- #1
issacnewton
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- Homework Statement
- Prove ##1 + a=s(a)=a+1## for ##a \in \mathbb{N}## where ##s## is a successor function from Peano postulates
- Relevant Equations
- Peano postulates
I have to prove that ##1 + a = s(a) = a + 1## using Peano postulates if ##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) ##
With this background, we now proceed to present my proof. Let us define a set
$$ G = \{ z \in \mathbb{N} | 1 + z = s(z) = z + 1 \} $$
We want to prove that ##G = \mathbb{N} ##.
Obviously, ## G \subseteq \mathbb{N} ##. Using definition of addition, ## 1 + 1 = s(1) = 1 + 1##. Since ##1 \in \mathbb{N} ##, it follows that ##1 \in G ##.
Now, suppose ## r \in G##. This means that ##r \in \mathbb{N}## and
$$ 1 + r = s(r) = r + 1 \cdots\cdots (1) $$
Since ##s: \mathbb{N} \rightarrow \mathbb{N} ## is a function, we have,
$$ 1 + r = s(r) $$
$$ s(1+r) = s(s(r)) $$
Using addition definition, part (b),
$$ 1 + s(r) = s(s(r)) $$
And from addition definition, part (a), we have ##s(s(r)) = s(r) + 1 ##. We got
$$ 1 + s(r) = s(s(r)) = s(r) + 1 $$
Since ## s(r) \in \mathbb{N} ##, we have ## s(r) \in G ##. So, using Peano postulates, part 3), we have ##G = \mathbb{N} ##. Now if ## a \in \mathbb{N} ## is arbitrary, ##a \in G ##. It follows that ##1 + a = s(a) = a + 1##. And since ##a## is arbitrary,
$$ \forall a \in \mathbb{N}\quad 1 + a = s(a) = a + 1 $$
Is the proof good enough ?
Thanks
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) ##
With this background, we now proceed to present my proof. Let us define a set
$$ G = \{ z \in \mathbb{N} | 1 + z = s(z) = z + 1 \} $$
We want to prove that ##G = \mathbb{N} ##.
Obviously, ## G \subseteq \mathbb{N} ##. Using definition of addition, ## 1 + 1 = s(1) = 1 + 1##. Since ##1 \in \mathbb{N} ##, it follows that ##1 \in G ##.
Now, suppose ## r \in G##. This means that ##r \in \mathbb{N}## and
$$ 1 + r = s(r) = r + 1 \cdots\cdots (1) $$
Since ##s: \mathbb{N} \rightarrow \mathbb{N} ## is a function, we have,
$$ 1 + r = s(r) $$
$$ s(1+r) = s(s(r)) $$
Using addition definition, part (b),
$$ 1 + s(r) = s(s(r)) $$
And from addition definition, part (a), we have ##s(s(r)) = s(r) + 1 ##. We got
$$ 1 + s(r) = s(s(r)) = s(r) + 1 $$
Since ## s(r) \in \mathbb{N} ##, we have ## s(r) \in G ##. So, using Peano postulates, part 3), we have ##G = \mathbb{N} ##. Now if ## a \in \mathbb{N} ## is arbitrary, ##a \in G ##. It follows that ##1 + a = s(a) = a + 1##. And since ##a## is arbitrary,
$$ \forall a \in \mathbb{N}\quad 1 + a = s(a) = a + 1 $$
Is the proof good enough ?
Thanks