Show that the set is inductive

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In summary, the problem asks you to PROVE that N is inductive. You need to show that 1 is in A- and that $A$ is inductive.
  • #1
mathmari
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Hey! 😊

I am looking the following:

Prove that for $m,n\in \mathbb{N}$, $m>n$ it holds that $m-n\in \mathbb{N}$.
Hint: Consider $A=\left \{n\in \mathbb{N}\mid \forall m\in \mathbb{N}, m>n:m-n \in \mathbb{N}\right \}$ and show that $A$ is inductive. I have done the following:

To show that $A$ is inductive we have to show that $1\in A$ and that if $n\in A$ then also $n+1\in A$, right?

We have that $1\in A$ since $\forall m\in \mathbb{N}$ with $m>1$ we have that $m-1 \in \mathbb{N}$. Can we just say that?

Let $n\in A$. That means that $ \forall m\in \mathbb{N}$ with $m>n$ we have that $m-n \in \mathbb{N}$

We have that $n+1\in A$ since $m-(n+1)=(m-n)+1$ We have that $m-n\in \mathbb{N}$ and the successor is also in $\mathbb{N}$.

Is that correct and complete? :unsure:
 
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  • #2
mathmari said:
Hey! 😊

I am looking the following:

Prove that for $m,n\in \mathbb{N}$, $m>n$ it holds that $m-n\in \mathbb{N}$.
Hint: Consider $A=\left \{n\in \mathbb{N}\mid \forall m\in \mathbb{N}, m>n:m-n \in \mathbb{N}\right \}$ and show that $A$ is inductive. I have done the following:

To show that $A$ is inductive we have to show that $1\in A$ and that if $n\in A$ then also $n+1\in A$, right?

We have that $1\in A$ since $\forall m\in \mathbb{N}$ with $m>1$ we have that $m-1 \in \mathbb{N}$. Can we just say that?
How does it follow that $m-1 \in \mathbb{N}$? That would be true if $1\in N$ but that is what you are trying to prove.

Let $n\in A$. That means that $ \forall m\in \mathbb{N}$ with $m>n$ we have that $m-n \in \mathbb{N}$

We have that $n+1\in A$ since $m-(n+1)=(m-n)+1$ We have that $m-n\in \mathbb{N}$ and the successor is also in $\mathbb{N}$.

Is that correct and complete? :unsure:
 
  • #3
Country Boy said:
How does it follow that $m-1 \in \mathbb{N}$? That would be true if $1\in N$ but that is what you are trying to prove.

Ok... So what I am supposed to do now? I got stuck right now. :unsure:
 
  • #4
Hey mathmari!

Which definitions and axioms do you have for $\mathbb N$, the operator $>$, and the operator $-$? 🤔
 
  • #5
Klaas van Aarsen said:
Which definitions and axioms do you have for $\mathbb N$, the operator $>$, and the operator $-$? 🤔

For the above proof do we need that these three sets are inductive? :unsure: There is also an other hint, to look at the idea of proof for $(n, n+1)\cap \mathbb{N}=\emptyset$.
 
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  • #6
mathmari said:
For the above proof do we need that these three sets are inductive? :unsure:
The problem asks you to PROVE that N is inductive!
Also you do NOT have to prove that 1 is in A- you were asked to prove that N is inductive, not A.

There is also an other hint, to look at the idea of proof for $(n, n+1)\cap \mathbb{N}=\emptyset$.
 
  • #7
Country Boy said:
The problem asks you to PROVE that N is inductive!
Also you do NOT have to prove that 1 is in A- you were asked to prove that N is inductive, not A.

I thought that we have to prove that $A$ is inductive, because of the given hint.

So we have to show that $\mathbb{N}$ is inductive?
 
  • #8
No, I misread the problem. It does, indeed, specifically say "prove A is inductive". Sorry about that.

One thing you will need, for x and y in N, exactly how is "n- m" defined?
 
  • #9
Country Boy said:
One thing you will need, for x and y in N, exactly how is "n- m" defined?

