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issacnewton
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- Homework Statement
- Prove ##a \cdot 1 = a = 1 \cdot a## for ##a \in \mathbb{N}## using Peano postulates
- Relevant Equations
- Peano postulates
I have to prove ##a \cdot 1 = a = 1 \cdot a## for ##a \in \mathbb{N}##.
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 ##
with this background, we proceed to the proof. Let us define a set
$$ G = \{ z \in \mathbb{N} | z \cdot 1 = z = 1 \cdot z \} $$
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} ##. From the multiplication definition given above, we have ## 1 \cdot 1 = 1 = 1 \cdot 1##. And since ## 1 \in \mathbb{N}##, it follows that ## 1 \in G##. Suppose ## r \in G##. This means that ## r \in \mathbb{N} ## and
$$ r \cdot 1 = r = 1 \cdot r $$
Now, from part a) of multiplication operation given above, it follows that
$$ s(r) \cdot 1 = s(r) \cdots\cdots (1)$$
Now from part b) of multiplication operation given above, it follows that
$$ 1 \cdot s(r) = (1 \cdot r) + 1 $$
And since ## r \in G##, we have ## 1 \cdot r = r ##. Hence
$$ 1 \cdot s(r) = r + 1 $$
And using addition operation definition, ## r + 1 = s(r) ##, so
$$ 1 \cdot s(r) = s(r) \cdots\cdots (2) $$
Collecting equation 1 and 2, we have
$$ s(r) \cdot 1 = s(r) = 1 \cdot s(r) $$
Since ##s(r) \in \mathbb{N} ##, it follows 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 \in \mathbb{N} ##, ##a \in G## and it follows that
$$ a \cdot 1 = a = 1 \cdot a $$
Is the proof correct ?
Thanks
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 ##
with this background, we proceed to the proof. Let us define a set
$$ G = \{ z \in \mathbb{N} | z \cdot 1 = z = 1 \cdot z \} $$
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} ##. From the multiplication definition given above, we have ## 1 \cdot 1 = 1 = 1 \cdot 1##. And since ## 1 \in \mathbb{N}##, it follows that ## 1 \in G##. Suppose ## r \in G##. This means that ## r \in \mathbb{N} ## and
$$ r \cdot 1 = r = 1 \cdot r $$
Now, from part a) of multiplication operation given above, it follows that
$$ s(r) \cdot 1 = s(r) \cdots\cdots (1)$$
Now from part b) of multiplication operation given above, it follows that
$$ 1 \cdot s(r) = (1 \cdot r) + 1 $$
And since ## r \in G##, we have ## 1 \cdot r = r ##. Hence
$$ 1 \cdot s(r) = r + 1 $$
And using addition operation definition, ## r + 1 = s(r) ##, so
$$ 1 \cdot s(r) = s(r) \cdots\cdots (2) $$
Collecting equation 1 and 2, we have
$$ s(r) \cdot 1 = s(r) = 1 \cdot s(r) $$
Since ##s(r) \in \mathbb{N} ##, it follows 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 \in \mathbb{N} ##, ##a \in G## and it follows that
$$ a \cdot 1 = a = 1 \cdot a $$
Is the proof correct ?
Thanks