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Math_history·Math Thematicbyxiao yun

Proof that 0 is infinity

1 = 1

1 = (-1) + 2

1 = (-2) + 3

1 = (-3) + 4

1 = (-4) + 5 ... And so on...

By adding term by term to all these equations, we get: 1 + 1 + 1 + 1 + 1 + 1 + ... = 1 + (-1) + 2 + (-2) + 3 + (-3) + 4 + (-4) + 5 + ...

In the expression on the right, all the terms cancel out in pairs, giving: 1 + 1 + 1 + 1 + 1 + 1 + ... = 0

The expression on the left, consisting of an infinite sum of terms equal to 1, tends to infinity. Thus, 0 is equal to infinity.

And yet 0 is not equal to infinity. So where is the mistake?

::: spoiler Before to answer In case that you decide to write it down please hide your answer to let others think about it 👍 :::


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3 replies

::: spoiler My thoughts...

Gonna have to say that, like in Cantor's Theorem, the fault lies in the handwaving phrase, "and so on." It's a mental shortcut that leads us astray. If it actually means going on to infinity, then the right side of the equation isn't zero. Rather, it never converges on a sum— for every negative number, there will always be a next term, a positive integer with greater magnitude.

:::

9

1 + 1 + 1 + 1 + 1 + 1 + … = 1 + (-1) + 2 + (-2) + 3 + (-3) + 4 + (-4) + 5 + …

You forgot -5+6 on purpose.

::: spoiler spoiler
"In the expression on the right, all the terms cancel out"

You forgot the last one which is equal to the sum on the left. :::

7

Not a mathematician but here's my guess:

::: spoiler spoiler The question assumes you can just cancel out all the values on the right side. However, by summing the first three terms we see: 1 + 1 + 1 = 1 + (-1) + 2 + (-2) + 3 3 = 3

This continues as the numbers get bigger and bigger, so no matter how large the sum of the ones on the LHS is, there will always be it's equivalent on the RHS to equal it, after the previous values are cancelled out. :::

Not sure how to prove rigorously, infinities always confuse me a bit when it comes to what is and isn't allowed.

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