1. ## [SOLVED] Natural Numbers/ Integers Closure of Addition/Multiplication

1. Let m and n be natural number. Prove that the sum and product of natural numbers are natural numbers.

2. Prove that the sum, difference, and product of integers are also integers.
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1. I plan to do this by induction. Let $\displaystyle S(n)=(m+n) \in N$

Base step: Fix m, let n = 1
$\displaystyle S(m+n)=(m+1) \in N$
The problem I have is that isn't this begging the question? I'm trying to prove that the sum of two natural numbers is a natural number, but since m is a natural number how can I just assume that m+1 is a natural number?

2. My attempt: Suppose that m, n,r,s are natural numbers and m > n. Then there exists an integer k such that n+k=m. Thus, k = m-n. Likewise, let l be an integer l = r - s. Again, isn't this begging the question?

2. Are you using the Peano axioms for the natural numbers?

3. Originally Posted by Paperwings
1. Let m and n be natural number. Prove that the sum and product of natural numbers are natural numbers.
I do not like this question.
Because we define $\displaystyle n+1 = n\cup \{ n \}$.
And now using the recursion theorem there is a function, $\displaystyle +:\mathbb{N}\times \mathbb{N}\to \mathbb{N}$
So that $\displaystyle +(m,0) = m$ and $\displaystyle +(m,n+1) = +(m,n)+1$.
Thus, there is nothing to prove.

You can look at this.

4. Originally Posted by JaneBennet
Are you using the Peano axioms for the natural numbers?
Hello, Jane. No, I was not using the Peano axioms; in my book, the natural numbers are defined as the intersection of all inductive sets, and that the number 1 is in the natural numbers.

I had to google the Peano axioms, which assumes that 0 is a natural number.

So, Since 0 is a natural number and let S(n) and S(m+n) be defined as a natural number. Fix m, then
m+0=m+S(0) = S(m+0) = S(m+0)

S(m+1)= S(m) + 1 = S(m) + S(0)
S(m+2)= S(m) + 2 = S(m) + S(S(0))...
Then by the induction axiom, this holds true for all natural numbers, yes?

5. Originally Posted by Paperwings
Hello, Jane. No, I was not using the Peano axioms; in my book, the natural numbers are defined as the intersection of all inductive sets, and that the number 1 is in the natural numbers.
Aha! I think I see what your book is doing. Your book first presents a list of axioms for the real numbers, and then defines an “inductive set” as a subset of $\displaystyle \mathbb{R}$ that (i) contains 1 and (ii) contains $\displaystyle n+1$ whenever it contains $\displaystyle n$; from this it follows that the intersection of two inductive sets is an inductive set, and $\displaystyle \mathbb{N}$ is thus defined as the intersection of all inductive subsets of $\displaystyle \mathbb{R}$. Is that right? Yes?

In that case, you’re on the right track with your inductive proof. Fix $\displaystyle m\in\mathbb{N}$ and let $\displaystyle S(n)=m+n$; then use induction to show that $\displaystyle S(n)\in\mathbb{N}$ for all $\displaystyle n\in\mathbb{N}$. First, $\displaystyle S(1)=m+1\in\mathbb{N}$ by the definition of an inductive set. If $\displaystyle S(k)\in\mathbb{N}$ for some $\displaystyle k\in\mathbb{N}$, then $\displaystyle S(k+1)=m+(k+1)=(m+k)+1=S(k)+1\in\mathbb{N}$ by the inductive hypothesis. Hence $\displaystyle S(n)\in\mathbb{N}$ for all $\displaystyle n\in\mathbb{N}$.

For multiplication of natural numbers, use the distributive property and the result above for addition.

The integers should be defined as $\displaystyle \mathbb{Z}=\mathbb{N}\cup\{0\}\cup\left(-\mathbb{N}\right)$ where $\displaystyle -\mathbb{N}=\{-n:n\in\mathbb{N}\}$, so just work on that.

6. Originally Posted by JaneBennet
Aha! I think I see what your book is doing. Your book first presents a list of axioms for the real numbers, and then defines an “inductive set” as a subset of $\displaystyle \mathbb{R}$ that (i) contains 1 and (ii) contains $\displaystyle n+1$ whenever it contains $\displaystyle n$; from this it follows that the intersection of two inductive sets is an inductive set, and $\displaystyle \mathbb{N}$ is thus defined as the intersection of all inductive subsets of $\displaystyle \mathbb{R}$. Is that right? Yes?
Yes, that's precisely right. Thank you Jane for your help.