Question

Prove the following statements. These exercises are cumulative, covering all techniques addressed in Chapters \(4-7\). Suppose \(a, b\) and \(c\) are integers. If \(a \mid b\) and \(a \mid\left(b^{2}-c\right),\) then \(a \mid c\).

Short answer

Yes, the proposition is true. If \(a\) divides both \(b\) and \((b^{2}-c)\), then \(a\) divides \(c\).

Step-by-step solution

Definition of Divisibility

The statement that \(a\) divides \(b\) (or \(b\) is divisible by \(a\)) means there exists an integer \(k\) such that \(b=ak\). Similarly, the statement \(a\) divides \((b^{2}-c)\) means there exists an integer \(m\) such that \(b^{2}-c=am\). So, express \(b\) and \(b^{2}-c\) in terms of \(a\).

Express \(c\) in terms of \(a\)

Since \(b=ak\) and \(b^{2}-c=am\), we can express \(c\) in terms of \(a\). Replace \(b\) with \(ak\) in the equation for \(b^{2}-c\). We obtain \((ak)^{2}-c=am\), which simplifies to \(a^{2}k^{2}-c=am\). Solving for \(c\), we obtain \(c=a^{2}k^{2}-am\).

Proving \(a\) divides \(c\)

Now, we have expressed \(c\) as \(c=a^{2}k^{2}-am\), which can be rewritten as \(c=a(ak^{2}-m)\). Here, \((ak^{2}-m)\) is an integer because the product or sum of integers is an integer. Therefore, by the definition of divisibility, \(a\) divides \(c\)

Key concepts

Definition of Divisibility

When we discuss divisibility in mathematics, we're talking about one number being able to divide another without leaving a remainder. More precisely, for two integers a and b, if a divides b, designated as a | b, there exists an integer k such that b = ak. This relationship is foundational to understanding various concepts in number theory.

For example, consider the statement 6 | 18. This can be visualized as dividing 18 into 6 equal parts, each part being 3, which confirms that indeed 6 divides 18 with no remainder. Likewise, in the given exercise where it’s stated that a | b and a | (b^2 - c), it implies the existence of integers k and m, satisfying b = ak and b^2 - c = am. This forms the underpinnings of the proof.

Mathematical Induction

Mathematical induction is a proof technique used to establish that a statement is true for all natural numbers. It works by first proving the base case, which is typically the statement applied to the smallest natural number. Then, you show that if the statement holds for an arbitrary natural number n, it also holds for n + 1. By connecting these two steps, you confirm the statement is true for all subsequent natural numbers.

Although the problem at hand does not directly involve proving a statement by mathematical induction, understanding this technique is vital for a well-rounded grasp of mathematical proof strategies. It’s like creating a domino effect; once the initial domino (the base case) topples over, it ensures the rest (succeeding cases) will follow suit.

Integer Properties

Integers are like building blocks in number theory and have specific properties that are widely utilized in proofs. For instance, integers are closed under addition, subtraction, and multiplication. This means that whenever you perform these operations on integers, the result will always be an integer. However, this is not the case with division, which can result in non-integers.

In the context of the given exercise, these properties assure us that ak^2 and m are integers since a, k, m are integers, and hence their sum or product is also an integer. This closure property under multiplication and addition plays a crucial role in proving that a divides c, as the expression for c is derived using these operations.

Proof Techniques

Proving mathematical statements requires strategic application of various proof techniques. Some common methods include direct proof, proof by contradiction, and proof by contraposition. In a direct proof, you start with known facts and use logical steps to arrive at the conclusion. Proof by contradiction involves assuming the opposite of what you want to prove and showing this assumption leads to a contradiction. Proof by contraposition means showing that if the conclusion is false, then the premises must also be false.

In our step by step solution, we use a direct proof technique to demonstrate that a divides c. We start with the known facts (the definitions and properties of integers), and we logically deduce that ak^2 and m being integers mean that their combination (ak^2 - m) is also an integer, thus fulfilling the conditions for divisibility.