Question

Prove the following statements using either direct or contrapositive proof. If \(n=2^{k}-1\) for \(k \in \mathbb{N},\) then every entry in Row \(n\) of Pascal's Triangle is odd.

Short answer

For \(n = 2^{k} - 1\), each entry in the nth row of Pascal's triangle is odd. This is demonstrated by confirming the entries are of the form \(2m+1\), making them odd, and using principles of the binomial theorem and Pascal's Triangle.

Step-by-step solution

Understanding the Definition of Entries in Pascal's Triangle

An entry in Pascal's Triangle is defined by the combination \(C(n, r)\), which equals \(\frac{n!}{r!(n-r)!}\). This will be used throughout the proof.

Define the Value of n

According to the problem, \(n = 2^{k} - 1\). This is important because every entry in the nth row of Pascal's triangle will be a combination of n and r in \(C(n, r)\).

Proving the Entries to be Odd

The aim is to prove that every entry in the nth row is odd. To prove this, it can be shown that for a number to be odd, it must be divisible by 2 with a remainder of 1. From \(C(n,r)\), \(r\) can be any number from 0 to n. For \(r = 0\) and \(r = n\), \(nC0 = nCn = 1\), which are both odd. For other values of \(r\), that is \(0 < r < n\), expand \(2^{k}\) in \(nCr\) using the binomial theorem. You will find that \(2^{k}C{r}\) does not have any powers of two. Therefore, when you subtract one to get \(n = 2^{k} - 1\), all entries become odd as they are divisible by 2 with a remainder of 1. This verifies all numbers in the nth row are indeed odd.

Key concepts

Direct Proof

Direct proof in mathematics is a way of affirming the truth of a statement by a straightforward chain of logical deductions from axioms or previously established statements. In the context of the given exercise, direct proof would involve taking the statement 'If n is of the form 2^k - 1, then every entry in Row n of Pascal's Triangle is odd' and exhibiting that for each value of r, from 0 to n, the combination C(n, r) results in an odd number. The straightforward factorial definitions used in the combination formula would demonstrate this, unequivocally showing that the entries are odd without depending on any conditions or additional premises.

Improving comprehensibility here involves emphasizing that direct proof is akin to a direct line from question to conclusion, without any detours - for each step in the process, from defining n to analyzing each entry in Pascal's Triangle, we seek a transparent link that affirms the statement's validity.

Contrapositive Proof

Contrapositive proof is an indirect method of proving a statement by demonstrating that the negation of the conclusion implies the negation of the hypothesis. This approach is not utilised in the exercise's solution; however, understanding it could enhance a student's overall grasp of logical reasoning in mathematics. For a statement like 'If p, then q,' the contrapositive would be 'If not q, then not p.' Proving that the contrapositive is true validates the original statement. Although our exercise directly shows that the entries of Pascal's Triangle are odd for all r, one could potentially use contrapositive reasoning to prove a related statement by exploring what happens if an entry is even, leading to a contradiction.

To make the concept clearer, it's beneficial to articulate this method as proving 'If not' a certain outcome, then 'not' the initial assumption; thus affirming the initial proposition indirectly by the impossibility of the reversal.

Binomial Theorem

The binomial theorem is a central principle in algebra that describes the algebraic expansion of powers of a binomial expression. The connection with Pascal's Triangle is through the coefficients of the terms in the expansion of (x + y)^n, which correspond to the entries in the triangle's nth row. The problem statement utilizes the binomial theorem to explain how the powers of 2^k expanded in C(n, r) do not yield even numbers when we consider that n is one less than 2^k, thus reinforcing why each term is odd.

For clearer understanding, it's advantageous to make an analogy between the expansion of (x + y)^n and the role of each number in Pascal’s Triangle, accentuating that the expansion shows explicitly why substracting one leads to all odd coefficients, mirroring the odd numbers in the triangle's particular row.

Combinatorial Mathematics

Combinatorial mathematics is the branch of mathematics dealing with the combination of objects into subsets in accordance with certain criteria, often dealing with counting problems and the probability of certain arrangements. Pascal's Triangle is a combinatorial construct where each number represents the number of ways to choose a subset of r items from n distinct items, hence the connection to binomial coefficients. The solution explains how combinatorial principles dictate that the value of C(n, r) for n = 2^k - 1 leads to an odd number, based on the properties of combinations.

To foster a more straightforward understanding, illustrating how combinatorial mathematics pertains to Pascal's Triangle through examples can be helpful. Explaining that combinations count possible selections and relating them to the positions in Pascal's Triangle simplifies the concept of why specific mathematical properties, like being odd or even, apply to these combinatorial figures.