.
THEOREM 1.2.2: If $a$ is a natural number, $a + 0 = a$
Proof:
When $a = 0$, it is obviously that $0 + 0 = 0$; (DEFINITION 1.2.1)
Inductive Hypothesis: let’s suppose that $a + 0 = a$ holds for $a = k$;
when $a = k^{++}$, then
$k^{++} + 0$
$= (k + 0)^{++}$ (DEFINITION 1.2.1)
$= k^{++}$ (Inductive Hypothesis)
This completes the proof.
THEOREM 1.2.3: If $a$ is a natural number, $1 + a = a^{++}$
Proof:
When $a = 0$, then
$1 + a$
= $0^{++} + a$ (DEFINITION 1.1.1)
= $(0 + a)^{++}$ (DEFINITION 1.2.1)
= $a^{++}$ (DEFINITION 1.2.1)
This completes the proof.
THEOREM 1.2.4: If $a$ and $b$ are natural numbers, $a + b^{++} = (a + b)^{++}$
Proof:
When $a = 0$, then
$0 + b^{++}$
$= b^{++}$ (DEFINITION 1.2.1)
$= (0 + b)^{++}$ (DEFINITION 1.2.1)
Inductive Hypothesis: suppose that $a + b^{++} = (a + b)^{++}$ holds for $a = k$;
when $a = k^{++}$, then
$k^{++} + b^{++}$
$= (k + b^{++})^{++}$ (DEFINITION 1.2.1)
$= ((k + b)^{++})^{++}$ (Inductive Hypothesis)
$= (k^{++} + b)^{++}$ (DEFINITION 1.2.1)
This completes the proof.
THEOREM 1.2.5: If $a$ and $b$ are natural numbers, $a + b = b + a$
Proof:
When $a = 0$, then
$0 + b$
$= b$ (DEFINITION 1.2.1)
$= b + 0$; (THEOREM 1.2.2)
Inductive Hypothesis: let’s suppose that $a + b = b + a$ holds for $a = k$;
when $a = k^{++}$, then
$k^{++} + b$
$= (k + b)^{++}$ (DEFINITION 1.2.1)
$= (b + k)^{++}$ (Inductive Hypothesis)
$= b + k^{++}$ (THEOREM 1.2.4)
This completes the proof.
THEOREM 1.2.6: If $a$ and $b$ are natural numbers, $(a + b) + c = a + (b + c)$
Proof:
When $c = 0$, then
$(a + b) + 0$
$= a + b$ (THEOREM 1.2.2)
$= a + (b + 0)$ (THEOREM 1.2.2)
Inductive Hypothesis: let’s suppose $(a + b) + c = a + (b + c)$ holds for $c = k$
when $c = k^{++}$, then
$(a + b) + k^{++}$
$= ((a + b) + k)^{++}$ (THEOREM 1.2.4)
$= (a + (b + k))^{++}$ (Inductive Hypothesis)
$= a + (b + k) ^{++}$ (THEOREM 1.2.4)
This completes the proof.
THEOREM 1.2.7: If $a$ and $b$ are natural numbers, $a + b^{++} = a^{++} + b$
Proof:
$a + b^{++}$
$= (a + b)^{++}$ (THEOREM 1.2.4)
$= a^{++} + b$ (DEFINITION 1.2.1)
This completes the proof.
Ordering
DEFINITION 1.3.1: Let $a$ and $b$ are natural numbers, if we can find a natural number $c$, let $a + c = b$, then we say $a \leq b$.
DEFINITION 1.3.2: If $a \leq b$, then $b \geq a$.
DEFINITION 1.3.3: Let $a$ and $b$ are natural numbers, if we can find a natural number $c$, and $c \neq 0$, let $a + c = b$, then we say $a < b$.
DEFINITION 1.3.4: If $a < b$, then $b > a$.
Mutiplication
DEFINITION 1.4.1: Let $a$ and $b$ are
Natural Numbers, the multiplication of $a$ and $b$ denoted by $a \times b$ and
- $0 \times a = 0$
- $a^{++} \times b = (a \times b) + b$
THEOREM 1.4.1: If $a$ and $b$ are natural numbers, then $a \times b$ has one and only one value.
Proof:
When $a = 0$, then
$0 \times b = 0$, (DEFINITION 1.4.1), and obviously, $a \times b$ exists and has only one value $0$;
Inductive Hypothesis: suppose that $a \times b$ has one and only one value holds for $a = k$;
when $a = k^{++}$, then
$k^{++} \times b$
$= (k \times b) + b$ (DEFINITION 1.4.1)
according to Inductive Hypothesis, we know that $(k \times b)$ has one and only one value,
so is $(k \times b) + b$ (THEOREM 1.2.1)
namely $(k^{++} \times b)$.
This completes the proof.
THEOREM 1.4.2: If $a$ is a natural number, $a \times 0 = 0$
Proof:
When $a = 0$, then
$0 \times 0 = 0$, (DEFINITION 1.4.1)
Inductive Hypothesis: suppose that $a \times 0 = 0$ holds for $a = k$;
when $a = k^{++}$, then
$k^{++} \times 0$
$= (k \times 0) + 0$ (DEFINITION 1.4.1)
$= 0 + 0$ (Inductive Hypothesis)
$= 0$ (DEFINITION 1.2.1)
This completes the proof.
