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issacnewton
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- Homework Statement
- Prove ##(a+b)\cdot c=a\cdot c+b\cdot c## using Peano postulates where ##a,b,c \in \mathbb{N}##
- Relevant Equations
- Peano postulates
I want to prove that ##(a+b)\cdot c=a\cdot c+b\cdot c## using Peano postulates where ##a,b,c \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
$$ G = \{ z \in \mathbb{N} | \mbox{if } x, y \in \mathbb{N} \; (x+y)\cdot z = x \cdot z + y \cdot z \} $$
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} ##. Now, I am going to use another result I have proven earlier. If ##a \in \mathbb{N}##, we have ##a \cdot 1 = a = 1 \cdot a##. So, if ##x, y \in \mathbb{N}## are arbitrary, it will follow that ##(x + y) \cdot 1 = (x + y) = x \cdot 1 + y \cdot 1##. Since ## 1 \in \mathbb{N}## from Peano postulates, it follows that ## 1 \in G##. Now suppose ## r \in G##. It means that ##r \in \mathbb{N}## and
$$ \forall x, y \in \mathbb{N} \bigl[ (x + y) \cdot r = x \cdot r + y \cdot r \bigr] \cdots\cdots (1)$$
Now, let ##x, y \in \mathbb{N}## be arbitrary. Now using multiplication operation part b), it will follow that
$$ (x + y) \cdot s(r) = ((x + y) \cdot r) + (x + y) $$
Using equation 1), it will follow that
$$ (x + y) \cdot s(r) = (x \cdot r + y \cdot r) + (x + y) $$
I have already proven associative law for addition. So, using that,
$$ (x + y) \cdot s(r) = ((x \cdot r + y \cdot r) + x) + y $$
$$ (x + y) \cdot s(r) = (x \cdot r + (y \cdot r + x)) + y $$
I have also proven commutative law for addition earlier. So, we use that now
$$ (x + y) \cdot s(r) = (x \cdot r + (x + y \cdot r)) + y $$
Using associative law for addition again,
$$ (x + y) \cdot s(r) = ((x \cdot r + x) + y \cdot r) + y $$
Now using part b) of the addition operation, ## x \cdot s(r) = x \cdot r + x ##. So, we have
$$ (x + y) \cdot s(r) = (x \cdot s(r) + y \cdot r) + y $$
Using associative law for addition again,
$$ (x + y) \cdot s(r) = x \cdot s(r) + (y \cdot r + y) $$
And, again using part b) of the addition operation, ## y \cdot s(r) = y \cdot r + y ##. So, we have
$$ (x + y) \cdot s(r) = x \cdot s(r) + y \cdot s(r) $$
Since ##x,y## are arbitrary, it follows that
$$ \forall x, y \in \mathbb{N} \bigl[ (x + y) \cdot s(r) = x \cdot s(r) + y \cdot s(r) \bigr] $$
And ##s(r) \in \mathbb{N}##. So, ##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 ##a,b,c, \in \mathbb{N}##. So, ##c \in G##. And it will follow that
$$ (a+b)\cdot c=a\cdot c+b\cdot c $$
Is this a valid proof ?
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
$$ G = \{ z \in \mathbb{N} | \mbox{if } x, y \in \mathbb{N} \; (x+y)\cdot z = x \cdot z + y \cdot z \} $$
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} ##. Now, I am going to use another result I have proven earlier. If ##a \in \mathbb{N}##, we have ##a \cdot 1 = a = 1 \cdot a##. So, if ##x, y \in \mathbb{N}## are arbitrary, it will follow that ##(x + y) \cdot 1 = (x + y) = x \cdot 1 + y \cdot 1##. Since ## 1 \in \mathbb{N}## from Peano postulates, it follows that ## 1 \in G##. Now suppose ## r \in G##. It means that ##r \in \mathbb{N}## and
$$ \forall x, y \in \mathbb{N} \bigl[ (x + y) \cdot r = x \cdot r + y \cdot r \bigr] \cdots\cdots (1)$$
Now, let ##x, y \in \mathbb{N}## be arbitrary. Now using multiplication operation part b), it will follow that
$$ (x + y) \cdot s(r) = ((x + y) \cdot r) + (x + y) $$
Using equation 1), it will follow that
$$ (x + y) \cdot s(r) = (x \cdot r + y \cdot r) + (x + y) $$
I have already proven associative law for addition. So, using that,
$$ (x + y) \cdot s(r) = ((x \cdot r + y \cdot r) + x) + y $$
$$ (x + y) \cdot s(r) = (x \cdot r + (y \cdot r + x)) + y $$
I have also proven commutative law for addition earlier. So, we use that now
$$ (x + y) \cdot s(r) = (x \cdot r + (x + y \cdot r)) + y $$
Using associative law for addition again,
$$ (x + y) \cdot s(r) = ((x \cdot r + x) + y \cdot r) + y $$
Now using part b) of the addition operation, ## x \cdot s(r) = x \cdot r + x ##. So, we have
$$ (x + y) \cdot s(r) = (x \cdot s(r) + y \cdot r) + y $$
Using associative law for addition again,
$$ (x + y) \cdot s(r) = x \cdot s(r) + (y \cdot r + y) $$
And, again using part b) of the addition operation, ## y \cdot s(r) = y \cdot r + y ##. So, we have
$$ (x + y) \cdot s(r) = x \cdot s(r) + y \cdot s(r) $$
Since ##x,y## are arbitrary, it follows that
$$ \forall x, y \in \mathbb{N} \bigl[ (x + y) \cdot s(r) = x \cdot s(r) + y \cdot s(r) \bigr] $$
And ##s(r) \in \mathbb{N}##. So, ##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 ##a,b,c, \in \mathbb{N}##. So, ##c \in G##. And it will follow that
$$ (a+b)\cdot c=a\cdot c+b\cdot c $$
Is this a valid proof ?
Thanks