We have to define - m and then we get the sum n+(-m), or not?

Or do we set n-m=s and so we consider n=s+m?
 
  • #10
I thought about that again and I did the following: $1\in A$ : Let $m > 1$ and $m-1=r \Rightarrow m = 1 + r$ for some $r \in \mathbb{N}$ . We have that since $\mathbb{N}$ is an inductive set and $r\in \mathbb{N}$ then $r+1\in \mathbb{N}$, i.e. $m\in \mathbb{N}$. Let $n\in A$. Then let $m > n$ and $m = n + r$ for some $r \in \mathbb{N}$ then $m \in \mathbb{N}$. We want to prove that $n+1\in A$, i.e. that if $m' > n+1$ and $m' = (n+1) + r$ for some $r \in \mathbb{N}$ then $m'\in \mathbb{N}$.

From $m'>n+1$ we have that $m'-1>n$. So we get that $m'-1\in A$. Then $m'-1=n+r$ for some $r\in \mathbb{N}$, so $m'=(n+1)+r$, for the same $r\in \mathbb{N}$ and so $n+1\in A$. I think that it is somehow confusing what I have done... :unsure:
 
  • #11
There are multiple definitions for the set of natural numbers $\mathbb N$.
The first formal set given on wiki is:
The five Peano axioms:
1. 0 is a natural number.
2. Every natural number has a successor which is also a natural number.
3. 0 is not the successor of any natural number.
4. If the successor of x equals the successor of y, then x equals y.
5 The axiom of induction: If a statement is true of 0, and if the truth of that statement for a number implies its truth for the successor of that number, then the statement is true for every natural number.

Some forms of the Peano axioms have 1 in place of 0. In ordinary arithmetic, the successor of x is x+1.
🧐

Furthermore, it defines a comparison operator:
A total order on the natural numbers is defined by letting $a\le b$ if and only if there exists another natural number $c$ where $a+c=b$.
The other comparison operators are then defined using this definition.
This definition assumes that $0$ is an element of $\mathbb N$. If it is excluded, then we must use a modified version of this definition.
🧐

This references addition ($+$) for which we need a definition as well, which is given as:
One can recursively define an addition operator on the natural numbers by setting a + 0 = a and a + S(b) = S(a + b) for all a, b. Here, S should be read as "successor".
🧐

I didn't immediately see a definition for the difference operator ($-$) for natural numbers.
Let's make it:
The difference $a-b$ is defined as the number $c$ such that $a=b+c$ provided that such a number $c$ exists in $\mathbb N$. Otherwise the difference is undefined.
Note that there is no such thing as $-b$, since it is not an element of $\mathbb N$ for positive $b$. So we cannot define the difference $a-b$ as the usual $a+(-b)$.
🧐

Can we find a proof in which we only use these definitions and axioms? 🤔
 

Related to Show that the set is inductive

1. What does it mean for a set to be inductive?

For a set to be inductive, it means that every element in the set follows a specific pattern or rule, and the set contains all the elements necessary to continue that pattern or rule infinitely.

2. How can I prove that a set is inductive?

To prove that a set is inductive, you must show that it satisfies two conditions: 1) the first element in the set follows the given pattern or rule, and 2) if an element is in the set, then the next element in the set also follows the given pattern or rule.

3. What is the purpose of proving that a set is inductive?

Proving that a set is inductive is important in mathematics and science as it allows us to make generalizations and predictions about the behavior of a system or phenomenon. It also helps us to understand and explain patterns and relationships within a set.

4. Can a set be inductive if it contains only a finite number of elements?

No, a set must contain an infinite number of elements to be considered inductive. This is because the inductive property requires that the pattern or rule continues infinitely.

5. Are all sets that follow a pattern or rule considered inductive?

No, not all sets that follow a pattern or rule are considered inductive. For a set to be inductive, it must also satisfy the two conditions mentioned earlier: 1) the first element in the set follows the given pattern or rule, and 2) if an element is in the set, then the next element in the set also follows the given pattern or rule.

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