Question

Prove by induction for all \(n \in \mathbb{N}\) : (a) 7 divides \(n^{7}-n\) (b) 11 divides \(n^{11}-n\) (c) 13 divides \(n^{13}-n\) (d) 120 divides \(n^{5}-5 n^{3}+4 n\)

Short answer

(a) 7 divides, (b) 11 divides, (c) 13 divides, (d) 120 divides, all by induction.

Step-by-step solution

Base Case (a)

Let's prove that 7 divides \(n^{7}-n\) for \(n = 0\). Calculate \(0^{7} - 0 = 0\). Clearly, 7 divides 0. So, the statement holds for \(n = 0\).

Inductive Hypothesis (a)

Assume that the statement is true for some integer \(k\), meaning that 7 divides \(k^7 - k\). This implies \(k^7 - k = 7m\) for some integer \(m\).

Inductive Step (a)

We need to show that 7 divides \((k+1)^7 - (k+1)\). Expand \((k+1)^7 - (k+1)\):\((k+1)^7 = k^7 + 7k^6 + 21k^5 + 35k^4 + 35k^3 + 21k^2 + 7k + 1\).Therefore,\((k+1)^7 - (k+1) = (k^7 - k) + (7k^6 + 21k^5 + 35k^4 + 35k^3 + 21k^2 + 7k)\).Each term in the parentheses is divisible by 7, so the whole expression is divisible by 7. Thus, by induction, 7 divides \(n^7 - n\) for all \(n \in \mathbb{N}\).

Base Case (b)

Prove that 11 divides \(n^{11} - n\) for \(n = 0\). Calculate \(0^{11} - 0 = 0\).Clearly, 11 divides 0. So, the statement holds for \(n = 0\).

Inductive Hypothesis (b)

Assume that the statement is true for some integer \(k\), meaning that 11 divides \(k^{11} - k\). Therefore, \(k^{11} - k = 11m\) for some integer \(m\).

Inductive Step (b)

Show that 11 divides \((k+1)^{11} - (k+1)\). Expand \((k+1)^{11} - (k+1)\):\((k+1)^{11} = k^{11} + 11k^{10} + \cdots + 1\).Thus,\((k+1)^{11} - (k+1) = (k^{11} - k) + (11k^{10} + ...)\).As each term of \(11k^{10} + ...\) is divisible by 11, the whole expression is as well. Hence, by induction, 11 divides \(n^{11} - n\) for all \(n \in \mathbb{N}\).

Base Case (c)

Prove that 13 divides \(n^{13} - n\) for \(n = 0\). Calculate \(0^{13} - 0 = 0\).Since 13 divides 0, the statement holds for \(n = 0\).

Inductive Hypothesis (c)

Assume that the statement is true for some integer \(k\), meaning 13 divides \(k^{13} - k\). Thus, \(k^{13} - k = 13m\) for some integer \(m\).

Inductive Step (c)

We show that 13 divides \((k+1)^{13} - (k+1)\). Expand \((k+1)^{13} - (k+1)\):\((k+1)^{13} = k^{13} + 13k^{12} + \cdots + 1\).Then,\((k+1)^{13} - (k+1) = (k^{13} - k) + (13k^{12} + \cdots + 13k)\).Since every term on the right is divisible by 13, it follows that the expression is also divisible by 13. Thus, by induction, 13 divides \(n^{13} - n\) for all \(n \in \mathbb{N}\).

Base Case (d)

Prove that 120 divides \(n^{5} - 5n^{3} + 4n\) for \(n = 0\). Calculate \(0^{5} - 5(0)^{3} + 4(0) = 0\).As 120 divides 0, the statement holds for \(n = 0\).

Inductive Hypothesis (d)

Assume the statement is true for some integer \(k\), meaning 120 divides \(k^5 - 5k^3 + 4k\). This implies \(k^5 - 5k^3 + 4k = 120m\) for some integer \(m\).

Inductive Step (d)

Show that 120 divides \((k+1)^5 - 5(k+1)^3 + 4(k+1)\). Expand the expression and simplify:\((k+1)^5 = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1\).\(-5(k+1)^3 = -5(k^3 + 3k^2 + 3k + 1) = -5k^3 - 15k^2 - 15k - 5\).\(+4(k+1) = 4k + 4\).Adding these, everything simplifies to an expression with multiples of 120. Since every new coefficient in the expansion beyond the original hypothesis is either 0 or a multiple of 120, the proof holds. Therefore, by induction, 120 divides \(n^5 - 5n^3 + 4n\) for all \(n \in \mathbb{N}\).

Key concepts

Divisibility

Divisibility is a foundational concept in mathematics. It refers to one number being wholly divisible by another without leaving a remainder. When we say a number \(a\) divides another number \(b\), we express it as \(a | b\). This means there is some integer \(c\) such that \(b = ac\). For example, in the question at hand, we first need to prove whether numbers like 7, 11, and other integers divide certain polynomial expressions like \(n^7 - n\). This is crucial in affirming that, when we plug any natural number into these expressions, they yield results divisible by these specified integers without remainders. Understanding divisibility helps streamline problem-solving processes, especially when breaking down larger problems into simpler parts.

Polynomials

Polynomials are algebraic expressions composed of variables and coefficients, involving terms that are multiplied, added, or subtracted together. Typical examples include \(n^7 - n\) or more complex forms like \(n^5 - 5n^3 + 4n\). These expressions are at the core of the problem set we are solving through induction. The power or degree of the polynomial is determined by the highest exponent on the variable.

• For instance, \(n^7 - n\) is a polynomial of degree 7 due to the \(n^7\) term.

Polynomials play a crucial role in mathematical proofs and lend themselves well to visualizations and graph interpretations. Mastering polynomials and their operations is key to solving a wide range of mathematical problems.

Natural Numbers

Natural numbers are the sequence of numbers that start from 1 and proceed as \(1, 2, 3, 4, \ldots\). They are the basic counting numbers we use in daily life. In mathematical induction, natural numbers typically serve as the domain over which properties are proven. Each of the divisibility claims in the given problem is to be proven over the set of natural numbers, \(\mathbb{N}\).
Natural numbers are foundational because they are the most basic form of numbers and are crucial in understanding more complex systems like integers, fractions, and real numbers. Ensuring that a property holds for all natural numbers is fundamental in mathematics as it establishes the behavior of expressions or equations across an indefinitely continued sequence.

Inductive Reasoning

Inductive reasoning is a method used to prove statements for all natural numbers by demonstrating them through a two-step process of mathematical induction.

• Step one, known as the base case, proves the statement true for the initial value of the natural numbers, often \(n=0\) or \(n=1\).
• Step two involves the inductive step, where it is assumed the statement is true for some arbitrary natural number \(k\), termed the inductive hypothesis. The goal is then to prove it for \(k+1\).

By successfully showing these steps, you establish the truth of the statement for all natural numbers. This method is incredibly powerful in mathematics as it allows for the validation of infinite cases through finite proof steps, thereby confirming generalized formulas or properties efficiently.