Problem 4 - A Gentle Introduction to the Art of Mathematics

Question

Prove that for all integers \(a, b,\) and \(c,\) if \(a \mid b\) and \(a \mid(b+c),\) then \(a \mid c\).

Step-by-Step Solution

Answer

If \(a \mid b\) and \(a \mid (b+c)\), then \(a \mid c\).

Step 1 - Understand the given conditions

We are given that for an integer divisor \(a\), it divides \(b\). This means there exists some integer \(k\) such that \[b = a \cdot k\]. Similarly, we are also given that \(a \mid (b+c)\), which means there exists some integer \(m\) such that \[b+c = a \cdot m\]. We need to prove that \(a \mid c\).

Step 2 - Substitute \(b\) in the second condition

Since \((b+c) = a \cdot m\), and we know that \(b = a \cdot k\) from the first condition, substitute \(b\) in the second condition: \[a \cdot k + c = a \cdot m\].

Step 3 - Simplify the equation

Rearrange the equation to solve for \(c\): \[c = a \cdot m - a \cdot k\]. This can be rewritten as \[c = a \cdot (m - k)\].

Step 4 - Conclude the proof

Because \(m - k\) is an integer, we have demonstrated that \(c\) is a multiple of \(a\). Therefore, \(a \mid c\). This completes the proof.

Key Concepts

Integer Division

In mathematics, the concept of 'integer division' is fundamental when dealing with divisors and multiples.
For example, when we say that an integer 'a' divides another integer 'b' (written as \(a \mid b\)), it means there is some integer 'k' such that \[b = a \cdot k\].
This k is known as the quotient, and 'a' is called a divisor of 'b'.
Consequently, 'b' is said to be divisible by 'a'. To internalize this, consider simple examples like 6 dividing 18 (\(6 \mid 18\)).
Here, there exists an integer quotient k (in this case, 3), such that \[18 = 6 \cdot 3\].
By unpacking such scenarios, understanding integer division becomes clearer and aligns with more complex mathematical operations. Integer division is the foundation for many proofs and algebraic manipulations.

Mathematical Proof

A mathematical proof is a logical argument demonstrating the truth of a given statement.
In mathematics, proving a statement involves deducing it based on previously established truths and axioms.
In the given exercise, we need to prove that for any integers 'a', 'b', and 'c', if \(a \mid b\) and \(a \mid (b+c)\), then \(a \mid c\).
To start, we interpret \(a \mid b\) as there being some integer 'k' such that \[ b = a \cdot k \].
Similarly, \(a \mid (b+c)\) implies there is some integer 'm' such that \[(b+c) = a \cdot m \].
By substituting our initial integer division into the new condition, we merge logic with algebra to draw our conclusions. Following consistent logical steps ultimately verifies the statement we set out to prove. Thus, understanding mathematical proofs emphasizes clear, logical thinking over simply getting the correct answer.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying algebraic expressions to solve problems or prove statements.
It is crucial in mathematical proofs where one needs to transition from one expression to another logically.
Consider the given step-by-step solution: we have \(b = a \cdot k\) and \(b + c = a \cdot m\).
Substituting \(b = a \cdot k\) into the second condition, the expression becomes \[a \cdot k + c = a \cdot m\].
To isolate 'c', we rearrange the equation: \[ c = a \cdot m - a \cdot k \].
By factoring out 'a', we get \[ c = a \cdot (m - k) \], showing 'c' as a product of 'a' and another integer, \(m - k\).
Hence, \(a \mid c\) is proven. This example highlights how purposeful manipulation allows for the extraction of critical insights, facilitating deeper comprehension and solution of mathematical challenges.