Question

Prove that \(\sum_{k=0}^{m}\left(\begin{array}{c}m \\\ k\end{array}\right)\left(\begin{array}{c}n \\\ p+k\end{array}\right)=\left(\begin{array}{c}m+n \\ m+p\end{array}\right)\) for non-negative integers \(m, n\) and \(p\). (This equation is from Exercise 8 in Section 3.10 . There we were asked to prove it by combinatorial proof. Here we are asked to prove it with induction.)

Short answer

After proving the base case, assuming that the identity is true for a particular p, and proving it for p+1 using the induction hypothesis, we can safely conclude via mathematical induction that the given combinatorial identity is true for all non-negative integers m, n, and p.

Step-by-step solution

Base Case

Start by proving the identity for the base case.You need to check the base case p=0. After simplifying the left hand side, you would have\[\sum_{{k=0}}^{m} \left( {m \choose k} \right)\left( {n \choose k} \right) = {m+n \choose m}\]which confirms that the base case holds.

Inductive Step

Now, assume the identity is true for p. That is,\[\sum_{k=0}^{m}\left( {m \choose k} \right)\left( {n \choose p+k} \right)= {m+n \choose m+p}\]We would use this assumption to prove the identity for p+1.

Prove for p + 1

Next, we aim to show that the identity holds for p+1. We should express \( {n \choose p+1+k} \) in terms of \( {n \choose p+k} \). We can do that by using the identity \( {n \choose p+1+k} = \frac{n-p-k}{p+1+k} {n \choose p+k} \). Then, substitute this in our original equation and simplify to get necessary formula valid for p + 1.

Summarize the Proof

Having proved the base case (step 1), and the inductive step (step 2 and 3), we can conclude that the given identity holds for all non-negative integer values of p. This completes the inductive proof.

Key concepts

Combinatorial Proof

A combinatorial proof is a technique for demonstrating the validity of an identity or mathematical statement using combinatorial arguments. This involves counting the number of ways to achieve a certain configuration or selection and showing that two different configurations lead to the same result.

For example, to prove the given identity, in a combinatorial context, we would interpret the binomial coefficients as ways to choose elements from a set. The left-hand side of the equation counts the number of ways to choose elements from two separate sets with certain restrictions, and shows that it must be equivalent to the binomial coefficient on the right-hand side, which counts the total number of ways to choose those elements from a combined set.

Using combinatorial reasoning, one provides an argument that essentially tells a story or draws upon a real-world scenario to establish the validity of the equation. This method can be quite intuitive and offers a different perspective than the formal algebraic manipulations used in induction.

Inductive Step

The inductive step is a critical phase in the process of mathematical induction, a fundamental proof technique in mathematics. After establishing the base case, the inductive step involves assuming that the statement to be proved holds for some integer, and then showing that if this assumption is true, then the statement must also hold for the next integer.

In the context of the problem we are examining, the inductive step entails demonstrating that, given the identity holds true for a particular non-negative integer value of p, it will also hold for p+1. In practice, this often involves manipulating the assumption to include p+1 and performing algebraic simplifications to arrive at the necessary result. Success in the inductive step usually depends heavily upon creative algebraic insight and a good grasp of the properties involved, in this case, those of binomial coefficients and summation notation.

Binomial Coefficient

A binomial coefficient, written as \( {n \choose k} \), is a fundamental concept in combinatorics that gives the number of ways to choose k elements out of a set of n distinct elements without considering the order of selection. It can be calculated using the formula: \[ {n \choose k} = \frac{n!}{k!(n-k)!} \] where n! denotes n factorial, which is the product of all positive integers up to n.

Binomial coefficients appear frequently in the expansion of binomial expressions and in various counting problems. Understanding their properties such as symmetry \( {n \choose k} = {n \choose n-k} \) or the Pascal's rule \( {n \choose k} + {n \choose k-1} = {n+1 \choose k} \) is important when using them in proofs, like the inductive proof for the provided exercise.

Proof Techniques in Mathematics

In the realm of proof techniques in mathematics, various methods exist to demonstrate the truth of statements and theorems. Some of the most commonly employed techniques include direct proofs, proof by contradiction, proof by contrapositive, and mathematical induction—which is used in the given exercise.

Each technique has its own set of rules and applications. Mathematical induction, for instance, is especially powerful when dealing with statements that assert something about an infinite series of natural numbers or configurations. It is composed of two parts: establishing a base case and executing the inductive step. When performed correctly, a proof by induction can establish the truth of a statement across an infinite number of cases, which is difficult to achieve through direct verification alone.

Summation Notation

Summation notation, represented by the Greek letter sigma (Σ), is a concise way to denote the sum of a sequence of numbers generated by an explicit formula. For example, the sum of the first m integers can be written as: \[ \sum_{k=1}^{m} k = 1 + 2 + 3 + ... + m \]In more complex scenarios, such as the one presented in the exercise we're exploring, summation notation helps to compactly express the sum of products of binomial coefficients.

Understanding the rules governing summation notation, such as distributive or index shifting, is critical when manipulating sums algebraically as part of an inductive proof. Additionally, recognizing certain series sums or properties can be key to simplifying and solving more complicated summation problems encountered in various mathematical contexts.