Question

Show that if \(a c \mid b c\), then \(a \mid b\).

Short answer

Question: Prove that if \(ac \mid bc\), then \(a \mid b\). Answer: Using the definition of divisibility and the cancellation law, we can show that if \(ac = kc\) for some integer \(k\), then it follows that \(a \mid b\) as \(b = lk\) for some integer \(l\). Therefore, if \(ac \mid bc\), then \(a \mid b\).

Step-by-step solution

Understand the definition of divisibility

Divisibility is the ability of one integer to be divided by another integer without leaving a remainder. In this problem, \(ac \mid bc\) can be written as \(bc = k(ac)\) for some integer \(k\). Similarly, if \(a \mid b\) it means that \(b = l(a)\) for some integer \(l\).

Use the cancellation law

Cancellation law states that if \(ab = ac\) and \(a \neq 0\), then \(b = c\). To apply the cancellation law, we need to show that \(ab = ac\) for some integers \(b\) and \(c\). We are given that \(bc = k(ac)\). If we rearrange the terms, we get \(acb = kbc\). If we let \(a \neq 0\) and \(b \neq 0\), we can cancel \(b\) from both sides of the equation. This gives us: \(ac = kc\).

Apply the definition of divisibility

Now that we showed \(ac = kc\), we can apply the definition of divisibility to show that \(a \mid b\). We know that \(b = l(a)\) for some integer \(l\). We also know that \(ac = kc\) from Step 2. So, we can write \(b = l(ac)/c = l(kc)/c\). By simplifying the equation, we get \(b = lk\). Since \(l\) and \(k\) are both integers, the product \(lk\) is also an integer. Thus, we can conclude that \(a \mid b\).

Key concepts

Cancellation Law

The cancellation law is a powerful tool in number theory. It allows us to simplify equations by removing common factors from both sides. Suppose we have an equation of the form \( ab = ac \), and we know that \( a eq 0 \). The cancellation law tells us that \( b = c \) if \( a \) is not zero. This principle is crucial because it provides a method to simplify expressions and solve problems involving divisibility more effectively.

In the context of the problem we are dealing with, the cancellation law is used to isolate terms and solve for divisibility. We start with the equation derived from the divisibility condition \( bc = k(ac) \). By rearranging, we achieve \( ab = ac \). With the non-zero condition \( a eq 0 \) \( b eq 0 \), we can cancel \( b \) to simplify it further to \( ac = kc \). This simplification, enabled by the cancellation law, is critical in proving subsequent divisibility properties.

Definition of Divisibility

Divisibility is a fundamental concept in mathematics, especially in number theory. It implies that one number can be divided by another without leaving a remainder. For instance, when we say \( a \mid b \), it means \( b \) is exactly divisible by \( a \), or that there exists an integer \( l \) such that \( b = l(a) \).

In more practical terms, this translates to performing division and getting a whole number without any leftover. For our original problem, understanding this definition allows us to translate \( ac \mid bc \) into the expression \( bc = k(ac) \) for some integer \( k \). This step of expressing divisibility conditions in equations is crucial for further mathematical operations, including employing laws like the cancellation law.

Proving Divisibility Properties

To prove properties of divisibility, we rely on definitions and mathematical laws like the cancellation law. The original exercise task was to prove that if \( ac \mid bc \), then \( a \mid b \). By restating the divisibility condition \( ac \mid bc \) as \( bc = k(ac) \), where \( k \) is an integer, we transform the problem into an equation.

After applying the cancellation law to simplify our equation to \( ac = kc \), we pivot back to the definition of divisibility. Since \( b \) can be expressed in terms of \( a \) such as \( b = l(a) \) for integer \( l \), and \( k \) is an integer, it implies \( b = lk \), which is an integer too. Hence, \( a \mid b \) is proven.

This step-by-step logical reasoning showcases the power of mathematical proofs in reinforcing the properties of integers and highlights how foundational concepts interconnect to solve complex problems.