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2021-01-04, 03:59   #78
SarK0Y

Jan 2010

2·43 Posts

Quote:
 Originally Posted by Dr Sardonicus (1) OP seems to be confounding "fractions" (rational numbers) and "decimal fractions," i.e. fractions that can be expressed with a power-of-ten denominator. Not all rational numbers are decimal fractions.
don't confuse numerical system & numbers itself == any proper numerical system can express any number. Another moment is how efficient that expression could be.
Quote:
 Originally Posted by Dr Sardonicus (2) OP also seems to think that invalidating a proof of A automatically proves ~A (not-A). It doesn't. (Here, A is "The square root of 2 is irrational.")
no, the very point is, you must uniform objects before you do operations on them. that proof messes w/ two different types of objects. for instance, 1/3 is odd or even? or let $p,q \in \mathbb{N}\; \frac{p}{q}=\epsilon,\; p=\epsilon\cdot q$ everything looks fine out there, right? But we have the damn grave problem, even two ones...
$p=\epsilon\cdot q$ exists everywhere for any p & q. second form has two troubling cases == 1st one w/ q = 0 and 2nd one is that..
$\displaystyle \lim_{q \to \infty}\lim_{p \to \infty}\frac{p}{q}\eq????$

2021-01-04, 04:00   #79
SarK0Y

Jan 2010

1268 Posts

Quote:
 Originally Posted by VBCurtis Neither side of your limit example exists, so your congruence is nonsensical- and irrelevant to whether 0.9-repeating is equal to 1. There is no sequence involved in the single number 0.9-repeating, either. I didn't ask about 0.9, nor 0.99. 0.9-repeating is neither of those numbers. Every member of your sequence is strictly less than 0.9-repeating, anyway. You might figure out the flaws in your reasoning if you used words properly- how do you define "continuous sequence"?
so $\ln(x)$ and $\frac{d\ln(x)}{x}$ do not exist, right?
Quote:
 Originally Posted by VBCurtis There is no sequence involved in the single number 0.9-repeating, either. I didn't ask about 0.9, nor 0.99. 0.9-repeating is neither of those numbers. Every member of your sequence is strictly less than 0.9-repeating, anyway. You might figure out the flaws in your reasoning if you used words properly- how do you define "continuous sequence"?
Oh, boy, really?
$\lim_{n \to \infty}\left(1-\frac{1}{10^{n}\right)\eq0.9999..99$

Last fiddled with by SarK0Y on 2021-01-04 at 04:10

 2021-01-04, 04:19 #80 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100100011101002 Posts The attempt to (and resurrect a dead thread) detected. Thread is closed. Mod warning: More wasting of people's time with BS will result in a ban.
 2021-01-04, 05:53 #81 LaurV Romulan Interpreter     Jun 2011 Thailand 52×7×53 Posts $$\lim\limits_{n \to \infty}\frac{n}{n+1}=0.9999..99$$; in fact, $$\lim\limits_{n \to \infty}\frac{n}{n+M82589933}=0.9999..99$$ too... hihi edit: whoops, sorry Serge, didn't see the thread is closed. (but this way I learned an important difference between \TeX and \MathJax (using the \limits when the formula is inline, otherwise the limits are placed as indices, and not compacted vertically). Last fiddled with by LaurV on 2021-01-04 at 05:59

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