0 = 1

Calculus Level 4

0 = ( 1 − 1 ) Step 1 = ( 1 − 1 ) + ( 1 − 1 ) Step 2 = ( 1 − 1 ) + ( 1 − 1 ) + ( 1 − 1 ) + ( 1 − 1 ) + ⋯ Step 3 = 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + ⋯ Step 4 = 1 + ( − 1 + 1 ) + ( − 1 + 1 ) + ( − 1 + 1 ) + ⋯ Step 5 = 1 Step 6 \begin{array} { l l r } 0 & = ( 1 -1 ) & \text{ Step 1 } \\\\ & = ( 1 -1 ) + ( 1 - 1) & \text{ Step 2 } \\\\ & = ( 1 -1 ) + ( 1 - 1) + (1-1) + (1-1) + \cdots & \text{ Step 3 } \\\\ & = 1 -1 + 1 - 1 + 1 - 1 + 1 - 1 + \cdots & \text{ Step 4 } \\\\ & = 1 + (-1 + 1) + (- 1 + 1) + (-1+1) + \cdots & \text{ Step 5 } \\\\ & = 1 & \text{ Step 6 } \end{array}

In which step above did we first introduce an error?

Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 No error was made

Calvin Lin by Calvin Lin

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1 solution

Calvin Lin Staff
Dec 8, 2016

The mistake was first made in Step 4, where we went from a sum with even number of terms, in which the sum is indeed 0, to a sum with arbitrary number of terms (which has no limit of partial sums).

Subsequently, in step 5, we went from a sum with an arbitrary number of terms, to a sum with odd number of terms, in which the sum is indeed 1.

Step 1 is justified since 0 = ( 1 − 1 ) 0 = ( 1- 1 ) .
Step 2 is justified since we're just adding 0.
Step 3 is justified as taking the limit, namely
∑ i = 1 ∞ ( 1 − 1 ) = lim ⁡ ∑ i = 1 n ( 1 − 1 ) = lim ⁡ 0 = 0 \sum_{i=1}^{\infty} (1-1) = \lim \sum_{i=1}^n (1-1 ) = \lim 0 = 0
Step 4 is wrong as explained above.
Step 5 is wrong as explained above.
Step 6 is justified.

16 Helpful 0 Interesting 0 Brilliant 0 Confused

Is it equivalent to saying that we are not allowed to remove or add brackets in case of an infinite sum?

William Nathanael Supriadi - 4 years, 6 months ago

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Generally speaking, essentially yes.

Strictly speaking, we can do always do certain things under "nice" conditions. For example, the statements would all be true if we replaced all the 1's with 0's.

So, we have to understand what these "nice conditions" are, in order to proceed. For example, absolute convergence is a sufficient condition to open up brackets and reorder terms.

Calvin Lin Staff - 4 years, 6 months ago

I think the problem started from step 1 is wrong because it possible to say

0 = 1 -1 = 2-2 = 3-3 = 4-4 = x -x .... etc.

This will lead to any answer you want !!!

Ossama Ismail - 4 years, 6 months ago

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I'm not sure of what you're claiming. Are you saying:

  • Step 1 is wrong because 0 ≠ ( 1 − 1 ) 0 \neq (1 - 1)
  • Step 2 is wrong because 0 ≠ ( 1 − 1 ) + ( 1 − 1 ) 0 \neq (1-1) + (1-1)
  • Step 3 is wrong because 0 ≠ ( 1 − 1 ) + ( 1 − 1 ) + … 0 \neq (1-1) + (1-1) + \ldots

Calvin Lin Staff - 4 years, 6 months ago

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