Are you using the Peano axioms for the natural numbers?
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
Base step: Fix m, let n = 1
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.
I do not like this question.
Because we define .
And now using the recursion theorem there is a function,
So that and .
Thus, there is nothing to prove.
You can look at this.
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.
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 that (i) contains 1 and (ii) contains whenever it contains ; from this it follows that the intersection of two inductive sets is an inductive set, and is thus defined as the intersection of all inductive subsets of . Is that right? Yes?
In that case, you’re on the right track with your inductive proof. Fix and let ; then use induction to show that for all . First, by the definition of an inductive set. If for some , then by the inductive hypothesis. Hence for all .
For multiplication of natural numbers, use the distributive property and the result above for addition.
The integers should be defined as where , so just work on that.