heyyy guysss....just trying to make a code to proove .999=1

any ideass?? :))
ty!!

I'm not sure that code can prove it; you're asking the machine to deal with numbers to infinite precision. As soon as you start hitting floating point limits, you can no longer trust the numbers.

Edited 5 Years Ago by Moschops: n/a

In my computer it takes 16 iterations from 0.9 to reach floating point limit of accuracy, if you mean that:

1 0.99
2 0.999
3 0.9999
4 0.99999
5 0.9999990000000001
6 0.9999999
7 0.9999999900000001
8 0.999999999
9 0.9999999999
10 0.99999999999
11 0.999999999999
12 0.9999999999999001
13 0.99999999999999
14 0.999999999999999
15 0.9999999999999999
16 1.0

The code was not in C++, but results should be similar in C++.

Idk if this works but maybe try out using this algebra:

1/9 = 0.1111....
9 * 1/9 = 9 * 0.1111
1 = 0.999....

See if you get similar result on a computer.

No code can do that..it's a totally crazy idea. I'd love to know if you find a way to do that.

No it is True with repeating decimal 9 of infinite precicion see previous post proving it simply and elegantly with 1 / 9 *9 (same could be done with decimal reprecentation of any number with inverse with repeating decimals)

Edited 5 Years Ago by pyTony: n/a

I'm not sure are you looking this?

LHS= .9999*
= ( 1.0 - 0.0*****1 )
= 1.0 - lim[n->infinity] 1/n
= 1.0 - 0
= 0

.: LHS = RHS

so proving this using code is simple. So you need a Automated Theorm Proving
math engine and in font end you can fetch the following rules.

I didn't see anything saying that the OP was asking to prove that 0.999<repeating> was equal to 1.0. The OP says, prove that 0.999 (which I interpreted as exactly 0.999 as in 0.999000).

Is this a sinister puzzle given by a computer science teacher or something? If so, I believe I see the trick to it.

The puzzle probably wants you to realize that floating point numbers are typically stored with a mantissa and an exponent. What I believe the puzzle is looking for is you to come up with a floating point representation format where the mantissa doesn't have enough bits to store 0.999 - the closest representation would be 1.0.

Comments
Good logic and well writeen answer with a start to solving it
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