Wikipedia:Reference desk/Archives/Mathematics/2006 July 23

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Quick Question

Is there any power of 3 that's one more than a power of 2? Black Carrot 00:48, 23 July 2006 (UTC)[reply]

9 = 8+1. Or do you mean other than this one? -- Rick Block (talk) 01:17, 23 July 2006 (UTC)[reply]

[after edit conflict] Yes there is. If we are talking about integer powers, 3^1 is one more than 2^1 (trivial case) and 3^2 = 9 is one more than 2^3 = 8. These are the only two solutions to the Diophantine equation 3^x - 2^y = 1. The proof that these were the only solutions was unsolved for over 100 years and was called Catalan's conjecture, but is now Mihăilescu's theorem. --AMorris (talk)●(contribs) 01:19, 23 July 2006 (UTC)[reply]

Cool. Are there any solutions to 3^ax - 2^bx = 1 ? Black Carrot 02:49, 23 July 2006 (UTC)[reply]

If x is supposed to be an integer, then it must be either 1 or -1. If it is 1, you get the previous case; ~~if it is -1, there are obviously no solutions.~~ -- Meni Rosenfeld (talk) 07:05, 23 July 2006 (UTC)[reply]

Except that 3⁰ - 2¹ = 3¹ - 2² = –1. --Lambiam ^Talk

I'm sorry, I wasn't thinking straight last night. I typed it in wrong. I meant 3^ax - 2^a+bx + 2^b = 1 Black Carrot 17:27, 23 July 2006 (UTC)[reply]

Other than x=0, b=0? Digfarenough 18:48, 23 July 2006 (UTC)[reply]

Yeah, for x>1, a>0, b>0. Black Carrot 20:31, 23 July 2006 (UTC)[reply]

I wasn't thinking straight myself. To your "original" question, there's Lambiam's solutions, and maybe others (I doubt it). About your revised question - A brute force search reveals that if there is a solution, at least one of x, a or b must be greater than 100. -- Meni Rosenfeld (talk) 05:52, 24 July 2006 (UTC)[reply]

Well, as far as the brute force approach, I know the answer is "no" up to 20-30 digits. I'm working on the Collatz conjecture for a little while, and this is to knock out a trivial loop - straight up followed by straight down. Black Carrot 14:33, 24 July 2006 (UTC)[reply]

zero times infinity, redux

I guess we get asked a couple times a month what zero times infinity is. The last time I got to thinking: are there any examples of, say, two sets, one of measure zero and one of infinite measure whose Cartesian product has nonzero measure in the product measure? I suspect that the real line cross the Cantor set has measure zero (Lebesgue measure). Is that correct? -lethe ^{talk +} 05:12, 23 July 2006 (UTC)[reply]

Your guess is correct. In fact, a necessary and sufficient condition for a measurable subset of a product

X\times Y

to have measure zero is that almost all X-sections (or almost all Y-sections) have measure zero. Madmath789 17:42, 23 July 2006 (UTC)[reply]

calculating standardized mortality/morbidity ratios

please could someone explain how to calculate standard mortality rates using either poisson or bayes methods to produce a more valid SMR. I discovered some studies used one of this approaches for the statistical smoothening of their SMR. however, they were abit vague in their explanation on how its done. I would like to know if this analysis can be done using excel or SPSS. —Preceding unsigned comment added by Saretin4life (talk • contribs)