I have to present this proof to my class on Monday. I was hoping if anyone would be so kind as to review it and tell me if I am making any fatal flaw errors. The Theorems I quote are from text materials and I give brief quote of them when first used here. My big concern is the induction for negative n's. Thank you for any help. Sorry for formatting, I seem to be losing all tabs when I preview it.
Prove that (m+n)a=ma+na for all m,n element of Z and a element of some ring R.
PROOF:
Let R be a ring. Let m be an element of Z and a be an element of R.
Let P(n) be the assertion that (m+n)a = ma+na for all n element of Z+.
Proving the base case we let n=1. It follows that:
(m+n)a = (m+1)a Substitution
=ma+1a Theorem 2.a.3 (m+1)a=ma+a for all m element Z, a element ring.
=ma+na Substitution
So, (m+n)a = ma+na for n=1.
Now we verify the induction step.
We need to show that (m+(n+1))a = ma+(n+1)a, for all n element of Z. Our induction hypothesis will be (m+n)a = ma+na for some n element of Z+.
Consider,
(m+(n+1)a = ((m+n)+1)a Associative prop on integers
=(m+n)a+a Theorem 2.a.3
=ma+na+a Induction Hypothesis
=ma+(n+1)a Theorem 2.a.3
Thus, by induction, we have shown that (m+n)a = ma+na, for all n element of Z+.
Next, we let n=0. Well,
(m+n)a=(m+0)a Substitution
=ma Integer Arithmetic
=ma+0R additive property of Ring
=ma+0a Defn 1.a 0a=0R 0R=additive identity for ring
=ma+na Substitution
So, we have shown that (m+n)a =ma+na for n=0.
Let P(n) be the assertion that (m+n)a = ma+na for all n element of Z-.
Proving the base case we let n=-1. It follows that:
(m+n)a = (m+(-1))a Substitution
=(m-1)a Integer Arithmetic
=ma-a Theorem 2.a.4 (m-1)a=ma-a for all m element Z, a element of ring.
=ma+(-1)a Theorem 2.a.1 (-1)a=-a, a element of ring
=ma+na Substitution
So, (m+n)a = ma+na for n=-1.
Now we verify the induction step.
We need to show that (m+(n-1))a = ma+(n-1)a, for all n element of Z-. Our induction hypothesis will be (m+n)a = ma+na for some n element of Z-.
Consider,
(m+(n-1))a = ((m+n)-1)a Associative prop of integers
=(m+n)a-a Theorem 2.a.4
=ma+na-a Induction Hypothesis
=ma+(n-1)a Theorem 2.a.4
Thus, by induction, we have shown that (m+n)a = ma+na, for all n element of Z-.
And we can conclude that (m+n)a=ma+na for all m,n element of Z
Prove that (m+n)a=ma+na for all m,n element of Z and a element of some ring R.
PROOF:
Let R be a ring. Let m be an element of Z and a be an element of R.
Let P(n) be the assertion that (m+n)a = ma+na for all n element of Z+.
Proving the base case we let n=1. It follows that:
(m+n)a = (m+1)a Substitution
=ma+1a Theorem 2.a.3 (m+1)a=ma+a for all m element Z, a element ring.
=ma+na Substitution
So, (m+n)a = ma+na for n=1.
Now we verify the induction step.
We need to show that (m+(n+1))a = ma+(n+1)a, for all n element of Z. Our induction hypothesis will be (m+n)a = ma+na for some n element of Z+.
Consider,
(m+(n+1)a = ((m+n)+1)a Associative prop on integers
=(m+n)a+a Theorem 2.a.3
=ma+na+a Induction Hypothesis
=ma+(n+1)a Theorem 2.a.3
Thus, by induction, we have shown that (m+n)a = ma+na, for all n element of Z+.
Next, we let n=0. Well,
(m+n)a=(m+0)a Substitution
=ma Integer Arithmetic
=ma+0R additive property of Ring
=ma+0a Defn 1.a 0a=0R 0R=additive identity for ring
=ma+na Substitution
So, we have shown that (m+n)a =ma+na for n=0.
Let P(n) be the assertion that (m+n)a = ma+na for all n element of Z-.
Proving the base case we let n=-1. It follows that:
(m+n)a = (m+(-1))a Substitution
=(m-1)a Integer Arithmetic
=ma-a Theorem 2.a.4 (m-1)a=ma-a for all m element Z, a element of ring.
=ma+(-1)a Theorem 2.a.1 (-1)a=-a, a element of ring
=ma+na Substitution
So, (m+n)a = ma+na for n=-1.
Now we verify the induction step.
We need to show that (m+(n-1))a = ma+(n-1)a, for all n element of Z-. Our induction hypothesis will be (m+n)a = ma+na for some n element of Z-.
Consider,
(m+(n-1))a = ((m+n)-1)a Associative prop of integers
=(m+n)a-a Theorem 2.a.4
=ma+na-a Induction Hypothesis
=ma+(n-1)a Theorem 2.a.4
Thus, by induction, we have shown that (m+n)a = ma+na, for all n element of Z-.
And we can conclude that (m+n)a=ma+na for all m,n element of Z