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# 0.9recurring equals 1.

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Round 1
Pro
Key:
* multiplied by
/ divided by
r = recurring
others are obvious

Third Proof
0.9r / 3 = 0.3r = x
x * 3 = 3x
3= 3 * 0.3r = 1

The 10xx proof

x = 0.9r
10x = 9.9r
10x - x = 9x = 9.9r - 0.9r
9.9r - 0.9r = 9
9x/9 = 9/9
x = 9/9 = 1

The 0.9r never has a last 9 to reach because it is inherently endless, the idea being it is endlessly approaching a destination value of 1.0

What I argue is that it's actually equal to 1.0

A common objection is that, while 0.999... "gets arbitrarily close" to 1, it is never actually equal to 1. But what is meant by "gets arbitrarily close"? It's not like the number is moving at all; it is what it is, and it just sits there, looking at you. It doesn't "come" or "go" or "move" or "get close" to anything.
On the other hand, the terms of the associated sequence, 0.9, 0.99, 0.999, 0.9999, ..., do "get arbitrarily close" to 1, in the sense that, for each term in the progression, the difference between that term and 1 gets smaller and smaller. No matter how small you want that difference to be, I can find a term where the difference is even smaller.
Con
Method 1: Subtraction Rule
The simplest way to determine if 0.99 repeating is the same as 1 is to subtract.
If a-b=0, the numbers are equal
If a-b does not equal to zero, the numbers are not equal.
1-0.99 repeating does not equal zero. The difference is an infinitely small number, but nevertheless, because the difference will never reach zero, 0.99 repeating does not equal one

Reason 2: 0.99 repeating gets very close, but never equals one.
We can visualize the equation quite easily.
1-0.9=0.1
1-0.09=0.01
1-0.009=0.001
As you can see, this will repeat itself. No matter how many nines you put, there will always be a one at the end. Yes, 0.99 repeating gets very close, but as long as that one at the end of the infinite zeros is there, the 2 numbers will never be equal.

If you want to use me as a win-farm, that's ok. I'm just here to have some fun. Enjoy your pointless wins on this meaningless website.

Round 2
Pro
Pro has not even touched on my proofs, both stand true and absolutely proven if you follow the algebra.

Both of Pro's arguments are the same argument flipped around

Pro is saying that a value of 0.0r with a '1' at the end separates 0.9r and 1 and would separate 0 from the difference between them. The problem with this is how infinite series work, meaning that the '1 at the end' is not even a value capable of being considered mathematically.

Let me put it differently, if something is infinitely approaching a value, this implies that it never stops that approach until it's reached the value. If you deny that it's approaching the value, you also deny that the 0.0r ever can have a 1 at the end of it. In other words, both ways around Con is defeated.

infinite

1extending indefinitely ENDLESSinfinite space
2immeasurably or inconceivably great or extensive

If we deem the numbers to be 'approaching', then we concede that they never ever stop that approach and can't ever stop until the 0.999... hits 1 (theoretically).

Alternatively, if we deem the number static and refuse to perceive 'movement towards' then we also need to understand that the number 0.0recurring with a 1 at the end is impossible to write as a number or mathematically express at all, defeating your static value as being plausible. The infinite 0's cannot ever stop being infinite zeros to then have a 1 at the end, they are infinitely close to 0.
Con
I'm sorry CON is confusing me. My head hurts.
50 gallons of Pepto Bismol later...

If you are saying that 0.99 repeating does not ever stop, then what the fuck is the point of this debate? How can a number equal a number if we cannot even agree on what one of the numbers is.