- #1

Goldenwind

- 146

- 0

## Homework Statement

This is not homework, this is me being stubborn. I have seen the proofs, I know that it is professionally accepted, but I still just don't agree.

## Homework Equations

1 = 2

1 + 1 = 3 for very large values of 1

Squareroot of anything will lead to very sharp trees.

## The Attempt at a Solution

One example of a proof:

1/3 = 0.333...

2/3 = 0.666...

3/3 = 0.999...

3/3 = 1

Therefore 0.999... = 1

My issue with this is, I don't consider 3/3 to be 0.999... I consider 3/3 to be 1/1 = 1. The reason this seems illusionary is because 2/3 isn't 0.666. THIS proof, I accept:

0.999 * (1/3) = 0.333

0.999 * (2/3) = 0.666

0.999 * (3/3) = 0.999

However, when you say that 2/3 = 0.666..., this is the same as 0.666 +(2/3)*0.001 (Or 0.667 if you round). However, when you denote the ellipsis after 0.666... to signify that the 6's go on forever, the fact is that 6 >= 5, therefore no matter how many 6's you have, it will still be over that (For example, 0.666... > 0.666). Therefore, 3/3 is 0.999+(3/3)*0.001, which = 1.

Another proof says that:

c = 0.999...

10c = 9.999... (My issue lies with this line right here, I'll explain below)

10c - c = 9.999... - 0.999...

9c = 9

c = 1

0.999... = 1

The problem here? Multiplying by 10 in this case is the same as just adding 9c. However, when you add 9c, you have already made the ASSUMPTION that c = 1 (Therefore when you add 9c, you just add 9), while first off assuming that c = 0.999...

To demonstrate how I see this as a flaw, here's a proof that 10 = 5

c = 10

10c = 55 (See the assumption here? I assumed that c already equaled 5, and then just added 9c to each side, but that contradicts the first statement, rendering the proof invalid.)

10c - c = 55 - 10

9c = 45

c = 5

10 = 5

__Edit:__And as a third note, this proof here:

My problem with this is that the limit of something IS NOT that something, unless it can be done directly.

The limit of 2n+5 as n approaches 10 = 25. This is fine.

However, say we have function F (Too lazy to find a valid function for this). F has the variable n in it, and when you plug in n = 0, the whole thing breaks apart (Divide by zero, or squareroot of negative thing). However, due to factoring, L'hopal's rule, or some other trick, we can figure out the limit anyway.

The limit of F as n approaches 0 = 2.

The problem here? As n APPROACHES 0, the value of the function APPROACHES 2.

The LIMIT equals 2. The value of F when n = 0, is not 2.

Unless the question can be done without limits at all (Such as 2n+5), the limit will only tell you what the value WOULD be, IF it existed. Using the proof above shows that as 0.999 adds on more 9s, the value APPROACHES 1.. this is fine. But it does not equal.

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