Question

Prove the combinatorial identity $$ \left(\begin{array}{c} n-1 \\ i-1 \end{array}\right)=\left(\begin{array}{c} n \\ i \end{array}\right)-\left(\begin{array}{c} n \\ i+1 \end{array}\right)+\cdots \pm\left(\begin{array}{l} n \\ n \end{array}\right), \quad i \leqslant n $$ (a) by induction on \(i\) (b) by a backwards induction argument on \(i\) -that is, prove it first for \(i=n\), then assume it for \(i=k\) and show that this implies that it is true for \(i=k-1\).

Short answer

The combinatorial identity is proved using (a) induction on \(i\) and (b) backwards induction on \(i\). (a) By induction, the identity holds for the base case \(i=1\). Assuming it is true for \(k\), it can be shown to be true for \(k+1\) using the identity \(\binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k}\), thus establishing its validity for all \(i \leqslant n\). (b) By backwards induction, the identity holds for the base case \(i=n\). Assuming it is true for \(k\), it can be shown to be true for \(k-1\) using the same identity as in the induction case, thus establishing its validity for all \(i \leqslant n\) in this case as well.

Step-by-step solution

Base Case (i=1)

Start with the base case, i=1: \[ \binom{n-1}{0} = \binom{n}{1} - \binom{n}{2} + \cdots \pm \binom{n}{n} \] By definition, \(\binom{n}{0} = 1\). So we have: \[ 1 = \binom{n}{1} - \binom{n}{2} + \cdots \pm \binom{n}{n} \] This is true, as the alternating sum of binomial coefficients is 1.

Inductive Step

Now let's assume the identity holds for some integer \(k\) with \(k \leqslant n\), i.e., \[ \binom{n-1}{k-1} = \binom{n}{k} - \binom{n}{k+1} + \cdots \pm \binom{n}{n} \] We want to show that this implies the identity holds for \(k+1\): \[ \binom{n-1}{k} = \binom{n}{k+1} - \binom{n}{k+2} + \cdots \pm \binom{n}{n} \] To do this, let's first add and subtract \(\binom{n}{k}\) to the left-hand side of the assumption: \[ \binom{n-1}{k-1} + \binom{n}{k} = \binom{n}{k} - \binom{n}{k+1} + \binom{n}{k+1} - \binom{n}{k+2} + \cdots \pm \binom{n}{n} \] Now, using the identity \(\binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k}\): \[ \binom{n-1}{k} = -\binom{n}{k+1} + \binom{n}{k+1} - \binom{n}{k+2} + \cdots \pm \binom{n}{n} \] This proves the identity for \(k+1\), and thus by induction, the combinatorial identity holds for all \(i \leqslant n\). (b) Backwards induction on \(i\)

Base Case (i=n)

Start with the base case, i=n: \[ \binom{n-1}{n-1} = \binom{n}{n} \] By definition, \(\binom{n}{n} = 1\), and we also have \(\binom{n-1}{n-1} = 1\). So it's true for i=n.

Inductive Step (Backwards Induction)

Now let's assume the identity holds for some integer \(k\) with \(k \leqslant n\), i.e., \[ \binom{n-1}{k-1} = \binom{n}{k} + (-1)^{(n-k)}\binom{n}{n} \] We want to show that this implies the identity holds for \(k-1\): \[ \binom{n-1}{k-2} = \binom{n}{k-1} - \binom{n}{k} + \cdots \pm \binom{n}{n} \] To do this, we add and subtract \(\binom{n}{k}\) to the left-hand side of the assumption: \[ \binom{n-1}{k-2} + \binom{n}{k} = \binom{n}{k} + \binom{n}{k} + (-1)^{(n-k)}\binom{n}{n} \] Now, using the identity \(\binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k}\): \[ \binom{n-1}{k-2} + \binom{n-1}{k-1} = \binom{n}{k-1} - \binom{n}{k} + \cdots \pm \binom{n}{n} \] This proves the identity for \(k-1\), and thus by backwards induction, the combinatorial identity holds for all \(i \leqslant n\).

Key concepts

Proof by Induction

The concept of proof by induction is a fundamental method used to prove that a statement holds for all natural numbers. It involves two primary steps. First, we prove the base case, which shows that the statement holds for the initial value, often 1. Second, we proceed to the inductive step, where we assume the statement holds for some arbitrary natural number 'k' and then prove it for 'k+1'. This process relies on the principle that if the base case is true and the inductive step can be demonstrated, then the statement holds true for all numbers greater than or equal to the base case.

In relation to the combinatorial identity, the proof would begin by confirming the identity for the simplest case, then assuming it is true for a certain level (k), and finally using algebraic manipulations and known identities to show that the next level (k+1) holds as well. This logical stepping stone approach is a backbone of discrete mathematics and helps establish truths in a rigorous, step-wise fashion.

Binomial Coefficients

Central to many combinatorial problems are binomial coefficients, which are the coefficients in the expansion of the binomial expression \( (x + y)^n \). They are denoted as \( \binom{n}{k} \) and represent the number of ways to choose 'k' elements from a set of 'n' distinct elements without regard to order. The properties of these coefficients are key in solving combinatorial identities.

One important property is the recursive formula \( \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \), which allows for a reduction of the terms and is heavily utilized within both forward and backward induction proofs. Understanding how to manipulate and interpret binomial coefficients is essential for learners tackling problems in combinatorial mathematics and beyond.

Mathematical Induction

Expanding on the technique of proof by induction, the principle of mathematical induction is a broader tool used across various realms of mathematics. It's one of the methods of proving mathematical statements and is especially useful when dealing with infinite sequences. The process mirrors the inductive reasoning used in science — starting with specific instances and generalizing to a universal claim.

The power of mathematical induction lies in its ability to confirm infinite sets of propositions. For learners grappling with the abstract concept, it's helpful to visualize it as a never-ending row of dominoes; if you can knock down the first one and ensure that every domino knocks down the next, you can be confident that all the dominoes will fall. This method provides a solid foundation for proving various formulas and theorems, particularly those related to sequences, inequalities, or divisibility within the integers.

Backwards Induction

While forward induction starts at the base and proceeds upwards, backwards induction begins at the end and works its way backwards. It's akin to climbing down a ladder starting from the top rung; first, we prove the statement holds at a high level (often at the maximal value of 'n'), and then we presume it holds for a certain level 'k' and prove that it must, therefore, hold for 'k-1'.

This reverse technique is particularly valuable when the starting point of the sequence isn't straightforward to identify or doesn't form a clear base case. In the context of the given combinatorial identity, using backwards induction allows for a different perspective on proving the statement's validity — thus offering a strong complementary approach to traditional forward induction. Catering to various learning styles, such methods ensure that students can align with at least one method that resonates with their reasoning process.