\[ n! = 1 \cdots k \cdots n \] therefore \[ \frac{n!}{k} = 1 \cdots \left( \frac{k}{k} \right) \cdots n \in \mathbb{ Z } \] so therefore \( k \mid n! \)

In the definition of divides take \( k = j \), then we can see that \( j = j \cdot 1 \) thus we can say that \( 1 | j \)

We have that \( 1 = k d \) but the only numbers in \( \mathbb{ Z } \) that have a multiplicative inverse are \( 1, -1 \) so \( d \) must be one of those or else we have a contradiction.

By definition we have \( d = k \cdot 0 = 0 \) so \( d = 0 \)

In the definition of divides take \( k = 0 \), then we can see that \( 0 = 0 \cdot d \) thus we can say that \( d | 0 \)

Clearly \( 1 \cdot a = a \) therefore \( a \mid a \)

We have that \( b = k _ a a \) and \( c = k _ b b \) therefore \( c = k _ b \left( k _ a a \right) \) therefore by taking \( k _ c = \left( k _ a k _ b \right) \) we see that \( a \mid c \)

Suppose that \( a \mid b \) therefore there is \( k \in \mathbb{ Z } \) such that \( b = ka \), therefore multiply both sides to get \( bc = \left( kc \right) a \) and therefore \( a \mid bc \) as needed.

We see that \( n = k d \) for some \( k \in \mathbb{ Z } \), since \( n \neq 0 \) then also we must have \( k, d \neq 0 \) specifically note that \( \lvert k \rvert \ge 1 \) so that we have \[ \lvert n \rvert = \lvert k d \rvert = \lvert k \rvert \cdot \lvert d \rvert \ge 1 \lvert d \rvert = \lvert d \rvert \] so that \( \lvert n \rvert \ge \lvert d \rvert \)

If \( a = 0 \) then because \( a \mid b \) then this forces \( b = 0 \) so that \( a = b = 0 \), also if we assumed \( b = 0 \) the same thing occurs.

In the case that neither of \( a, b = 0 \) then recall that if \( a \mid b \) we have \( \lvert a \rvert \le \lvert b \rvert \) also since \( b \mid a \) then also \( \lvert a \rvert \le \lvert b \rvert \) so that \( \lvert a \rvert = \lvert b \rvert \).

It is reflexive because we know that everything divides itself, we know it is transitive, and we also know that it is anti-symmetric, because for any \( n \in \mathbb{ N } _ 0, \lvert n \rvert = n \) and using this proposition.

For the base case of \( n = 2 \) if \( a \mid b _ 1 \) and \( a \mid b _ 2 \) then we know that there are \( k _ 1, k _ 2 \in \mathbb{ Z } \) such that \( b _ 1 = k _ 1 a \) and \( b _ 2 = k _ 2 a \) so that \( c _ 1 b _ 1 + c _ 2 b _ 2 = c _ 1 \left( k _ 1 a \right) + c _ 2 \left( k _ 2 a \right) = a \left( c _ 1 k _ 1 + c _ 2 k _ 2 \right) \).

For the induction step, assume the statement holds true for \( k \in \mathbb{ N } _ 1 \) and we'll show that it holds true for \( k + 1 \) so suppose that \( a \mid b _ 1, \ldots b _ { k + 1 } \) by the induction hypothesis we know that \( a \mid c _ 1 b _ 1 + \cdots + c _ k b _ k \) so we get some \( j ^ \prime \in \mathbb{ Z } \) such that \( c _ 1 b _ 1 + \cdots + c _ k b _ k = j ^ \prime a \) and also some \( j ^ { \prime \prime } \in \mathbb{ Z } \) such that \( b _ { k + 1 } = j ^ { \prime \prime } a \) and therefore \[ c _ 1 b _ 1 + \cdots + c _ k b _ k + c _ { k + 1 } b _ { k + 1 } = j ^ \prime a + j ^ { \prime \prime } a = a \left( j ^ \prime + j ^ { \prime \prime } \right) \] so therefore \( a \mid c _ 1 b _ 1 + \cdots + c _ { k + 1 } b _ { k + 1 } \)

If \( n \) is even then there is some \( k \in \mathbb{ Z } \) such that \( n = 2k \) and therefore \( n ^ m = \left( 2k \right) ^ m = 2 ^ m k ^ m = 2 \left( 2 ^ { m - 1 } k ^ m \right) \) so that \( n ^ m \) is also even.

