Since i+j=j+i we only have half to prove (commutativity was proven last time). But that is one of the Peano axioms nothing to prove.Originally Posted by jenjen
0 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, for every natural number k, ( i + j ) + k = i + ( j + k )
2) For every natural number i, i + 0 = i and 0 + i = i
Use this peano axiom.Originally Posted by jenjen
Then (by axiom) it is necessay and sufficient to prove,1) For every natural number i, for every natural number j, for every natural number k, ( i + j ) + k = i + ( j + k )
But then by top axiom ya have,
But then by top axiom ya have,
from the other thread :Originally Posted by jenjenWe will prove that for all that
Lemma A: for all natural numbers
To be proven later
Lemma B: for all natural numbers
To be proven later.
by induction on .
Base case:
For all
, by Lemma A
, again by Lemma A.
So the base case is true.
Induction step:
For any , suppose there exists a such that:
Now look at the successor of the LHS:
double application of Lemma B
Now look at the successor of the RHS:
again by Lemma B
So: , so we have proven asociativity
for from its assumption for which completes the induction step.
Hence for all we have proven:
by induction on
RonL