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Math Help - [SOLVED] Natural Numbers/ Integers Closure of Addition/Multiplication

  1. #1
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    [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.
    =====================================
    1. I plan to do this by induction. Let S(n)=(m+n) \in N

    Base step: Fix m, let n = 1
    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?



    Thank you for your time.
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  2. #2
    Senior Member JaneBennet's Avatar
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    Are you using the Peano axioms for the natural numbers?
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    Quote Originally Posted by Paperwings View Post
    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 n+1 = n\cup \{ n \}.
    And now using the recursion theorem there is a function, +:\mathbb{N}\times \mathbb{N}\to \mathbb{N}
    So that +(m,0) = m and +(m,n+1) = +(m,n)+1.
    Thus, there is nothing to prove.

    You can look at this.
    Last edited by ThePerfectHacker; August 5th 2008 at 07:43 PM.
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    Quote Originally Posted by JaneBennet View Post
    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?

    Thanks TPH for the link.
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  5. #5
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Paperwings View Post
    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 \mathbb{R} that (i) contains 1 and (ii) contains n+1 whenever it contains n; from this it follows that the intersection of two inductive sets is an inductive set, and \mathbb{N} is thus defined as the intersection of all inductive subsets of \mathbb{R}. Is that right? Yes?

    In that case, you’re on the right track with your inductive proof. Fix m\in\mathbb{N} and let S(n)=m+n; then use induction to show that S(n)\in\mathbb{N} for all n\in\mathbb{N}. First, S(1)=m+1\in\mathbb{N} by the definition of an inductive set. If S(k)\in\mathbb{N} for some k\in\mathbb{N}, then S(k+1)=m+(k+1)=(m+k)+1=S(k)+1\in\mathbb{N} by the inductive hypothesis. Hence S(n)\in\mathbb{N} for all 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 \mathbb{Z}=\mathbb{N}\cup\{0\}\cup\left(-\mathbb{N}\right) where -\mathbb{N}=\{-n:n\in\mathbb{N}\}, so just work on that.
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    Quote Originally Posted by JaneBennet View Post
    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 \mathbb{R} that (i) contains 1 and (ii) contains n+1 whenever it contains n; from this it follows that the intersection of two inductive sets is an inductive set, and \mathbb{N} is thus defined as the intersection of all inductive subsets of \mathbb{R}. Is that right? Yes?
    Yes, that's precisely right. Thank you Jane for your help.
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