Show that if $a$ and $b$ are positive integers and $a\mid b$ , then $a \leq b$.

Step 1: Define Divisibility Recall the definition of divisibility: A positive integer $a$ divides another positive integer $b$ (denoted as $a\mid b$ ) if there exists a non-negative integer $k$ such that $b=a\cdot k$ . Step 2: Use the definition of Divisibility Given that $a\mid b$, we know that there exists some non-negative integer $k$ such that $b=a\cdot k$ . Step 3: Consider the different cases for k We need to check both cases for the value of $k$ : $k=0$ and $k>0$ . Case 1: $k=0$ In this case, $b=a\cdot k=a\cdot 0=0$ . However, this contradicts the given condition that $b$ is a positive integer. So, this case is not possible. Case 2: $k>0$ In this case, we have $b=a\cdot k$, where $k$ is a positive integer. Since $a$ is also a positive integer, their product ($a\cdot k$) must be greater than or equal to $a$ (i.e., $a\cdot k\ge a$ ). Therefore, $b\ge a$ , which implies that $a\le b$ . Step 4: Conclusion Since we have shown that $a\le b$ for any positive integers $a$ and $b$ with $a\mid b$, the statement is proven to be true.