# Math Help - helpppppp!!!!!! math is difficulttttt

1. ## helpppppp!!!!!! math is difficulttttt

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

2. Originally Posted by jenjen
2) For every natural number i, i + 0 = i and 0 + i = i
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.

3. Originally Posted by jenjen
For every natural number n, for every natural number m, n + (Sm) = S(n + m);
Use this peano axiom.

1) For every natural number i, for every natural number j, for every natural number k, ( i + j ) + k = i + ( j + k )
Then (by axiom) it is necessay and sufficient to prove,
$S((i+j)+k)=S(i+(j+K))$
But then by top axiom ya have,
$i+j+S(k)=i+S(j+k)$
But then by top axiom ya have,
$i+j+S(k)=i+j+S(k)$

4. Hi,

Thanks for your reply, but then my professor told me I can't do it that way. From the axioms, I know that for every natural number n, n + 0 = n but not for every natural number n, 0 + n = n. That is why I am having a hard time.

5. Then you can do it by induction.
$S=\{x\in \mathbb{N}| 0+x=x\}$

For $n=0$,
$0+0=0$
Thus, $n\in S$

If, $k\in S$ then,
$0+k=k$
If and only if,
$S(0+k)=S(k)$
Thus,
$0+S(k)=S(k)$
Thus, $S(k)\in S$

Conditions for induction are satisfied, thus
$S=\mathbb{N}$

6. so you mean after that is proven by induction, like how you did it, then I can say that "For every natural number i, i + 0 = i and 0 + i = i?"

7. Originally Posted by jenjen
so you mean after that is proven by induction, like how you did it, then I can say that "For every natural number i, i + 0 = i and 0 + i = i?"
Yes, but i+0 does not need to be proven (one of axioms).

For every natural number i, for every natural number j, for every natural number k, ( i j ) k = i ( j k ). Which axioms can I use??

9. Originally Posted by jenjen
2) For every natural number i, i + 0 = i and 0 + i = i
This is Lemma A in the other thread

RonL

10. 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 )

Lemma A: for all natural numbers $x,\ x+0=x=0+x$
To be proven later

Lemma B: for all natural numbers $x, y,\ x+S(y)=S(x+y)=S(x)+y$
To be proven later.
We will prove that for all $x,\ y,\ z\ \in \mathbb{N}$ that

$
(x+y)+z=x+(y+z)
$

by induction on $x$.

Base case:

For all $y,\ z \in \mathbb{N}$

$
(0+y)+z=y+z$
, by Lemma A
$
=0+(y+z)
$
, again by Lemma A
.

So the base case is true.

Induction step:

For any $y,\ z \in \mathbb{N}$, suppose there exists a $x$ such that:

$
(x+y)+z=x+(y+z)
$

Now look at the successor of the LHS:

$
S((x+y)+z)=S(x+y)+z=(S(x)+y)+z
$
double application of Lemma B

Now look at the successor of the RHS:

$
S((x+(y+z))=S(x)+(y+z)
$
again by Lemma B

So: $(S(x)+y)+z=S(x)+(y+z)$, so we have proven asociativity
for $S(x)$ from its assumption for $x$ which completes the induction step.

Hence for all $x,\ y,\ z \in \mathbb{N}$ we have proven:

$
(x+y)+z=x+(y+z)
$

by induction on $x$

RonL