Problem 16 - Algebra and Trigonometry

Question

Show that \(n^{3}-n+3\) is divisible by 3 for all natural numbers \(n .\)

Step-by-Step Solution

Answer

\(n^3 - n + 3\) is divisible by 3 for all natural numbers \(n\) by induction.

Step 1: Understand what is being asked

We need to prove that the expression \(n^3 - n + 3\) is divisible by 3 for all natural numbers \(n\). This means that \(n^3 - n + 3\) when divided by 3 should give a remainder of 0.

Step 2: Use mathematical induction

We can use mathematical induction to prove that the statement is true for all natural numbers \(n\). With induction, first prove the base case for \(n=1\), and then show that if the statement is true for \(n=k\), it must also be true for \(n=k+1\).

Step 3: Base case

Check the base case where \(n = 1\).\(1^3 - 1 + 3 = 3.\)3 is divisible by 3. Hence, the base case holds.

Step 4: Inductive step assumption

Assume that the expression \(k^3 - k + 3\) is divisible by 3 for some natural number \(k\), i.e.,\(k^3 - k + 3 = 3m\)for some integer \(m\).

Step 5: Inductive step - check \(n = k+1\)

Now, check the expression for \(n = k+1\):\((k+1)^3 - (k+1) + 3.\)Calculate \((k+1)^3\):\((k+1)^3 = k^3 + 3k^2 + 3k + 1.\)Substitute to form the expression:\((k+1)^3 - (k+1) + 3 = k^3 + 3k^2 + 3k + 1 - k - 1 + 3.\)Simplify this:\(k^3 + 3k^2 + 2k + 3.\)Recollect from our assumption:\(k^3 = 3m + k - 3,\)Substitute back:\(3m + k - 3 + 3k^2 + 2k + 3.\)Combine:\(3m + 3k^2 + 3k,\)which factors to:\(3(m + k^2 + k).\)This expression is clearly divisible by 3.

Step 6: Conclusion

Since both the base case and the inductive step have been verified, by mathematical induction, \(n^3 - n + 3\) is divisible by 3 for all natural numbers \(n\).

Key Concepts

Divisibility

Divisibility is an essential concept in mathematics and aids in understanding how certain numbers can be evenly divided by others without leaving a remainder. In this context, it is about proving that certain expressions are divisible by a specified number, which here is 3.

To check if an expression like \(n^3 - n + 3\) is divisible by 3, we need to see if, upon division, the remainder is zero.
This ensures the expression can fit into exact groups of 3, without anything left over.

In each calculation step, whether it's for the base case with a small number, or a generalized number \(n\), ensuring divisibility guides the overall proof.
Involving properties of numbers and simplifying complex expressions to show clean division is the crux of divisibility in algebraic proofs.

Natural Numbers

Natural numbers are the set of positive integers starting from 1, extending infinitely. They form the basic counting numbers we use daily.
The natural numbers are \(1, 2, 3, \ldots\) without end. In mathematical problems, these numbers come in handy for building simple and relatable proofs.

When dealing with mathematical induction, the principle applies to natural numbers, ensuring that a base case starts from the smallest natural number, typically 1 or 0, when applicable.

• These numbers are innate for constructing proofs on statements that apply infinitely, as in this problem, where we prove something for all natural \(n\).
• Since the behavior of our function as \(n\) progresses is crucial, verifying our expression's behavior for all natural numbers is integral to proving continuous properties like divisibility.

Algebraic Proofs

Algebraic proofs are logical sequences of statements used to demonstrate mathematical truths. They rely on established algebraic identities and properties to show that an expression holds universally for all values in consideration.

In this case, we are using mathematical induction to make an algebraic proof: proving a base case, then using a logical chain that moves from one case \(n=k\) to the next \(n=k+1\).

This approach ensures that once a base case is proven, each subsequent case holds as well, forming a domino effect of sorts. Using algebraic manipulations during the proof, such as expanding \((k+1)^3\), abstract thought turns into verifiable facts.

• Algebraic proofs not only show one-time truths but provide a structure to understand the continuity across a sequence of numbers.
• They allow us to take specific step-based approaches like simplifying expressions and rearranging terms to reveal inherent characteristics such as divisibility.

Algebraic proofs, especially in the context of induction, are powerful because they convert small-scale observations into broad, universal rules.