Question

The following argument, that \(0=1,\) is incorrect. Describe the error. $$ \begin{aligned} 0 &=0+0+0+\cdots \\ &=(1-1)+(1-1)+(1-1)+\cdots \\ &=1+(-1+1)+(-1+1)+\cdots \\ &=1+0+0+\cdots \\ &=1 \end{aligned} $$

Short answer

The mistake in the argument lies in rearranging the terms of an infinite series without acknowledging the lack of a universal sum property for challenging sequences such as this one. The rearrangement leads to the incorrect conclusion that \(0=1\).

Step-by-step solution

Understanding the Infinite Series

A basic comprehension of infinite series is necessary. An infinite series is the sum of the terms of an infinite sequence. Not all infinite series converge to a finite value, and the infinite series here is not convergent.

Reflecting on the Equivalent Transformations

While the first transformation from \(0+0+0+...\) to \((1-1)+(1-1)+(1-1)+...\) is accurate, the subsequent transformation is not. Here, the distributive law (elementary algebra) is applied inappropriately to an infinite series.

Identifying the Error

The error lies in the step when the expression \((1-1)+(1-1)+(1-1)+...\) is rearranged to \(1+(-1+1)+(-1+1)+...\). The equivalent transformation changes the sum of infinite terms, but it does not account for the fact that the sequence doesn't possess a universal sum property. Instead of grouping \(1-1\) together as was done in the initial series, the groupings are now rearranged arbitrarily, which does not hold in infinite series due to lack of universal sum property. This is what leads to the faulty conclusion that \(0=1\).