Prove ##(a\cdot b)\cdot c =a\cdot (b \cdot c)## using Peano postulates

[issacnewton]

Homework Statement

Prove ##(a\cdot b)\cdot c =a\cdot (b \cdot c)## using Peano postulates given that ##a,b,c \in \mathbb{N}##

Relevant Equations

The book I am using ("The real numbers and real analysis" by Ethan Bloch) defines Peano postulates little differently.

Following is a set of Peano postulates I am using. (Axiom 1.2.1 in Bloch's book)


There exists a set ##\mathbb{N}## with an element ##1 \in \mathbb{N}## and a function ##s: \mathbb{N} \rightarrow \mathbb{N} ## that satisfy the following three properties.

1) There is no ##n \in \mathbb{N}## such that ##s(n) = 1##
2) The function ##s## is injective.
3) Let ##G \subseteq \mathbb{N}## be a set. Suppose that ##1 \in G##, and that if ##g \in G## then ##s(g) \in G##. Then ## G = \mathbb{N} ##

And addition operation is given in Theorem 1.2.5 as follows

There is a unique binary operation ##+: \mathbb{N} \times \mathbb{N} \to \mathbb{N} ## that satisfies the following two properties for all ##n, m \in \mathbb{N} ##

a. ## n + 1 = s(n) ##
b. ## n + s(m) = s(n + m) ##

Multiplication operation is given in Theorem 1.2.6 as follows

There is a unique binary operation ## \cdot: \mathbb{N} \times \mathbb{N} \to \mathbb{N} ## that satisfies the following two properties for all ##n, m \in \mathbb{N} ##

a. ##n \cdot 1 = n ##
b. ## n \cdot s(m) = (n \cdot m) + n ##

I am also going to assume following results for this proof.
Identity law for multiplication for ##a \in \mathbb{N} ##
$$ a \cdot 1 = a = 1 \cdot a $$

Distributive law

$$ c \cdot (a + b) = c \cdot a + c \cdot b $$

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with this background, we proceed to the proof. Let us define a set

$$ G = \{ z \in \mathbb{N} | \; x, y \in \mathbb{N}\; (x \cdot y) \cdot z = x \cdot (y \cdot z) \} $$

We want to prove that ##G = \mathbb{N} ##. For this purpose, we will use part 3) of Peano postulates given above. Obviously, ## G \subseteq \mathbb{N} ##. Let ##x, y \in \mathbb{N}## be arbitrary. Using Identity law for multiplication, we have ##(x \cdot y) \cdot 1 = (x \cdot y) ## and ##x \cdot (y \cdot 1) = x \cdot y = (x \cdot y) ##. Since ##1 \in \mathbb{N}## from Peano postulates, it follows that ##1 \in G##. Suppose ## r \in G##. So ##r \in \mathbb{N}## and

$$ \forall x,y \in \mathbb{N} \; \bigl[ (x \cdot y) \cdot r = x \cdot (y \cdot r) \bigr] \cdots\cdots (1) $$

Now, let ##x, y \in \mathbb{N}## be arbitrary. Now, using addition definition part a), ##(x \cdot y) \cdot s(r) = (x \cdot y) \cdot (r+1)##. Now using Distributive law, we have ##(x \cdot y) \cdot s(r) = (x \cdot y) \cdot r + (x \cdot y) \cdot 1##. Since ##r \in G##, we have , from equation (1), that ##(x \cdot y) \cdot r = x \cdot (y \cdot r) ##. So, we have ##(x \cdot y) \cdot s(r) = x \cdot (y \cdot r) + (x \cdot y) \cdot 1##. And using Identity law for multiplication, ##(x \cdot y) \cdot 1 = (x \cdot y)##. It means that ##(x \cdot y) \cdot s(r) = x \cdot (y \cdot r) +(x \cdot y)##. Now, using distributive law, ##(x \cdot y) \cdot s(r) = x \cdot (y \cdot r + y)##. And, using definition of multiplication part b), ##y \cdot r + y = y \cdot s(r)##. So, we have ##(x \cdot y) \cdot s(r) = x \cdot (y \cdot s(r))##. And since ##x, y \in \mathbb{N}## are arbitrary, it proves that

$$ \forall x,y \in \mathbb{N} \; \bigl[ (x \cdot y) \cdot s(r) = x \cdot (y \cdot s(r)) \bigr] $$

Since ##s(r) \in \mathbb{N}##, this proves that ##s(r) \in G##. So, ##r \in G## implies that ## s(r) \in G##. Using part 3) of the Peano postulates, it follows that ## G = \mathbb{N}##.

Now, if ##a,b,c \in \mathbb{N}##, we have ##c \in G## and it follows that ##(a\cdot b)\cdot c =a\cdot (b \cdot c)##.

Is this a valid proof ?

Thanks

 

[PeroK]

Looks good.