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We had allowed two operations of division and subtraction on the set of natural numbers they could lead to fractional numbers or negative numbers, both not defined in the set of natural numbers. However sometimes these operations can be performed without causing any issues. For example one can divide $6$ by $2$ to get $3$; or one can subtract $2$ from $7$ to get $5$. In both of these cases the operations were borrowed from the rational numbers or from the set of integers. If we are tempted to use these operations sometimes, we must be conscious of the following discussions.

$\color{orange}{a)}$ Given natural numbers $a,b$ and $c$, if $a+b=a+c$ then $b=c$.

$\color{orange}{b)}$ Given two natural numbers $a < b$, we define $b−a$ to be a natural number that satisfies $$a+ (b−a) =b$$ Prove that this number is unique.

That is, if some other number $k$ claims that $a+k=b$, then $k=b−a$.

$\color{orange}{c)}$ What would you say about two natural numbers $b$ and $c$ if for some natural number $a$ we knew $ab=ac$. Prove your claim.)

(Remember we don’t have division, we have not proven any cancellation law, nor do we have $0$ when we deal with natural numbers.)

$\color{orange}{d)}$ When $a\mid b$ then we can define $\frac{b}{a}$ as the $\underline{\text{unique}}$ natural number $k$ that satisfies $a·k=b$.

Why is this $k$ unique?

(Note: if we didn’t have this uniqueness then we were not allowed to write $\frac{b}{a}$.)

$\color{orange}{e)}$If $a\mid b$ then what is $a\cdot\frac{b}{a}$?

Prove your claim.

(Warning: this is really trivial; so don’t overlook some important idea.)


List of properties that we can use:

  1. reflexivity of equality: $\forall a, a=a$

  2. symmetry of equality: $\forall a,b\in\mathbb{R},a=b\rightarrow b=a$

  3. transitivity of equality: $\forall a,b,c\in\mathbb{R},(a=b \wedge b=c)\rightarrow a=c$

  4. reflexivity of total ordering: $\forall a\in\mathbb{R},a\le a$

  5. anti symmetry of total ordering: $\forall a,b\in\mathbb{R}, (a\le b\wedge b\le a)\rightarrow a=b$

  6. transitivity of total ordering: $\forall a,b,c\in\mathbb{R},(a\le b \wedge b\le c)\rightarrow a\le c$

  7. Trichotomy: $\forall a,b\in\mathbb{R},a<b \vee a>b \vee a=b$

  8. R1) of ordering: $\forall a,b,c\in\mathbb{R}, (a<b\wedge c>0)\rightarrow ac<bc$

  9. R2) of ordering: $\forall a,b,c\in\mathbb{R}, a<b \rightarrow a+c<b+c$


Proof.

(Note: $\mathbb{N}=\mathbb{Z^+}$)

$\color{orange}{a)}$

Assume the negation is true, have $$a+b=a+c \text{ and }b\neq c$$

$\color{red}{\text{(here i want $a+b=b+a$ and $a+c=c+a$ implies $b+a=c+a$ and $c+a=b+a$)}}$

By Trichotomy we consider two cases:

Case1: $b>c$

By $R2)$ that $c<b$ implies $c+a<b+a$, contradiction!

Case2: $b<c$

By $R2)$ that $b+a<c+a$, contradiction!$\tag*{$\square$}$

$\color{orange}{b)}$

$$\text{WTS }\forall a\in\mathbb{N},b\in(a,\infty)\cap\mathbb{N},(a+(b-a)=b\rightarrow \forall k\in\mathbb{N}, (a+k=b\rightarrow k=b-a))$$

Have $$a+(b-a)=a+k$$

Apply $\color{orange}{a)}$, then we have:

$$b-a=k$$

By symmetry of equality that

$$k=b-a$$ $\tag*{$\square$}$

$\color{orange}{c)}$

I would say $\forall b,c\in\mathbb{N},(\exists a\in\mathbb{N},s.t.ab=ac)\rightarrow b=c$

Proof.

Prove by contradiction again, if not then we have:

$$b\neq c$$

$\color{red}{\text{(here i want $ab=ba$ and $ac=ca$ implies $ba=ca$ and $ca=ba$)}}$

By Trichotomy we consider two cases:

Case1: $b<c$

By $R1)$, $b<c$ and $a>0$ implies $$ba<ca$$

Contradiction!

Case1: $c<b$

By $R1)$, $c<b$ and $a>0$ implies $$ca<ba$$

Contradiction!

The statement hold since negation is false $\tag*{$\square$}$

$\color{orange}{d)}$

$$\text{WTS }(\forall a,b\in\mathbb{N}, \exists k\in\mathbb{N}, s.t. a\cdot k=b)\rightarrow \forall k_2\in\mathbb{N},(a\cdot k_2=b\rightarrow k=k_2)$$

By symmetry of equivality that:

$a\cdot k_2=b$ implies $b=a\cdot k_2$

By transitivity of equality we have:

$a\cdot k=b$ and $b=a\cdot k_2$ implies $a\cdot k=a\cdot k_2$

Apply $\color{orange}{c)}$ we have

$k=k_2\tag*{$\square$}$

$\color{orange}{e)}$

My claim: $a\cdot\frac{b}{a}=b$ where $b\in\mathbb{N}$

Apply $\color{orange}{d)}$, we have:

$\exists !k\in\mathbb{N},s.t. a\cdot\frac{b}{a}=a\cdot k=b\tag*{$\square$}$


From the listed nine properties, is it possible to prove commutative law for multiplication/addition? Which I think is requested in the $\color{red}{\text{red}}$ part of my proof, or is there a way to do the proof without using them?

Any help or hint or suggestion would be appreciated.

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  • $\begingroup$ Please don't delete then repost your prior question. Instead, edit it. $\endgroup$ Sep 22, 2019 at 2:22

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If you aren't limited to solve it by order, you can solve (b) first, and then I think you may use substraction in (a): you can subtract a from both sides, still having integers, and then from anti-symmetry of order (in opposite way) you get a=b. in (c), you can subtract ac from both sides, getting that ab-ac=0, which tells that a(b-c)=0, a is positive so b-c=0, and therefore b=c

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  • $\begingroup$ which i am not sure about your question cause its a bit trivial laws about natural numbers... If you were asking about natural numbers or integers as a ring there would be place for your question, but as i know for number theory commutative for those operations is well-defined $\endgroup$
    – friedvir
    Sep 22, 2019 at 1:05

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