The result is true for,
Thus there is some such as,
If and only if,
If and only if,
If and only if,
From the two previous result we have induction.
Proved.
O is a natural number
For every natural number n, its sucessor Sn is a natural number.
For every natural number n, Sn is not equal to 0;
For every natural number n, for every natural number m, if Sn = Sm, then n = m;
The principle of mathematical induction;
For every natural number n, for every natural number m, their sum n + m is a natural number;
For every natural number n, n + 0 = n;
For every natural number n, for every natural number m, n + (Sm) = S(n + m);
For every natural number n, for every natural number m, their product nm is a natural number;
For every natural number n, n0 = 0; and
For every natural number n, for every natural number m, n(Sm) = (nm) + n.
Using these axioms (and logic), prove the following law of algebra:
1) For every natural number i, for every natural number j, i + j = j + i
Hi ThePerfectHacker,
Thank you so much for helping me out on this problem, but then I can only use the axioms given to proof the law. Therefore, now I have to proof how you get from to and then I have to proof how you get from to but I do not know how. Will you please show me?
This last "axiom" is redundant "addition" is defined by the next two "axioms"Originally Posted by jenjen
This last "axiom" is redundant "product" is defined by the next two "axioms"For every natural number n, n + 0 = n;
For every natural number n, for every natural number m, n + (Sm) = S(n + m);
For every natural number n, for every natural number m, their product nm is a natural number;
For every natural number n, n0 = 0; and
For every natural number n, for every natural number m, n(Sm) = (nm) + n.
Using these axioms (and logic), prove the following law of algebra:
1) For every natural number i, for every natural number j, i + j = j + i
Lemma A: for all natural numbersOriginally Posted by jenjen
To be proven later
Lemma B: for all natural numbers
To be proven later.
For any we proceed by induction on . The base case is:
,
which is true by Lemma A.
Now suppose for some that:
Now:
by Lemma B, also:
by assumption, and:
from Lemma B with and interchanged.
Hence we have proven that:
which allows us to conclude that for any :
for all
RonL
Proofs of the lemma to follow as I find time
We proceed by induction on .Originally Posted by CaptainBlack
Base case: - trivially true
Now suppose that for some natural number :
.
Now:
from the axiom , but:
from the same axiom, so:
by assumption, and:
from the axiom n+(Sm)=S(n+m), so we have proven:
which allows us, with the base case, to conclude by induction that for
all natural numbers:
.
RonL
For any natural number x we proceed by induction on y.Originally Posted by CaptainBlack
Base case:
by axiom:
by axiom:So the base case: is true.
by axiom:
Now suppose for natural number there exists a natural number such that:
Now:
by axiom: .
Also:
, by supposition.
, by axiom.
So we have proven that:
which proves the induction step, hence0 with the base case the lemma is proven by induction.
RonL
(You should check this carefully as there is a lot of transcription from scraps
of paper to TeX, which is very prone to error)