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issacnewton

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- Homework Statement
- Prove ##a + b = b + a## using Peano postulates

- Relevant Equations
- Peano postulates

Following is a set of Peano postulates I am using as defined in the book "Th real numbers and real analysis" by Ethan Bloch.

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} ##I have to prove the Commutative Law for addition ## a + b = b + a## using Peano postulates, given that ##a, b\in \mathbb{N}##. Now define the set

$$ G = \{ x \in \mathbb{N} |\forall\; y \in \mathbb{N} \quad (x + y) = (y + x) \} $$

I have proven previously, that for ## a \in \mathbb{N}##, we have,

$$ 1 + a = s(a) = a + 1 \cdots\cdots (1) $$

So, for some ##y \in \mathbb{N} ##, we get ## 1 + y = y + 1 ##. That proves that ## 1 \in G ##. Now, suppose that ## r \in G##. This means that

$$ \forall\; y \in \mathbb{N} \quad (r + y) = (y + r) \cdots\cdots (2) $$

We need to prove that

$$ \forall\; y \in \mathbb{N} \quad s(r) + y = y + s(r) $$

Let ## y \in \mathbb{N} ## be arbitrary. From ##(2)##, we get ## (r + y) = (y + r)##. It follows that ##s(r + y) = s(y + r) ##. Now, addition function is defined as follows in this book

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} ##

$$ n + 1 = s(n) $$

$$ n + s(m) = s(n + m) $$

Using this, we get, ## s(r + y) = y + s(r)##. Now, using ##(1)##, we have, ## 1 + (r + y) = y + s(r) ##. I have also previously proven Associative Law for Addition. So, using that, we get, ## (1 + r) + y = y + s(r) ##. Again using ##(1)##, this becomes ## s(r) + y = y + s(r) ##. Since ## y \in \mathbb{N} ## is arbitrary, ## s(r) \in G##. So, for some ## r \in G## it implies that ## s(r) \in G##. From Peano postulates, this means that ## G = \mathbb{N} ##.

Now let ##a, b \in \mathbb{N}## be arbitrary. So, ## a \in G##. It follows that ## (a + b) = (b + a) ##

Is this a sound proof ?

Thanks

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} ##I have to prove the Commutative Law for addition ## a + b = b + a## using Peano postulates, given that ##a, b\in \mathbb{N}##. Now define the set

$$ G = \{ x \in \mathbb{N} |\forall\; y \in \mathbb{N} \quad (x + y) = (y + x) \} $$

I have proven previously, that for ## a \in \mathbb{N}##, we have,

$$ 1 + a = s(a) = a + 1 \cdots\cdots (1) $$

So, for some ##y \in \mathbb{N} ##, we get ## 1 + y = y + 1 ##. That proves that ## 1 \in G ##. Now, suppose that ## r \in G##. This means that

$$ \forall\; y \in \mathbb{N} \quad (r + y) = (y + r) \cdots\cdots (2) $$

We need to prove that

$$ \forall\; y \in \mathbb{N} \quad s(r) + y = y + s(r) $$

Let ## y \in \mathbb{N} ## be arbitrary. From ##(2)##, we get ## (r + y) = (y + r)##. It follows that ##s(r + y) = s(y + r) ##. Now, addition function is defined as follows in this book

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} ##

$$ n + 1 = s(n) $$

$$ n + s(m) = s(n + m) $$

Using this, we get, ## s(r + y) = y + s(r)##. Now, using ##(1)##, we have, ## 1 + (r + y) = y + s(r) ##. I have also previously proven Associative Law for Addition. So, using that, we get, ## (1 + r) + y = y + s(r) ##. Again using ##(1)##, this becomes ## s(r) + y = y + s(r) ##. Since ## y \in \mathbb{N} ## is arbitrary, ## s(r) \in G##. So, for some ## r \in G## it implies that ## s(r) \in G##. From Peano postulates, this means that ## G = \mathbb{N} ##.

Now let ##a, b \in \mathbb{N}## be arbitrary. So, ## a \in G##. It follows that ## (a + b) = (b + a) ##

Is this a sound proof ?

Thanks