Show that if $a\mid b$ and $b\mid a$ then $a=\pm b$ .

Step 1: Definition of Divisibility Divisibility property is defined as follows: For two integers $a$ and $b$, $a$ is said to divide $b$ (written as $a\mid b$) if there exists an integer $k$ such that $b=a\cdot k$. Step 2: Use the Given Conditions The given information states that $a\mid b$ and $b\mid a$. We can use the definition of the divisibility property to obtain the following equations:

1. $b=a\cdot k_1$ where $k_1$ is an integer.
2. $a=b\cdot k_2$ where $k_2$ is an integer.

Step 3: Substitution Substitute the expression for $b$ from equation 1 into equation 2 to get:
$a=(a\cdot k_1)\cdot k_2$ Step 4: Simplification and Derivation Simplify the equation obtained from Step 3 to derive the result:
$a=a\cdot k_1\cdot k_2$
Dividing both sides by $a$ gives:
$1=k_1\cdot k_2$
As both $k_1$ and $k_2$ are integers, there are only two possibilities: either $k_1=1$ and $k_2=1$, or $k_1=-1$ and $k_2=-1$. Using these values, we can find the relationship between $a$ and $b$.

1. If $k_1=1$ and $k_2=1$, using equation 1
   $b=a\cdot k_1$
   we get
   $b=a\cdot 1$ or $b=a$.
2. If $k_1=-1$ and $k_2=-1$, using equation 1
   $b=a\cdot k_1$
   we get
   $b=a\cdot (-1)$ or $b=-a$

Therefore, we have shown that if $a\mid b$ and $b\mid a$, then $a=\pm b$. Step 5: Final Result If $a\mid b$ and $b\mid a$, then $a=\pm b$.