THEOREM 1.4.3: If $a$ and $b$ are natural numbers, $a \times b^{++} = (a \times b) + a$
Proof:
When $a = 0$, then
$0 \times b^{++}$
$= 0$ (DEFINITION 1.4.1)
$= (0 \times b)$ (DEFINITION 1.4.1)
$= (0 \times b) + 0$ (DEFINITION 1.2.1)
Inductive Hypothesis: suppose that $a \times b^{++} = (a \times b) + a$ holds for $a = k$;
when $a = k^{++}$, then
$k^{++} \times b^{++}$
$= (k \times b^{++}) + b^{++}$ (DEFINITION 1.4.1)
$= (k \times b) + k + b^{++}$ (Inductive Hypothesis)
$= (k \times b) + k^{++} + b$ (THEOREM 1.2.7)
$= (k \times b) + b + k^{++}$ (THEOREM 1.2.5)
$= (k^{++} \times b) + k^{++}$ (DEFINITION 1.4.1)
This completes the proof.
THEOREM 1.4.2: If $a$ is a natural number, $1 \times a = a$
Proof:
When $a = 0$, then
$1 \times 0 = 0$, (THEOREM 1.4.2)
Inductive Hypothesis: suppose that $1 \times a = a$ holds for $a = k$;
when $a = k^{++}$, then
$1 \times k^{++}$
$= (1 \times k) + 1$ (THEOREM 1.4.3)
$= k + 1$ (Inductive Hypothesis)
$= 1 + k$ (THEOREM 1.2.5)
$= k^{++}$ (THEOREM 1.2.3)
This completes the proof.
THEOREM 1.4.3: If $a$ and $b$ are natural numbers, $a \times b = b \times a$
Proof:
When $a = 0$, then
$0 \times b$
$= 0$ (DEFINITION 1.4.1)
$= b \times 0$ (THEOREM 1.4.2)
Inductive Hypothesis: suppose that $a \times b = b \times a$ holds for $a = k$;
when $a = k^{++}$, then
$k^{++} \times b$
$= (k \times b) + b$ (DEFINITION 1.4.1)
$= (b \times k) + b$ (Inductive Hypothesis)
$= b \times k^{++}$ (THEOREM 1.4.3)
This completes the proof.
THEOREM 1.4.4: If $a$, $b$ and $c$ are natural numbers, $(a \times b) + (a \times c) = a \times (b + c)$
Proof:
When $a = 0$, then
$(0 \times b) + (0 \times c)$
$= 0 + 0$ (DEFINITION 1.4.1)
$= 0$ (DEFINITION 1.2.1)
$= 0 \times (b + c)$ (DEFINITION 1.4.1)
Inductive Hypothesis: suppose that $(a \times b) + (a \times c) = a \times (b + c)$ holds for $a = k$;
$(k^{++} \times b) + (k^{++} \times c)$
$= ((k \times b) + b) + ((k \times c) + c)$ (DEFINITION 1.4.1)
$= ((k \times b) + (k \times c)) + (b + c)$ (THEOREM 1.2.5, 1.2.6)
$= (k \times (b + c)) + (b + c)$ (Inductive Hypothesis)
$= k^{++} \times (b + c)$ (DEFINITION 1.4.1)
This completes the proof.
THEOREM 1.4.5: If $a$, $b$ and $c$ are natural numbers, $(a \times b) \times c = a \times (b \times c)$
Proof:
When $c = 0$, then
$(a \times b) \times 0$
$= 0$ (THEOREM 1.4.2)
$= a \times 0$ (THEOREM 1.4.2)
$= a \times (b \times 0)$ (THEOREM 1.4.2)
Inductive Hypothesis: suppose that $(a \times b) \times c = a \times (b \times c)$ holds for $c = k$;
when $c = k^{++}$, then
$(a \times b) \times k^{++}$
$= ((a \times b) \times k) + (a \times b)$ (THEOREM 1.4.3)
$= (a \times (b \times k)) + (a \times b)$ (Inductive Hypothesis)
$= a \times ((b \times k) + b)$ (THEOREM 1.4.4)
$= a \times (b \times k^{++})$ (THEOREM 1.4.3)
This completes the proof.
DEFINITION 1.4.2: The symbol $\times$ in $a \times b$ is usually omitted.
So, some theorems we proved above can be written like $ab = ba$ and $(ab)c = a(bc)$.
THEOREM 1.4.6: If $a$, $b$, $c$ and $d$ are natural numbers, $(a + b)(c + d) = ac + ad + bc + bd$
If you are curious about this, you can take some time to prove it!
References
- John Stillwell. (2013). The Real Numbers, An Introduction to Set Theory and Analysis. Springer.
- Edmund Landau. (1966). Foundations Of Analysis. Chelsea Publishing Company.
- Terence Tao. (2015). Analysis Ι, Third Edition. Springer.