Pigeon hole principle for odd or even product

**oldguynewstudent** · June 25th, 2010, 05:20 AM

Let n be odd and suppose $(x_{1},x_{2},\ldots,x_{n})$ is any permutation of [n]. Prove that the product $(x_{1}-1)(x_{2}-2)\cdots(x_{n}-n)$ is even. Is the result necessarily true if n is even? Give a proof or counterexample.

If the $x_{i}'s$ are written in ascending order, then each is paired with its additive inverse and each factor is zero, so the product is zero and we are done.

Since the product of any two even numbers is even and the product of an even number and an odd number is even, we need all the terms to be odd if the product is to be odd. Since n is odd, the first term $x_{1}$ and the last term $x_{n}$ will be odd. The sum (or difference) of any two numbers is even if they are both even or both odd and it is odd if one is even and one is odd. Starting with the $x_{i}'s$ in ascending order, we notice that each term has the two numbers paired so that each even $x_{i}$ is paired with an even number and each odd $x_{i}$ is paired with an odd number. We also notice that we can rearrange n-1 $x_{i}'s$ so that each even $x_{i}$ is paired with an odd number and each odd $x_{i}$ is paired with an even number. But by the pigeon hole principle, there is one $x_{i}$ left that is paired so that its term is even. Therefore with n being an odd number, the product of the terms must be even.

By a similar argument, if n is even, there is a way to rearrange the terms so that the product can be odd. Example: (2-1)(1-2)(4-3)(3-4).

Is this correct?

**undefined** · June 25th, 2010, 05:37 AM

Originally Posted by oldguynewstudent

Let n be odd and suppose $(x_{1},x_{2},\ldots,x_{n})$ is any permutation of [n]. Prove that the product $(x_{1}-1)(x_{2}-2)\cdots(x_{n}-n)$ is even. Is the result necessarily true if n is even? Give a proof or counterexample.

If the $x_{i}'s$ are written in ascending order, then each is paired with its additive inverse and each factor is zero, so the product is zero and we are done.

Since the product of any two even numbers is even and the product of an even number and an odd number is even, we need all the terms to be odd if the product is to be odd. Since n is odd, the first term $x_{1}$ and the last term $x_{n}$ will be odd. The sum (or difference) of any two numbers is even if they are both even or both odd and it is odd if one is even and one is odd. Starting with the $x_{i}'s$ in ascending order, we notice that each term has the two numbers paired so that each even $x_{i}$ is paired with an even number and each odd $x_{i}$ is paired with an odd number. We also notice that we can rearrange n-1 $x_{i}'s$ so that each even $x_{i}$ is paired with an odd number and each odd $x_{i}$ is paired with an even number. But by the pigeon hole principle, there is one $x_{i}$ left that is paired so that its term is even. Therefore with n being an odd number, the product of the terms must be even.

By a similar argument, if n is even, there is a way to rearrange the terms so that the product can be odd. Example: (2-1)(1-2)(4-3)(3-4).

Is this correct?

Looks right. A somewhat abbreviated paraphrase: In the set [n]={1,2,...,n} with n odd, There are $\displaystyle \left\lfloor \frac{n}{2} \right\rfloor$ even numbers and $\displaystyle \left\lfloor \frac{n}{2} \right\rfloor+1$ odd numbers. In order for the product $(x_{1}-1)(x_{2}-2)\cdots(x_{n}-n)$ to be odd, each multiplicand $(x_{i}-i)$ must be odd. That means that for even $i$ , $x_{i}$ must be odd, and vice versa. But the number of odd $x_i$ is greater than the number of even $i$ , therefore by the pidgeonhole principle there is at least one odd $i$ for which $x_i$ is odd. Thus for n odd, the product $(x_{1}-1)(x_{2}-2)\cdots(x_{n}-n)$ is even.

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