# helpppppp!!!!!! math is difficulttttt

• Jul 25th 2006, 04:21 PM
jenjen
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
• Jul 25th 2006, 08:25 PM
ThePerfectHacker
Quote:

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 :eek: nothing to prove.
• Jul 25th 2006, 08:56 PM
ThePerfectHacker
Quote:

Originally Posted by jenjen
For every natural number n, for every natural number m, n + (Sm) = S(n + m);

Use this peano axiom.

Quote:

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,
$\displaystyle S((i+j)+k)=S(i+(j+K))$
But then by top axiom ya have,
$\displaystyle i+j+S(k)=i+S(j+k)$
But then by top axiom ya have,
$\displaystyle i+j+S(k)=i+j+S(k)$
• Jul 25th 2006, 08:58 PM
jenjen
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.
• Jul 25th 2006, 09:05 PM
ThePerfectHacker
Then you can do it by induction.
$\displaystyle S=\{x\in \mathbb{N}| 0+x=x\}$

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

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

Conditions for induction are satisfied, thus
$\displaystyle S=\mathbb{N}$
• Jul 25th 2006, 09:10 PM
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?"
• Jul 25th 2006, 09:13 PM
ThePerfectHacker
Quote:

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).
• Jul 25th 2006, 09:20 PM
jenjen

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??
• Jul 26th 2006, 04:42 AM
CaptainBlack
Quote:

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
• Jul 26th 2006, 11:50 AM
CaptainBlack
Quote:

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 )

Quote:

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

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

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

by induction on $\displaystyle x$.

Base case:

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

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

So the base case is true.

Induction step:

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

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

Now look at the successor of the LHS:

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

Now look at the successor of the RHS:

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

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

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

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

by induction on $\displaystyle x$

RonL