I think Fermet's Last Theorem works with any infinite number. Read details: What do you think?

Question

I just asked a question that proved 0.99999... = 1.

The concept is true, but I thought the math might be an anomale:

x = 0.99999...

so 10x = 9.99999...

then 10x - x = 9x and 9.99999... - 0.99999... = 9

so 9x = 9

and then x = 1

----- ----- ----- ----- ----- -----

But I tried it with 0.33333...

x = 0.33333...

10x = 3.33333...

so 10x - x = 9x and 3.33333... - 0.33333... = 3

so 9x = 3

x = 3/9 or 1/3

1/3 = 0.33333...

So this is even more proof that 1 = 0.99999...

Answer 1

What do I think?

1 - this has nothing to do with Fermat's Last Theorem

2 - You don't understand maths very well

Answer 2

The problem with Fermet's theorem, IMO, is that it ignores that inifinity has "size".

I recently stubbed my toe on this fact in a math forum.

The important thing here is that:

9.999... is a "larger" infinity than 0.999...

I know, it is counterintuitive, yet try to ignore this in the presence of math gurus at your own peril!

The result, as I see it, is that the proof you use above is not valid. As I mentioned in a realted question you asked, this would also ignore the phenomena of the Aysmptote. By its very definition, it represents something that ever approaches a value but can never reach it.

It's a hairy, rather philosophical issue. I think to be extremely rigorous, one can only say that 1 is equivalent to 0.999... Not truly equal.

Answer 3

are you even considering the number of digits thatis in place after the decimal point.... that is the most important factor....

i totally disagree with u that .99999=1 because the .1 or .01 or .00000000001 is very different from each other...
unless u do consider the fact that each numeric after the decimal point in its right place then such a situation will not arise ...
please check the earlier question ...i think i have answered it there.

Answer 4

I think you need to brush up on your maths.