Add Me!Close Menu Navigation

What are the Odds?

Let k be an even number. Is it possible to write 1 as the sum of the reciprocals of k odd integers?

The answer to the problem is no, and we will prove it by creating a contradiction when assuming that the above claim is true.

Assume for now that it is possible to write 1 as the sum of the reciprocals of an even number of odd integers.

(1)   \begin{align*} 1 = \frac{1}{n_1} + \stackrel{k}{\ldots} + \frac{1}{n_k} \hspace{4mm} n_i,k \in \mathbf{Z} \wedge k \mbox{ even}, n_i \mbox{ odd} \end{align*}

In middle school arithmetic, we learned that \frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab}. Generalized, suppose we want to evaluate \sum_{i} \frac{1}{n_i}, then letting N = n_1 \times \stackrel{k}{\ldots} \times n_k

    \[\sum_i^k \frac{1}{n_i} = \frac{\frac{N}{n_i} + \stackrel{k}{\ldots} + \frac{N}{n_k}}{N}\]

For example, \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{bc + ac + ab}{abc}

Now, because each of the k n_i‘s are odd, then N = n_1 \times \stackrel{k}{\ldots} \times n_k must also be odd. Furthermore, each of \frac{N}{n_i} are also odd. We now rewrite the above equation (1) as

    \begin{align*} 1 &= \frac{\frac{N}{n_i} + \stackrel{k}{\ldots} + \frac{N}{n_k}}{N} \\ N &= \frac{N}{n_i} + \stackrel{k}{\ldots} + \frac{N}{n_k} \end{align*}

Now, the left hand side N is already shown to be odd. The right hand side however is the sum of k also odd terms. However, because k is an even number, the sum of an even number of odds becomes even. Therefore, if 1 can be written as the sum of the reciprocals of an even number of odd integers, then there exists some odd number N than is equal to another even number. Because this creates a contradiction, then 1 cannot be expressed as above. \blacksquare

Posted By Lee

Woosh

Recent Comments