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Chapter 1: Problem 10

Let \(n\) be an integer \(\geq 2\) Show by examples that the bionomial coefficient \(\left(\begin{array}{l}p^{n} \\ k\end{array}\right)\) need not be divisible by \(p^{*}\) for \(1 \leqq k \leqq p^{n}-1 .\)

Short Answer

Expert verified
For the given example where \(p = 3\) and \(n = 2\), we calculated the binomial coefficients \(\left(\begin{array}{l}3^{2} \\ k\end{array}\right)\) for \(1 \leqq k \leqq 8\). We observed that the coefficient for \(k=1\) is 9, which is not divisible by \(3^{2}\). Thus, based on this example, it can be concluded that the binomial coefficient \(\left(\begin{array}{l}p^{n} \\ k\end{array}\right)\) does not necessarily need to be divisible by \(p^{*}\) for \(1 \leqq k \leqq p^{n} - 1\).

Step by step solution

01

Understanding the Binomial Coefficient

The binomial coefficient is denoted by \(\left(\begin{array}{l}n \\ k\end{array}\right)\) and is defined as the number of ways to choose k elements from n elements, which is calculated by the formula: \(\left(\begin{array}{l}n \\ k\end{array}\right) = \dfrac{n!}{k!(n-k)!}\)
02

Choose an Example

Pick a prime number p and an integer n such that \(n \geq 2\). For this demonstration, select \(p = 3\) and \(n = 2\). The given inequality becomes: \(1 \leqq k \leqq p^{n} - 1 = 3^{2} - 1 = 8\)
03

Calculate Example

Calculate the binomial coefficient for the selected example, \(\left(\begin{array}{l}3^{2} \\ k\end{array}\right)\) for \(1 \leqq k \leqq 8\), using the binomial coefficient formula: For \(k=1\), \(\left(\begin{array}{l}9 \\ 1\end{array}\right) = \dfrac{9!}{1!(9-1)!} = 9\) For \(k=2\), \(\left(\begin{array}{l}9 \\ 2\end{array}\right) = \dfrac{9!}{2!(9-2)!} = 36\) Continue calculating the binomial coefficients for all the values of k within the given range.
04

Verify Divisibility by \(p^{*}\) for Example

Verify if binomial coefficients calculated above are divisible by \(p^{*}\). In the selected example, where \(p = 3\), consider whether the coefficients are divisible by any power of 3: - The coefficient for \(k=1\) is 9, which is divisible by \(3^{1}\) but NOT by \(3^{2}\). - The coefficient for \(k=2\) is 36, which is divisible by \(3^{2}\).
05

Conclude Proof by Example

Based on the example above, it can be observed that the binomial coefficient \(\left(\begin{array}{l}p^{n} \\ k\end{array}\right)\) does not always need to be divisible by \(p^{*}\) for \(1 \leqq k \leqq p^{n} - 1\). The coefficient for \(k=1\) was not divisible by \(3^{2}\), thus demonstrating it through an example.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Theorem
The binomial theorem is a fundamental principle in algebra that describes the expansion of powers of binomial expressions. A binomial is an algebraic expression containing two terms. The theorem provides a formula to expand expressions of the form \( (a + b)^n \) where \( n \) is a non-negative integer. This expansion results in a sum involving terms of the form \( a^{n-k}b^k \) multiplied by a binomial coefficient \( \left(\begin{array}{l}n \ k\end{array}\right) \).

For instance, \( (a + b)^2 = a^2 + 2ab + b^2 \) could be seen as a result of the binomial theorem. Each term has a corresponding binomial coefficient, which indicates the number of ways we can pick \( k \) elements out of \( n \) positions. In our textbook exercise, binomial coefficients play a crucial role and understanding how to compute them is essential for solving divisibility problems.
Factorial Notation
Factorial notation is essential when dealing with binomial coefficients and combinatorics. The factorial of a non-negative integer \( n \) is the product of all positive integers less than or equal to \( n \) and is denoted by \( n! \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow at a very fast rate as \( n \) increases. They are used in the formula for binomial coefficients, which we can see in the exercise: \( \left(\begin{array}{l}n \ k\end{array}\right) = \frac{n!}{k!(n-k)!} \). Understanding factorials is vital, as they are fundamental in calculating probabilities, permutations, combinations, and in this particular scenario, determining the divisibility properties of the binomial coefficients.
Prime Numbers
Prime numbers are the building blocks of arithmetic. They are natural numbers greater than one that have no positive divisors other than one and themselves. For example, the first few prime numbers are 2, 3, 5, 7, and 11. A key property of prime numbers is that each one has a unique factorization.

When we talk about the divisibility of binomial coefficients by prime powers (noted as \( p^{*} \) in our exercise), we're considering the impact of prime numbers on the factors of binomial coefficients. Prime factorization plays a crucial role in evaluating the divisibility of expressions by prime powers, especially when using factorials within the coefficients. Understanding prime numbers is therefore crucial when studying properties such as divisibility.
Combinatorics
Combinatorics is the branch of mathematics that studies counting, both as a means and an end in arriving at results, and certain properties of finite structures. It is foundational in the derivation of binomial coefficients, which can represent combinations in probability and combinatorial problems.

In the context of our exercise, the combinatorial interpretation of the binomial coefficient \( \left(\begin{array}{l}n \ k\end{array}\right) \) is the number of ways to choose \( k \) elements from a set of \( n \) distinct elements. However, not all combinatorial numbers (binomial coefficients) display simple divisibility properties, especially when considering powers of prime numbers. This observation is key for students to understand that although combinatorial numbers may seem formulaic, their properties can be quite intricate and demand a deeper understanding of number theory and divisibility.

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Most popular questions from this chapter

(a) Prove that a positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3 . (b) Prove that it is divisible by 9 if and only if the sum of its digits is divisible by \(9 .\) (c) Prove that it is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. In other words. let the integer be $$ n=a_{k} a_{i-1} \cdots a_{0}=u_{0}+a_{1} 10+a_{2} 10^{2}+\cdots+a_{k} 10^{4}, \quad 0 \leqq a_{i} \leq 9. $$ Then \(n\) is divisible by 11 if and only if \(a_{0}-a_{1}+a_{2}-a_{3}+\cdots+(-1)^{4} a_{k}\) is divisible by 11

A positive integer is called palyndromic if its digits from left to right are the same as the digits from right to left. For instance, 242 and 15851 are palyndromic. The integers \(11,101,373,10301\) are palyndromic primes. Observe that except for 11, the others have an odd number of digits. (a) Is there a palyndromic prime with four digits? With an even number of digits fexcept for \(111 ?\) (b) Are there infinitely many palyndromic primes? (This is an unsolved problem in mathematics.)

L.et \(x\) be a real number. Prove that there exists an integer \(q\) and a real number \(x\) with \(0 \leqslant s<1\) such that \(x=q+s\), and that \(q, x\) are uniquely determined. Can you deduce the euclidean algorithm from this result without using induction?

Let \(n, d\) be positive integers and assume \(1

Show that any rational number \(a \neq 0\) can be written in the form $$ a=\frac{x_{1}}{p_{1}^{n}}+\cdots+\frac{x_{n}}{p_{n}^{p_{n}}}, $$ where \(x_{1+\ldots,} x_{n}\) are integers, \(p_{1}, \ldots, p_{n}\) are distinct prime numbers, and \(r_{1}, \ldots, r_{n}\) are integers \(\geq 0 .\)
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