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.
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.
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.
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.
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.