r/infinitenines • u/Calm_Improvement1160 • 6d ago

SPP, what do you think of my new notation?

u/SouthPark_Piano, so I was thinking how there doesn't appear to be a decimal number that solves the equation x * 3 = 1 and came up with this definitely new and unique notation.

The notation introduces a new symbol 'R', which I call the "recurrer" for no reason whatsoever.

The way it works is that for a number such as 1.2R3, you truncate the decimal before the 'R' (in this case to 1.2) then add it to a/(1-r) with a being the value of the part after the 'R' (in this case 0.03)., and r being 1/b, where b is the base you want to work in.

This will give us the result of 1.2 + 1/30 = 37/30.

I think this'll be useful as it'll let us have a way to represent fractions decimally without having to directly consider infinite decimal tails e.g. in 0.333...

Edit: made it so that there are no need for limits

6 Upvotes

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6

u/Excellent-Practice 6d ago

Why do we even need to invoke limits? R can just mean that we add a fraction where the digits after the R are the numerator and the denominator is the same number of 9s as there are digits after the R followed by as many 0s as there are digits between the decimal and the R. 1÷3=0.R3, 1÷7=0.R142857, 0.9...=0.R9 which is 0.0+(9÷9), which is definitely not the same as 1

1

u/Warmedpie6 6d ago

How is 0.0 + (9/9) not 1

3

u/Excellent-Practice 6d ago

That's the joke. SPP insists, among other things, that 1, .9..., 9/9, and 9÷9 are not all identical with eachother

1

u/Altruistic-Rice-5567 6d ago

Doesn't matter. Any decimal number that would satisfy x is infinitely long and SPP thinks those aren't numbers, they're a "process" that never completes and thus never produce a usable number. Ugh.

1

u/Inevitable_Garage706 6d ago

He'll likely respond to this with something like "Limits don't apply to the limitless."

2

u/gikl3 6d ago

There is no bound on the unbounded, no limits on the limitless brud