Base case, set \( m = 0 \) then \( n ^ m = 1 \) which is odd. For the induction step assume it holds true for \( k \in \mathbb{ N } _ 0 \) so that \( n ^ k \) is odd, which means that there is some \( j \in \mathbb{ Z } \) such that \( n ^ k = 2j + 1 \) therefore \( n ^ { k + 1 } = \left( 2 j + 1 \right) n \) but \( n \) is odd so there is some \( j ^ \prime \in \mathbb{ Z } \) such that \( n = 2 j ^ \prime + 1 \) and thus \[ n ^ { k + 1 } = \left( 2 j + 1 \right) \left( 2 j ^ \prime + 1 \right) = 4j j ^ \prime + 2j + 2 j ^ \prime + 1 = 2 \left( 2 j j ^ \prime + j + j ^ \prime \right) + 1 \] hence \( n ^ { k + 1 } \) is prime so the statement follows from the principle of induction.

Suppose for the sake of contradiction that it was not true, so that we knew that \[ \left( b \le \sqrt{ a } \lor c \ge \sqrt{ a } \right) \land \left( b \ge \sqrt{ a } \lor c \le \sqrt{ a } \right) \] Suppose that \( b \le \sqrt{ a } \) and \( b \ge \sqrt{ a } \) holds true, then \( b = \sqrt{ a } \) since \( c \neq b \) then if \( c \lt b \) we see that \( b c \lt \sqrt{ a } \sqrt{ a } = a \) which is a contradiction since we know that \( b c = a \), if it turns out that \( c \le \sqrt{ a } \) and \( c \ge \sqrt{ a } \) it is handeled symetrically.

On the other hand if we know that \( b \le \sqrt{ a } \) and that \( c \le \sqrt{ a } \), since \( b \neq a \) then we know that \( b \neq \sqrt{ a } \), therefore more specifically we see that \( b \lt \sqrt{ a } \) this implies that \( a = b c \lt \sqrt{ a } \sqrt{ a } = a \) which is a contradiction. In the case of \( c \ge \sqrt{ a } \) and \( b \ge \sqrt{ a } \) it follows in a symmetrical manner.

We have that \( n = mk \) for some \( k \in \mathbb{ Z } \) and we want to prove that \( a ^ { mk } - 1 = \left( a ^ m - 1 \right) j \) for some \( j \in \mathbb{ Z } \), a good guess would be that \( j = a ^ { \left( m - 1 \right) k } \) which produces \( \left( a ^ m - 1 \right) \left( a ^ { \left( m - 1 \right)k } \right) = a ^ { mk } - a ^ { \left( m - 1 \right) k } \) and so we have to get rid of the \( a ^ { \left( m - 1 \right)k } \) term, which inspires us to try \( j = a ^ { \left( m - 1 \right)k } + a ^ { \left( m - 2 \right) k } \) which yields \[ \left( a ^ m - 1 \right) \left( a ^ { \left( m - 1 \right)k } + a ^ { \left( m - 2 \right) k } \right) = a ^ { km } + a ^ { \left( m - 1 \right)k } - \left( a ^ { \left( m - 1 \right) k } + a ^ { \left( m - 2 \right) k } \right) = a ^ { km } - a ^ { \left( m - 2 \right) k } \]

As we've just run into the same problem as last time we see that adding \( a ^ { \left( m - 3 \right) k } \) would remove the most recent residue term, but then leave the new residue term \( a ^ { \left( m - 3 \right) k } \) . Therefore formally using induction we can show that \[ \left( a ^ m - 1 \right) \left( \sum _ { i = 0 } ^ { m - 1 } a ^ { ik } \right) = a ^ { mk } - 1 = a ^ n - 1 \] where \( 1 \) is really the residue term \( a ^ { \left( 0 \right)k } \)