# resscueeee meee

• July 26th 2006, 07:20 AM
jenjen
resscueeee meee
will you pleaseeee show me how to use the axioms given to prove this :

1) For every natural number i, i1 = i and 1i = i
2) For every natural number i, i0 = 0 and 0i = 0

I just think that the number multiply 1 is always the number and anything multiple by 0 is 0 but my professor wants me to prove them...i really dont' know how..
• July 26th 2006, 08:35 AM
Soltras
I always thought that the first one was a field axiom already (i.e. considered true implicitly, but not provable).

The second one can be shown this way:

$i0 = i(0 + 0) = i0 + i0$.

Since we now have $i0 = i0 + i0$, there is an axiom (if $x+y=x$ then $y=0$). So $i0 = 0$

Use that knowledge and the commutative axiom to show $0i$.
• July 26th 2006, 11:19 AM
CaptainBlack
Quote:

Originally Posted by Soltras
I always thought that the first one was a field axiom already (i.e. considered true implicitly, but not provable).

The second one can be shown this way:

$i0 = i(0 + 0) = i0 + i0$.

Since we now have $i0 = i0 + i0$, there is an axiom (if $x+y=x$ then $y=0$). So $i0 = 0$

Use that knowledge and the commutative axiom to show $0i$.

I think these are to be proven from what are essentially the axioms of Peano
Arithmetic given in the other thread, namely:

Quote:

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.