1

Introduction

I saw a video of a guy talking about proving that a even number squared is still even. It was something like this:

Let n be a natural/whole number, and 2n defines all even numbers:

2n * 2n = 4n² = 2(2n²)

So, 2(2n²) is an even number.

When I saw it, I thought: assuming that n² and 2n are natural numbers, right? People told me that it's unnecessary to prove that those are natural numbers. It goes out of the scope of the question, and it's very easy to see that n² and 2n are natural numbers.

It left me thinking. How to prove it? I recently asked a question about rigour of a proof, and someone showed me the proof of infinite many primes. It has "...". I always thought that "..." was prohibited in proving, like a lazy way to write. But it apparently ain't. Then I thought "it makes things easier", so I decided to give it a try. Here's my proof that the result of a sum of two natural numbers is a natural number:

My try

Definitions

It'll be shown that a number x is a natural number using the notation N(x). [I've already seen people using it, so I ain't inventing notation].

Natural Number:

  1. N(0)
  2. ~[N(-1)]
  3. [N(x)] → [N(x+1)]

"Which I think it means: -1 ain't a natural number, 0 is a natural number, and if a number is a natural number, this number plus one is also a natural number".

"A → B → C" means "A implies B, which implies C". This part is because that's not right/common notation, I think.

Proof

Since N(0) and [N(x)] → [N(x+1)]:

N(0) → N(0+1) → N(0+2) ... → N(0+a)

This chain reaction only works because the first number is a natural number, so we can generalize it:

N(b) → N(b+1) → N(b+2) ... → N(b+a)

We know that N(a) because N(0+a), which can be simplified to N(a).

So it can be said that [N(a) ^ N(b)] → N(a+b). Which hopefully means: if a and b are natural numbers, their sum is also a natural number.

Ending

How accurate or how right is my proof?

And thanks for reading this Bible-long text.

P.S.: I wrote this down at 0500 and in my cellphone, so I am 99% sure I made grammar and spelling mistakes. Also, if this text is too long or not focusing on the matter, please, tell me or, if you want, edit it.

1 Answers1

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Your definition of natural numbers is problematic. For example this: $[N(x)]\to[N(x+1)]$. What is $x+1$ anyway? What is $1$? And in this: $\sim [N(-1)]$, what is $-1$? You have circular references here, you define naturals using naturals (or even integers) and operations on naturals.

So that won't do. In mathematics we typically treat naturals axiomatically via Peano axioms. With that we have an axiomatic $0$ and the successor function $S(n)$. One of the axioms is that $0\in\mathbb{N}$, the other is that for any $n\in\mathbb{N}$ we have $S(n)\in\mathbb{N}$. We also have that $S:\mathbb{N}\to\mathbb{N}$ is injective and that there is no $n\in\mathbb{N}$ with $S(n)=0$.

In that setup we first define some constants: $1:=S(0)$, $2:=S(1)$, $3:=S(2)$, etc. Then we define addition recursively via

$$a+0:=a$$ $$a+S(b):=S(a+b)$$

analogously we define multiplication:

$$a\cdot 0:=0$$ $$a\cdot S(b):=a+(a\cdot b)$$

and then exponentiation:

$$a^1:=a$$ $$a^{S(b)}:=a\cdot (a^b)\text{ for }b\neq 0$$

Note that addition, multiplication and exponentiation produce new naturals by definition.

So all of that guarantees that:

  1. $2\in\mathbb{N}$
  2. $2\cdot n\in\mathbb{N}$
  3. $n^2\in\mathbb{N}$
freakish
  • 42,851
  • But isn't S(n) = n + 1? And saying that "there is no n∈N with S(n)=0" the same thing of saying ~[N(-1)]? I mean, I think it may be less formal, but I think that my definition is enough for what I did. I am no mathematician, so I can be 100% wrong. Sorry, if I am. –  Sep 16 '20 at 08:27
  • Yes, $S(n)=n+1$ but note that in the Peano axioms there are only two primitives: the zero $0$ and the successor function $S$. We define $1$ as $S(0)$. We also define "$+$" in terms of $S$ function. So $S(n)=n+1$ is not a definition, it is a theorem. The right side doesn't make sense on its own. Also $N(-1)$ doesn't make sense because there is no such thing as $-1$ inside naturals. To define $-1$ we would have to have integers. But to create integers we need naturals. That is circular. Do not treat mathematics naively, this leads to all sort of errors. – freakish Sep 16 '20 at 08:30
  • to be honest, I wasn't thinking about creating the natural numbers. I was thinking about classifying them. I had never thought about the natural number like you said. –  Sep 16 '20 at 08:35
  • Would my proof work if my definition were correct? I mean, waa my thought correct? –  Sep 16 '20 at 08:40
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    Well, you want to prove something about naturals. So you need to know precisely what naturals are. For example, if I ask you whether $\Xi\circ\Theta\in \Gamma$? How would you answer that? Well, you would first ask about every symbol and its exact meaning. Otherwise the question is meaningless. – freakish Sep 16 '20 at 08:43
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    Your proof is technically correct, but only when every symbol is well defined. And in the chain $N(0)\to N(0+1)\to\cdots\to N(0+a)$ if the symbol $0+a$ is well defined then it has to be natural (btw we don't really know what "$\cdots$" means). And so no further proof is needed. You are walking around things that are already either proven or axiomatically defined. You are circulating around well established things, you don't actually prove anything. It's like saying "grass is green, because it is green". Well, yeah... – freakish Sep 16 '20 at 08:44
  • thanks a lot. Should I edit the question or should I keep it as it is, since the question is answered? I am asking also because it would invalidate your answer, which ain't something nice. –  Sep 16 '20 at 08:46
  • No, leave the question as it is. It is a fine question in my opinon. – freakish Sep 16 '20 at 08:47